How to return a file in java. Read the file and loop over its content line by line.

How to return a file in java Leave off the With Spring 3. Then split each line on whitespace using String. First, we’ll learn how to load a file from the classpath, a URL, or from a JAR file using standard Java classes. getEncoding()); But on a file which I know to be encoded with ISO8859_1 the above code yields ASCII, which is not correct, and does not allow me to correctly render the content of the file back to the console. The Java program is run in a Unix shell script, which later goes on to call another java program Java Program_2 (say). nio. ArrayList; import java. class file's debug info. Reader — JAX-RS will do the right thing. Commented Jun 21, import java. Consider this: you upload file from Internet Explorer and it has the path "C:\\Hello\\AnotherFolder\\The File Name. How can I achieve this? Skip to main content here Execute another jar in a java program depending on your runtime environment you may need to specify a path to java and/or the jar file. conf, sample_b. a,b,c 1,2,3 4,5,6 No matter what I have tried, the newlines come out as literally '\n' in the response content, and if I double escape them as in "\n", the response just contains the double backslashes too. To use the File class, create an object of the class, Returns the size of the file in bytes: list() String[] Returns an array of the files in the directory: mkdir() Boolean: Creates a directory: When I save file I use: File file = new File(filename); But, since I no longer have privileges to write to folders, I would like rather to save it to memory and then read file to FileOutputStrea Unix gurus! I have a Java program which passes some parameters to a Servlet. Java- Searching for a string in text file, then print line. At this point, an obvious problem becomes visible. lang. any kind of help or suggestion is appreciated. Any extension is fine, The list() method that you're using lists the file names, not their absolute paths. APPLICATION_OCTET_STREAM) public Response downloadExportedFile(@PathParam("loginId") String loginId) { File file I have to run executable jar file from shell script to get a string value. What I need to figure out is how to return the three values on separate lines, something like: public String toString() { return String. Here's an example assuming you're using Google Gson to convert a Java object to a JSON string:. web. How to implement a method to return Iterator of files recursively in Java. The data should be available in the . apache. here, here and here and there are lots of possible solutions. note : the file format can be anything (txt, pdf , xls and etc ). Basically like the window which pops up in every standard texteditor after you hit the "OPEN" button to choose the file to open in the editor. toString(new File("file. For that I have written the following code: Str. You can store the file anywhere you like. Is there a way to In short: Files. @Override protected void doGet(HttpServletRequest request, HttpServletResponse respo You should set the media type on basis of file content type. How do I mock a method that returns a file in java? The code looks something along the lines of private File baseDirec I am trying to return a list of files from a directory. Improve this question. I'm such a dufus. read(chars); // now chars contains the whole file -- or does it? It will probably work most of the time for small files on a lightly loaded system. com and this service brings requested file from b. conf, refund_b. There is no point in doing all that twice, and several reasons not to, such as: In Java, I have text from a text field in a String variable called "text". Since you also tagged ioutils one possible way is to: File file = new File("Data. txt"); if (myObj. I am unable to do it because as soon I write the file creation logic in java, rather than returning it, it creates a file on my host. Load Comments. exists()) { System. However, this is not a static method: public String toString() { If you're trying to call . file; import java. RandomAccessFile. Java's File. println("Absolute path: " + Java's File Class, part of the java. The problem is with a extension controller, I ccheck if a file is really zip file and I throw an exception if isn't so. ok((Object) file); response. The size may differ from the actual size on the file system due to compression, support for sparse files, or // internal method that lists files, converts them into a List and makes sure it does not return null private static List<File> listFilesWithExtension(File directory, ExtensionFilter filter) { File[] files = directory. My issue is this - I need to pass the ID we It seems like if the resource you are trying to load is much bigger than 1MiB the InputStream you get from getResourceAsStream stops to retrieve the bytes of the resource after that size and instead returns 0 iff it is contained in a compressed filesystem like a jar. delete() to delete a file, it will return a boolean value to indicate the delete operation status; true if the file is deleted; false if failed. GET) public ResponseEntity download() throws IOException { String filePath = "PATH_HERE"; InputStream inputStream = new The static files should be served by the container if the files are in the webapp folder of your project (assuming you're using Maven project setup). io. renameTo() is problematic, especially on Windows, it seems. My requirement is t Return file with filename. Instead the JVM ends when no more non-daemon threads are running or when System. txt"), Charsets. There is no built-in way to get the size of a folder, you are going to have to walk the directory tree recursively (using the listFiles() method of a file object that represents a directory) and accumulate the directory size for yourself: Return a file using Java Jersey; file downloading in restful web services; However, it's not solving my purpose. Instead of returning the zip file that I have created, it returns an html file. How can I achieve this? Search string in file and return next line (in Java) 1. This method returns the file size, in bytes by taking the path of the file as a parameter. Then parse the lines and create a new Employee() in each iteration. Don't use JSONObject to map to JSON and don't read String directly. Alternatively, BufferedReader does support mark(). As you say, the outermost layer of your JSON blob is an array. format( firstName + " " + lastName + " " + Time ); } That's what I have right now. Files; import java The return value of a Java application is not the return value of it's main method, because a Java application doesn't necessarily end when it's main method has finished execution. I'm having a problem handling the returns and I don't know how to fix I'm looking for a way to get a list of files that match a pattern (pref regex) in a given directory. And even then, you should rather use target/classes/abc. List; public class GetFilesRecursive { public static List <String> getFilesRecursively(File dir Yes, I do have enough permissions. Add a using apache commons io for java copy/replace file operation . InputStream or java. Why the image has a weird file name - byteimage and has no file extension that would indicate the format?byteimage is just the last part of In this episode find out how to return a file from a spring controller. The place to start is the javadoc - it contains quite a lot of useful information. The benefits of this approach over using an array are type safety and it will make your program much easier to understand. working again with Java EE. How are you serving the rest of your static files? If you are making a login, you should consider redirecting with Response. txt via getResource I've written a download Servlet to return a file based on a messageID parameter. If you use this, then your controller does not need a HttpServletResponse object, and therefore it is easier to test. I have a data base table where the files have been stored as a BLOB. RESOURCE_PATH + "/file", headers = "content-type=application/json" ) public Possible Duplicate: How to create a Java String from the contents of a file Is it possible to process a multi-lined text file and return its contents as a string? If this is possible, please You can just return an instance of java. FileOutputStream; import java. Now I want to enhance the response. Now, I need to have a spring boot java API which will download the file and return to front end of the application (node js). java in command prompt: run javac Maintu. move(source, source. StackTraceElement but I'm not sure if it's possible to force Java to create a java. toString() on a specific instance of Files then you need to call it on that instance, not on the class itself. For files in src/main/resources they will become resources that you should read from the class path using something like In my @ActionMapping I create a PDF file for the user. If you know that the resource is a writeable file, though, you can jump through some hoops, e. Anything in src will not be there unless you configure your deployment to include it. // Read data from file try (BufferedReader br = new BufferedReader(new FileReader(file))) { // List to collect Don't. Follow While downloading a file , your code needs to stream a file chunk by chunk - thats what Java streams are for. of() (from JDK 11) to build a Path with the resource's URI. name = name; } public String toString() { return name; } } readLine() Returns: A String containing the contents of the line, not including any line-termination characters, or null if the end of the stream has been reached see how to read File from a particular line till the end of the File in java. @GET @Produces({MediaType. getResourceAsStream() to access it. of was called Paths. java : search for a word in a text and give its line. Class object, to get the source file name and the line number at which the class was declared?. 3. TEXT_HTML}) public InputStream viewHome() { File f = getFileFromSomewhere(); return new FileInputStream(f); } is a better option then using a java. public static byte[] getBytesFromFile(File file) throws IOException { // Get the size of the file long length = file. Files; import java. And place your html files in the src/main/webapp/static/ folder. FUNCTION get_file (p_dir in varchar2, p_file in varchar2) RETURN VARCHAR2 AS LANGUAGE JAVA NAME 'FilesFromUnix. demo. txt). Once you have a valid non-null URL, you can attempt to create a File object from it. public class FilePanel /* extends not sure */ { private File file; I had a similar requirement, but what I did was just to serialized it from the Gson to the entity to handle it internally, and when print it or save it, just serializes it back to GSON, the logic is something like this (by the way, I used Gson for both ways):import com. html (note the . // Returns the contents of the file in a byte array. 8 million lines. The java. Recursively list files in Java. If I manually change the file extension afterwards, I can see that the contents of the file are still the search results that I need. getAbsolutePath(); how can i read this text file into String ? UPDATE i found the solution , what do you think? The while loop exists when the line is equal to null i. Jan Nielsen. exit(). temporaryRedirect(). getAbsolutePath()); The File class has several more methods you might find useful. Paths; import I'd like to mock up some JSON (that I'm reading from a file), and return it as a result of some Spring Controller. The File class from the java. FormDataContentDisposition; import com. You want to return a value from a process, the process is a java virtual machine. Either way, returning a null item (File or otherwise) is perfectly acceptable within Java. (It's a fabulous library. html file located in static folder. /Test1/ Possible Duplicate: List all files from a directory recursively with Java. So when you return the string line after the while loop, obviously it will return null. System. println("File name: " + myObj. Creating a zip response without actually creating all the files in the machine. First, you can read the file with Files. The file contains various names, one name per line. build(); // Obviously I can't do this but at this point I need to delete the file! Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company size() method of java. While the OP and I are contemplating ways to turn a ByteArrayOutputStream to an InputStream, I missed the simplest solution, which is simply @GET @Path("/{loginId}") @Produces(MediaType. io package, allows us to work with files. The common methods getTargetFiles and printFiles are used to This should be really simple. What this means in plain English is that the resource bundle must be on the classpath and that baseName should be the package containing the bundle plus the bundle name, mybundle in your case. 0 you can use the HttpEntity return object. The whole file is an array and there are objects and other arrays (e. newLine()) . If I have a String like this: . The most preferable solution is to use InputStreamResource with ResponseEntity. Same goes for InputStream, etc. Then one day when you least expect it, read() will only read part of the file. Share. getResource("sample. This uses the decorate-sort-undecorate pattern so that the last-modified date of each file is fetched only once rather than every time the sort algorithm compares two files. do something like this: while((line = br. Obviously PDFBox cannot return a meaningful File object without saving a PDDocument to the file system. header("Content-Disposition", "attachment; filename=\"the_new_file\""); return response. 352. readLine()) != null) { System. getProperty("java. So the problem just lies in returning the proper extension to the response. In that case you can declare a boolean variable and set it appropriately inside the conditional blocks. Java read text file and search for a line. I've done this where i take an uploaded file and replace the file name with a guid and no extension and browsers / smart phones are able to load the image no issues. To write a ZIP file, you use a ZipOutputStream. String'; But what I want to do is not return it as a VARCHAR2 and a String but use it as a Blob instead, but when I try to do this like with this code snippet: In Java, when we write text to a file, we must provide the character encoding to create a file with a specified charset. txt"); file. You can obviously use a different folder or a more sofiticated directory structure. FilenameFilter to filter the file by specific condition. g. docx file into the client system. getFileName() will return the whole path, not just The File Name. getBundle(String baseName):. 7k 14 14 gold list of config files: sample_a. google. ( If the second record failed during file upload the response file should be having the failed reason and failed record information in it. File does not provide an abstraction which allows you to manipulate in-memory files. Commented Jul 15, 2012 at In my Project, I want to open the windows file explorer with java, in which you can select a file or a folder and click the "OK" button. Java File Handling. I've found a tutorial online that uses apache's commons-io package with the following code: I have read a file into a String. I would like to return that file from Spring controller. You are already planning to use Apache libraries to read your uploaded file, so I recommend using apache. As the API documentation says: Many aspects of the behavior of this method are inherently platform-dependent: The rename operation might not be able to move a file from one filesystem to another, it might not be atomic, and it might not succeed if a file with the destination abstract pathname I'm trying to return a file stream from Rest web service deployed on server, and handle this stream from Rest web service on client. Here is what I've tried but it doesn't seem to work. toString() Because Files is the name of the class. In this example we will show how to read a PDF from the classpath simulating a read from a database or a file here is a sample: @RequestMapping( method = RequestMethod. How can I save the contents of the "text" variable to a file? Is it possible, given a java. 54. Which one to use depends on the Java version you're working with and whether you need to read bytes or characters, and the size of the file/lines etc. Gson; import com. I'm just learning and I used the server from small correction on @PetrosTsialiamanis post , new File( multipart. Why is BufferedReader readLine reading past EOF. setContentType("Content-type: text/zip"); response. fileupload to upload your file as well. IOException; //*This is a program where we are reading and Constructing a path object or resolving a child, does not mean the file or directory actually exists. io package, enables file handling operations such as creating, reading, writing, and deleting files, utilizing streams for input and output operations. length(); // Convert the bytes to Kilobytes (1 KB = 1024 Bytes) long fileSizeInKB = fileSizeInBytes / 1024; // Convert the KB to MegaBytes (1 MB = An advice of limited use, as it relies on the working directory to be the maven root. println(r. home). tmpdir") will create temp directory where your Java Files Java Create/Write Files Java Read Files Java Delete Files Java How To's Add Two Numbers Count Words Reverse a String Sum of Array Elements Convert String to Array Sort an Array Find Array Average Find Smallest Element ArrayList Loop HashMap Loop Loop Through an Enum Area of Rectangle Even or Odd Number Positive or Negative Square This is a static reference: Files. Spring Boot In this episode find out how to return a file from a spring controller. 11. This service needs to request the file from external domain. However, you can return json object with csv inside one of the properties and on the frontend parse that property, create csv and /** * Get ZIP file containing datafiles for a given request id * * @param id of the request * @param responseOutputStream for streaming the zip results * @return ZipOutputStream a ZIP file stream for the contents * @throws AccessDeniedException if user does not have access to this function * @throws UnauthenticatedException if user is not In my unit testing, I want to mock a method which returns a file in java. In the following example, we use the FileWriter class together with its write() method to write some text to the file we created in the example above. Just to be clear, I'm not looking for the MIME type. html), it will return the test. I documented 15 ways to read from a file in Java and these are from that article. String) return java. getDefaultFileTypeMap(). You seem to be forced to use getResource and load the file independently from this in that case. data. Files help us to get the size of a file (in bytes). java in command prompt: run java Main. exit() will end the process (the virtual machine that you executed as "java -jar ") and return the %errorlevel% back to the OS (in a portable way) – Edit: Sometimes you can't return early because there's more work to be done. About us; Blog; Style Guide; Exercises; Examples; Tutorials I have this code which load the File from class path a=its text file and i want to read it to string what im using is : File file = new File(classLoader. Here are 3 ways to read a text file in one line, without requiring a loop. Entity; import Im making a java method for an android app that returns the directory the app should be working in in the external directory. So your code lists all file names in sourcedir (not their full paths), then looks for those same file names in the current directory you're running your program from. All you need is set Content-Length manually: @RequestMapping(value = "/download", method = RequestMethod. The executable jar can't return a value as the main returns void. /Test?/sample*. Could somebody show me on example how add file as array of bytes to zip and return on servlet as atachment? Thanks for any tips and advices The below Sample code worked for me and might help someone. This way you don't have to create a controller. length(); This returns the length of the file in bytes or 0 if the file does not exist. Note that you still have to loop through the list that's returned, even though the actual call to read the contents of the file requires just 1 line, without looping. And System. POST, value = DataController. println(line); } I am trying to write an app to use Files. conf, refund_a. URL resourceUrl = getClass(). baseName - the base name of the resource bundle, a fully qualified class name. File instance. Tagged: java, java-tutorial, spring, and rest. The file system has to test whether the file exists anyway. The problem is caused by Spring trying to encode the response as Json. find method in it. How to achieve this in java ? I have a method to download a zip file and at the moment I use FileSystemResource to do it. getName()); System. cars) in the whole array of the file. Follow edited Sep 1, 2017 at 13:03. br. The simplest fix to this is to add a produces to your request mapping, so your code becomes: @RestController I am trying to retrieve a File or InputStream instance from PDDocument without saving a PDDocument to the file system. txt for the extension. } } } return files. header. Your request probably specifies Accepts = "*/*" and since Spring ignores the ResponseEntity's ContentType, the best encoding is found to be application/json. class and what exactly does, it receives text from the client and then sends that exact text back to the client. Please suggest. I have the following questions: Is it possible to send message data (like subject, message, message id, etc) along with Just put the two files in the same directory. I suggest to always use those Files methods which returns Stream in try-with-resources blocks: otherwise you'll keep the resource opened – riccardo. Java/Android recursive file null pointer exception. lines(), which returns a Stream How is this done in Java? I wish to create an array of fileNames to output in a program, the files will be the ones in my home directory. String, java. Is there any method in PDFBox which returns File or InputStream from an existing PDDocument?. exit(int) as jar has to return value of String type. isDirectory() to see if you need to recurse further is a good start. conf; Need to check whether the file exists and then return the list of sample. It simply reads the bytes into a buffer and counts the \n characters. If you request some/path/test. Load File from Resource Folder as java. Next time you need a reader, you can check the "Direct known subclasses" of the Reader class. multipart. File; import java. Now I was wondering how I can return this pdf to the user in the form of a save/open file dialog box? I'd prefer this over showing a download link if the generation was succesful. To get the number of characters you could either look at the size of each line or of each split word (depending of if With Java 8, the best approach to standard problems like reading files, comparing, grouping, collecting is to use the streams api, since it is much more concise to do that in that way. All those files will get put into the same location as your classes when the project is built. println("File path: " + new File("Your file name"). Below is the doGet method. get, which you’ll need to use if you’re stuck on older Java versions or building a library that needs some backward compatibility. Improve this answer From the JavaDocs for ResourceBundle. Then just write it out to the response writer along with content type of application/json and character encoding of UTF-8. readLine(). The program below prints true, illustrating the effect. The only place I know of, where the JDK returns such debug info is in java. json"). Please help me to get this done. gson. Except this, this answer is relative equals to the one of Infeligo. Search for string then print the next line from text file. util. The domain for this service is a. At least when the files is only a few KB, then there will be no problems with that. StringReader: return new StringReader(string);. FilenameUtils; String fileNameWithoutExt = FilenameUtils. Create new POJO class for the response and let RestTemplate do the work. the second is to serve a file to be downloaded. pdf"); FileOutputStream out = new FileOutputStream(file) "\r\n" Java only knows about the platform it is currently running on, so it can only give you a platform-dependent output on that platform (using bw. I'm making spring boot web service that receives the request for file download. Hot Network Questions Who can be a primary supervisor for a PhD student? File in = new File(args[0]); InputStreamReader r = new InputStreamReader(new FileInputStream(in)); System. getResourceAsStream() or ClassLoader. sun. As I said, I can write to the file, but the delete and replace functions return false. zip"); // Get length of file in bytes long fileSizeInBytes = file. Share on: Twitter Facebook Google+. Let's say I have the following input: /path/to/file/foo. PDF. getForObject(uri, Item. How can i return a file array that include all files on the folder and also sub folders my method just work for folder and it doesn't include sub folders . Can you guys give me some code snippet? Great thanks. Now I want to have the path of the selected file in my Javacode. StackTraceElement instance I'm trying to write a REST API (java) that can receive a file. getResource(path); File file = new File(resourceUrl. This is a program where we are reading and printing array of bytes offset and length using String Builder and Writing the array of bytes offset length to the new file. println("Read have "+read()); its invoking and storing the returned string value in array and in the second time I am storing to the "tt" String variable and its appending the last returned string from the read() function to the "tt" String variable. // Get the file as a steam for the entity File file = new File("the_new_file"); ResponseBuilder response = Response. This is my controller code @RequestMapping(valu So I'm trying to write a java application where I have a text file containing some data like: id, username, password and each record is on a new line. I wanna know how to return response to the client for the file download request in this case. api. private static void copyFileUsingApacheCommonsIO(File source, File dest) throws IOException the way I did it was: write code for files in command prompt: run javac Maintu. getContentType(filePath); byte[] byteContent = ; //Content final HttpHeaders headers = new HttpHeaders(); I want to return a file list from a function which loops through all the files/directories on SD card on android and which confirm the specific data type(mp3, mp4, png). File file = new java. Url has a query parameter named format which can be either text or file. It's much better to use the abc. This might be faster if you have many files. Give the class a meaningful name. GsonBuilder; import com. txt then what is a generally-accepted way to get a list of files that match this pattern? (e. resolveSibling("newname")); More detail: import java. file. ByteArrayResource; import org. Pre Java-11, Path. What type of resource can I use to return the generated excel file? java; excel; spring; spring-boot; apache-poi; Share. If it's on the file system it's slightly easier. I have to create a rest web service where in user access data using url. Reader reader = new FileReader(file); char[] chars = new char[file. Just catch the FileNotFoundException. tasso. File("myfile. java; javascript; System. ) String content = Files. To use the File class, create an object of the In this tutorial, we’ll explore different ways to read from a File in Java. – Sean Reilly. e line == null. 1. The record can be written in the file, but is append in one Line, each time I call this method example: Hello WorldHello WorldHello WorldHello W Basically you want to read from an InputStream and then write to an OutputStream. springframework. exit() is also the only way to specify the return value: the I am using Apache POI to generate . file. You might have to check that the file is also non-Null (or perhaps this method is allowed to return null items). Second, we’ll see how to read the content Return value: The function returns a File array, or null value if the file object is a file. conf files and list of refund. So, you’ll have to separately verify its existence. FileInputStream; import java. java. public static void main(String[] args) { File myObj = new File("filename. getBaseName(fileName); OR String fileNameWithOutExt = FilenameUtils. UPDATE. it should match . commons. reading a file into a byte[]: example here; I'd suggest adding a random string to the temporary PDF file name inside showHelp() to avoid overwriting the file if two users send a request at the same time Even in Postman, I see the same, and it only works when I use the 'Send & Download' option when making the request, which prompts me to save the response and I get the zip file. FormDataP I would change your approach a bit. After "save as" is performed store the File as a field then for "save" write to the existing file. Now the problem is that I want those names in a String array. Note: There are many available classes in the Java API that can be used to read and write files in Java: FileReader, BufferedReader, Files, Scanner, FileInputStream, FileWriter, BufferedWriter, FileOutputStream, etc. I used the code below to write record into a File. java in command prompt: run javac Main. Here's my code: package com. Any help on how to hit the API and download the returned zip file through Java? Thank you in advance. 6× faster than the top rated answer when tested on a file with 13. Classes to generate Java Files with Specified CharsetUsing a FileOutputStream and the OutputStreamWriter How can I use java to get a js file located on a web server, then execute the function in the js file and get the result and use the result in java. getFile()); i don't want to use: file. I'm not sure I understand you completely, but if you wish to get the absolute file path provided that you know the relative file name, you can always do this: System. The Servlet enters the info into a DB and returns back the ID of a row created back to the java program that calls it. List of File to array in Java. Using apache commons. listFiles(filter); return files != null ? java. Improve this answer. conf files. public class Person { public String name; public Person(String name) { this. Below program works perfectly : package ehsan; /* I have removed imports for code brevity */ public class Main { public static v I don't get the point why you don't want to use System. exit() is called. import java. readLine() reads only one line at time. protected void doXxx(HttpServletRequest request, HttpServletResponse response) { // Take a look at java. txt to reference the file, as this is the canonical place where Maven puts resource files after processing (for example, the maven-resources plugin might have done property substitution on abc. Exception: This method throws Security Exception if the function is not allowed read Return file content via REST API Normally, if I wanted to have an HTTP GET endpoint, I would use a Spring @RestController annotation on a class and a @GetMapping The Spring Framework provides powerful tools to handle these requirements efficiently, allowing developers to return images and files directly from endpoints. Path; import java. PDF" but your code is working on a Unix/Linux machine then p. It does not offer a method providing Yes, Path copes with the platform dependent problem of slash/backslash, but only if the file path is from the same machine (or platform). All the code on this page is available on github: View the source. This potentially reduces the number of I/O calls from O(n log n) to O(n). Path. Java has several methods for creating, reading, updating, and deleting files. Hot Network Questions UK Masters Application: UG Exams missed due to illness: concerned about low degree grade along with a downloadable file with failed record information. It allows reading a file from an arbitrary position, but its constructor requires a filename or a File object, not just an input stream as your method currently receives. In the java File API here. Path, you can employ the static method Path. My REST API looks as follows: import com. Using this along with . You could play with the buffer size, but on my machine, anything above 8KB didn't make the code faster. It returns an array of all the files in a directory. If user chooses text as format then I have to return text data to browser or if user chooses file as format then return a file for user to download. File contains of course correct JSON data format inside, like: {"country":"","ci Write To a File. com domain. FileReader would appear to inherit mark() Unfortunately, markSupported() returns false. I'm using JEXcelApi. Improve this answer Let say I have CSV data in a string and want to return it from a Spring controller. UTF_8); There are similar methods for reading any Readable, or loading the entire contents of a binary file as a byte array, or reading a file into a list of strings, etc. ) The Java program A needs to call another java program B(jar), passes arguments to it and receives the return value from B. It's the Post/Redirect/Get pattern – In my cache class, I'm going to have a function which writes the serialized version of an object (undecided type) to a file, something like this (generic function): public &lt;O&gt; void write(O o So I have a Spring controller, and I'd like to create an Excel file and return it so that it is downloaded by the browser. Note that when you are done writing to the file, you should close it with the close() method: Not directly, no - getResourceAsStream() is intended to return a view on read-only resources. Internally Spring will use suitable converter to map the objects: public class Item { int id; String name; String description; // other fields // getters and setters } Item result = restTemplate. Here's what I've done so far: Controller: @RequestMapping(method = RequestMethod. That's why length() returns 0 for all those files (as per the docs, it'll return 0 if the file doesn't exist. The path is merely a reference to a potential file. Commented Sep 4, 2016 at 16:02. 0. Please suggest how to return File[] array as my code below is giving NullPointerException when I am trying to add all text files from "D:\" to File[] Java adding file contents to array list. On the server I use this code: @Override @RequestMapping(value Not the built-in API - but Guava does, amongst its other treasures. core. – Romulus3799. Instead, if you need an instance of java. println("Result returned by Javascript is: " + result); } Read the file and loop over its content line by line. toURI()); OutputStream output = new FileOutputStream(file); First Java file: public class Swordsman extends Beginner { public String attFunc; public String attSkill; private String nickname; private int power; private int result; p However, this currently doesn't work quite right. [HttpGet] public IHttpActionResult Test() { var stream = new MemoryStream(); // var content = new In case you don't know the file/mime type you can do this. Resource; import First convert the JSON object to String. // Get file from file name File file = new File("U:\intranet_root\intranet\R1112B2. Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company I am developing a rest controller to download a . ) As @Clarke said, you can use java. jersey. . For each entry that you want to place into the ZIP file, you create a ZipEntry object. Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company Here is the consolidated list order by my preference. You pass the file name to the ZipEntry constructor; it sets the other parameters such as file date and decompression method. It's common, but bad code style: Whenever you catch an exception, the right move is to either deal with that exception, or, to ensure that you (re)throw some exception. Set your values, such as id and name. renameTo(file2) to rename a file, it will return a boolean value to indicate the rename operation status; true You have to read the lines in the file repeatedly in the loop using br. length(); // You cannot create an array using a long type. length()]; reader. File For the specific question of returning to a previous point in a file, java. repository; import com. import org. In example On servlet I have byte[] fileContent ---->(file. FilenameFilter to search file in current directory and its subdirectory. Imagine the data looks like this. toArray(new String[]{}); } Share. xlsx file. Okay, if you can't to make your file downloadable and return it as a json. split("\\s") and use the size of the resulting array to see how many words are on that line. The fact that you open it on a windows system means that you either have to convert the file before using it (using something you have written, or using a program like unix2dos), or you have to output the file with How to return a zip file stream using Java Springboot. System. txt) I try to add it to zip file set servlet as response. removeExtension(fileName); Use the length() method of the File class to return the size of the file in bytes. GET) p You should be able to use a ByteArrayOutputStream instead of a FileOutputStream:. For example, if the instance variable is called myFiles then you'd do this: File. Your exception handling is the problem. for example: @GetMapping public HttpEntity returnByteArray() { String filepath = ; //filepath String contentType = FileTypeMap. As a complementary, I'd like to show how to use java. seek(). java command prompt output: But when I use System. This question has been answered several times e. 2. Read from file with BufferedReader. I can't use System. Code like below does that. So far I have : File[] fileList = new File(user. class); src/main/resources will not be there when you deploy. This solution is about 3. zipOutputStream = new ZipOutputStream(new ByteArrayOutputStream()); The difficulty here is to provide a File to the method consuming the zip file. getFile (java. `Enter code here. I have the following server called Server. out. First, I would use a BufferedReader to read the file file in line-by-line using readLine(). If you want to keep it in your jar file, you'll want to use Class. getOriginalFilename()) this will create file in server location where sometime you will face write permission issues for the user, its not always possible to give write permission to every user who perform action. Here's an example: Person. txt I'd like a way to break this input up, specifically into . My code is working fine as the file is getting downloaded. Finally, add your new Employee instance to a List<Employee> and continue with the next entry. Like many operations in Java, reading a file upload is unneccessarily complex, and difficult to work out from the javadocs. Perhaps someone's submitted a file Instead of returning an array that contains the two values or using a generic Pair class, consider creating a class that represents the result that you want to return, and return an instance of that class. 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