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store file in spring boot resource folder after deployment

I have deployed a spring-boot application JAR file. Now, I want to upload the image from android and store it in the myfolder of resource directory. But unable to get the path of resource directory.
Error is:
java.io.FileNotFoundException: src/main/resources/static/myfolder/myimage.png
(No such file or directory)
This is the code for storing the file in the resource folder
private final String RESOURCE_PATH = "src/main/resources";
String filepath = "/myfolder/";
public String saveFile(byte[] bytes, String filepath, String filename) throws MalformedURLException, IOException {
File file = new File(RESOURCE_PATH + filepath + filename);
OutputStream out = new FileOutputStream(file);
try {
out.write(bytes);
} catch (Exception e) {
e.printStackTrace();
} finally {
out.close();
}
return file.getName();
}
UPDATED:
This is what I have tried
private final String RESOURCE_PATH = "config/";
controller class:
String filepath = "myfolder/";
String filename = "newfile.png"
public String saveFile(byte[] bytes, String filepath, String filename) throws MalformedURLException, IOException {
//reading old file
System.out.println(Files.readAllBytes(Paths.get("config","myfolder","oldfile.png"))); //gives noSuchFileException
//writing new file
File file = new File(RESOURCE_PATH + filepath + filename);
OutputStream out = new FileOutputStream(file); //FileNotFoundException
try {
out.write(bytes);
} catch (Exception e) {
e.printStackTrace();
} finally {
out.close();
}
return file.getName();
}
Project structure:
+springdemo-0.0.1-application.zip
+config
+myfolder
-oldfile.png
-application.properties
+lib
+springdemo-0.0.1.jar
+start.sh
-springdemo-0.0.1.jar //running this jar file

Usually when you deploy an application (or start it using Java), you start a JAR file. You don't have a resource folder. You can have one and access it, too, but it certainly won't be src/main/resources.
When you build your final artifact (your application), it creates a JAR (or EAR or WAR) file and your resources, which you had in your src/main/resources-folder, are copied over to the output directory and included in the final artifact. That folder simply does not exist when the application is run (assuming you are trying to run it standalone).
During the build process target/ is created and contains the classes, resources, test-resources and the likes (assuming you are building with Maven; it is a little different if you build using Gradle or Ant or by hand).
What you can do is create a folder e.g. docs next to your final artifact, give it the appropriate permissions (chmod/chown) and have your application output files into that folder. This folder is then expected to exist on the target machine running your artifact, too, so if it doesn't, it would mean the folder does not exist or the application lacks the proper permissions to read from / write to that folder.
If you need more details, don't hesitate to ask.
Update:
To access a resource, which is bundled and hence inside your artifact (e.g. final.jar), you should be able to retrieve it by using e.g. the following:
testText = new String(ControllerClass.class.getResourceAsStream("/test.txt").readAllBytes());
This is assuming your test.txt file is right under src/main/resources and was bundled to be directly in the root of your JAR-file (or target folder where your application is run from). ControllerClass is the controller, which is accessing the file. readAllBytes just does exactly this: read all the bytes from a text file. For accessing images inside your artifact, you might want to use ImageIO.
IF you however want to access an external file, which is not bundled and hence not inside your artifact, you may use File image = new File(...) where ... would be something like "docs/image.png". This would require you to create a folder called docs next to your JAR-artifact and put a file image.png inside of it.
You of course also may work with streams and there are various helpful libraries for working with input- and output streams.
The following was meant for AWT, but it works in case you really want to access the bytes of your image: ImageIO. In a controller you usually wouldn't want to do that, but rather have your users access (and thus download) it from a given available folder.
I hope this helps :).

Proper packaging of runnable Jar project in netbeans

So my task is to create a small program that displays a list of media files and run these media files with default OS media player separately.
My current solution was to create a package that holds all media files, something like:
-com.media
|_a.mp4
|_b.mp4
The following code copies to a temp dir the selected mp4, then runs the default os media player:
public File copyTempMedia(File tempAppFolder, String videoName) {
URL f = getClass().getResource(String.format("%s/%s", Constants.MEDIA_LOCATION, videoName));
File from = new File(f.getPath());
File to = new File(tempAppFolder.getAbsolutePath());
try {
FileUtils.copyFileToDirectory(from, to);
} catch (IOException ex) {
Logger.getLogger(MediGUIModel.class.getName()).log(Level.SEVERE, null, ex);
}
System.out.println("Temp video copied: " + to.getAbsolutePath() + "/" + to.getName());
return to;
}
public void triggerMediaPlayer(String fileLocation) {
System.out.println("Triggering media player: " + fileLocation);
try {
if (OPERATIN_SYSTEM.contains("Linux")) {
Runtime.getRuntime().exec("sh -c " + fileLocation);
} else if (OPERATIN_SYSTEM.contains("Windows")) {
Runtime.getRuntime().exec("cmd /c " + fileLocation);
}
} catch (IOException ex) {
Logger.getLogger(MediGUIModel.class.getName()).log(Level.SEVERE, null, ex);
ex.printStackTrace();
}
}
When I run the program through Netbeans it works as espected, but when I do a clean/build the run the .jar created from the build, the media file doesn't seem to be read, so my questions are:
Why does it work through Netbeans and not through build .jar ?
Is this the best solution to this problem ?
Should I package the media differently ?
Thanks in advance.
Edit
So after running through console instead of double clicking jar, is get a null pointer exception in the line where I read the file:
URL f = getClass().getResource(String.format("%s/%s", Constants.MEDIA_LOCATION, videoName));
Why does it work in Netebeans but not on build/jar ?
Is there another place in the jar I could place the media files, so that they are read with no problem through getResource or getResourceAsStream ?

When you run the project in NetBeans, it isn't running the executable jar like java -jar yourproject.jar. Instead it sets the classpath to build/classes sort of like java -cp build/classes com.media.YourMainClass. This means your video files are actual files located in yourproject/build/classes/com/media, and they can be accessed as normal files in the filesystem and copied like a normal file. When you run from the jar, the files are packed in the jar file and can't be copied using simple file copy commands.
Instead of getting the URL by calling getClass().getResource(), try getting an InputStream by calling getClass().getResourceAsStream(). You can then write a simple loop to copy the bytes from the input stream to your temporary file.

This snippet may be helpful:
BufferedInputStream result = (BufferedInputStream) getClass().getResourceAsStream("/com/media/a.mp4");
byte[] bytes = new byte[4098];
try {
result.read(bytes);
} catch (IOException e) {
e.printStackTrace();
}
System.out.println(new String(bytes));
You'll need to read the bytes in a loop or something but that should work without needing a separate jar.

I think it's not a good idea to put your media files in the jar because you need to rebuild the project if you want to change one and the jar size will grow.
Use:
File from = new File(String.format("%s/%s", Constants.MEDIA_LOCATION,videoName));
To load your files from the same folder as the jar.
If you want to keep the medias in the jar, create a Maven project and put the mp4 in src/main/resources.
Use maven to create a fat jar and the src/main/resources will be included in the jar.
See 'maven-shade-plugin' to configure the pom.xml and https://www.mkyong.com/maven/create-a-fat-jar-file-maven-shade-plugin/
Then you can use the others maven's great properties!
See Reading a resource file from within jar
Edit
After some tries, i can't get it right with 'getResource' from the jar.
The path you get from within the jar is like:file:/C:/.../JavaApplication4/dist/JavaApplication4.jar!/test.txt
and not recognized as a valid filesystem path.
You can use 'getResourceAsStream' and copy the file from the jar to the local folder.
InputStream in;
OutputStream out;
IOUtils.copy(in,out);
in.close();
out.close();

Ok so I found a solution:
Create a separate project with media.*.mp4.
Export as Jar library.
Import library to desktop app.
Make sure library is in classpath.
This solution works for me...
If anyone has a better solution, happy to hear, hopefully before bounty is up :)

How to create executable file in Java using Netbeans

I made a cache simulator program for a homework, I decided to use java. I want to create an executable jar file that will work on any system, but the problem is that my program gathers data from an external text file. How can I include that text file inside the jar so that there won't be any problem when executing file? By the way, I am using NetBeans IDE.

If you don't need to write to the file, copy into the src directory. You will no longer be able to access like a File, but instead will need to use Class#getResource, passing it the path from the top of the source tree to where the file is stored.
For example, if you put it in src/data, then you'd need to use getClass().getResource("/data/..."), passing it what ever name the file is...
Clean and build...

Yes and I said Yes. To really make your jarfiles along with the text files. Please ensure that links to the folder on which the text files is where properly coded and well linked.
The three Examplary Method below should get you working irrespective of any IDEs. Please rate this and give me a shout if you still need further help.......Sectona
Method 1
Step 1:- Locate your folder that contain your java file by using cd command.
Step 2:- Once your enter your folder location then view your java file by dir
command.
Step 3:- Compile your java file using
javac file.java
Step 4:- view class file by type dir command.
Step 5:- Now you want to create a manifest file.
I)Go to folder<br>
II)Right-click->New->Text Document
III)open text document. Type main class name ,
Main-Class: main-class-name
IV)Save this file your wish like MyManifest.txt
Step 6:- To create executable jar file type
jar cfm JarFileName.jar MyManifest.txt JavaFike1.class JavaFile2.class
Step 7:- Now you see the Executable jar file on your folder. Click the file to
Run.
Step 8:- To run this file via command prompt then type
java -jar JarFileName.jar
Step 9:- You done this..........Sectona
Method 2
The basic format of the command for creating a JAR file is:
jar cf jar-file input-file(s)
The options and arguments used in this command are:
The c option indicates that you want to create a JAR file.
The f option indicates that you want the output to go to a file rather than to stdout.
jar-file is the name that you want the resulting JAR file to have. You can use any filename for a JAR file. By convention, JAR filenames are given a .jar extension, though this is not required.
The input-file(s) argument is a space-separated list of one or more files that you want to include in your JAR file. The input-file(s) argument can contain the wildcard * symbol. If any of the "input-files" are directories, the contents of those directories are added to the JAR archive recursively.
Method 3
import java.io.*;
import java.util.jar.*;
public class CreateJar {
public static int buffer = 10240;
protected void createJarArchive(File jarFile, File[] listFiles) {
try {
byte b[] = new byte[buffer];
FileOutputStream fout = new FileOutputStream(jarFile);
JarOutputStream out = new JarOutputStream(fout, new Manifest());
for (int i = 0; i < listFiles.length; i++) {
if (listFiles[i] == null || !listFiles[i].exists()|| listFiles[i].isDirectory())
System.out.println();
JarEntry addFiles = new JarEntry(listFiles[i].getName());
addFiles.setTime(listFiles[i].lastModified());
out.putNextEntry(addFiles);
FileInputStream fin = new FileInputStream(listFiles[i]);
while (true) {
int len = fin.read(b, 0, b.length);
if (len <= 0)
break;
out.write(b, 0, len);
}
fin.close();
}
out.close();
fout.close();
System.out.println("Jar File is created successfully.");
} catch (Exception ex) {}
}
public static void main(String[]args){
CreateJar jar=new CreateJar();
File folder = new File("C://Answers//Examples.txt");
File[] files = folder.listFiles();
File file=new File("C://Answers//Examples//Examples.jar");
jar.createJarArchive(file, files);
}
}

You can keep any file in classpath and read as class path resource. Sample code is given below.
InputStream in = this.getClass().getClassLoader().getResourceAsStream("yourinputFile.txt");
Your jar will be class path, that means you can keep your file in root folder of java source which will get added to jar file while building it.

Java Jar file: use resource errors: URI is not hierarchical

I have deployed my app to jar file. When I need to copy data from one file of resource to outside of jar file, I do this code:
URL resourceUrl = getClass().getResource("/resource/data.sav");
File src = new File(resourceUrl.toURI()); //ERROR HERE
File dst = new File(CurrentPath()+"data.sav"); //CurrentPath: path of jar file don't include jar file name
FileInputStream in = new FileInputStream(src);
FileOutputStream out = new FileOutputStream(dst);
// some excute code here
The error I have met is: URI is not hierarchical. this error I don't meet when run in IDE.
If I change above code as some help on other post on StackOverFlow:
InputStream in = Model.class.getClassLoader().getResourceAsStream("/resource/data.sav");
File dst = new File(CurrentPath() + "data.sav");
FileOutputStream out = new FileOutputStream(dst);
//....
byte[] buf = new byte[1024];
int len;
while ((len = in.read(buf)) > 0) { //NULL POINTER EXCEPTION
//....
}

You cannot do this
File src = new File(resourceUrl.toURI()); //ERROR HERE
it is not a file!
When you run from the ide you don't have any error, because you don't run a jar file. In the IDE classes and resources are extracted on the file system.
But you can open an InputStream in this way:
InputStream in = Model.class.getClassLoader().getResourceAsStream("/data.sav");
Remove "/resource". Generally the IDEs separates on file system classes and resources. But when the jar is created they are put all together. So the folder level "/resource" is used only for classes and resources separation.
When you get a resource from classloader you have to specify the path that the resource has inside the jar, that is the real package hierarchy.

If for some reason you really need to create a java.io.File object to point to a resource inside of a Jar file, the answer is here: https://stackoverflow.com/a/27149287/155167
File f = new File(getClass().getResource("/MyResource").toExternalForm());

Here is a solution for Eclipse RCP / Plugin developers:
Bundle bundle = Platform.getBundle("resource_from_some_plugin");
URL fileURL = bundle.getEntry("files/test.txt");
File file = null;
try {
URL resolvedFileURL = FileLocator.toFileURL(fileURL);
// We need to use the 3-arg constructor of URI in order to properly escape file system chars
URI resolvedURI = new URI(resolvedFileURL.getProtocol(), resolvedFileURL.getPath(), null);
File file = new File(resolvedURI);
} catch (URISyntaxException e1) {
e1.printStackTrace();
} catch (IOException e1) {
e1.printStackTrace();
}
It's very important to use FileLocator.toFileURL(fileURL) rather than resolve(fileURL)
, cause when the plugin is packed into a jar this will cause Eclipse to create an unpacked version in a temporary location so that the object can be accessed using File. For instance, I guess Lars Vogel has an error in his article - http://blog.vogella.com/2010/07/06/reading-resources-from-plugin/

I got a similiar issues before, and I used the code:
new File(new URI(url.toString().replace(" ","%20")).getSchemeSpecificPart());
instead of the code :
new File(new URI(url.toURI())
to solve the problem

While I stumbled upon this problem myself I'd like to add another option (to the otherwise perfect explanation from #dash1e):
Export the plugin as a folder (not a jar) by adding:
Eclipse-BundleShape: dir
to your MANIFEST.MF.
At least when you export your RCP app with the export wizard (based on a *.product) file this gets respected and will produce a folder.

In addition to the general answers, you can get "URI is not hierarchical" from Unitils library attempting to load a dataset off a .jar file. It may happen when you keep datasets in one maven submodule, but actual tests in another.
There is even a bug UNI-197 filed.

How to pass a text file as a argument?

Im trying to write a program to read a text file through args but when i run it, it always says the file can't be found even though i placed it inside the same folder as the main.java that im running.
Does anyone know the solution to my problem or a better way of reading a text file?

Do not use relative paths in java.io.File.
It will become relative to the current working directory which is dependent on the way how you run the application which in turn is not controllable from inside your application. It will only lead to portability trouble. If you run it from inside Eclipse, the path will be relative to /path/to/eclipse/workspace/projectname. If you run it from inside command console, it will be relative to currently opened folder (even though when you run the code by absolute path!). If you run it by doubleclicking the JAR, it will be relative to the root folder of the JAR. If you run it in a webserver, it will be relative to the /path/to/webserver/binaries. Etcetera.
Always use absolute paths in java.io.File, no excuses.
For best portability and less headache with absolute paths, just place the file in a path covered by the runtime classpath (or add its path to the runtime classpath). This way you can get the file by Class#getResource() or its content by Class#getResourceAsStream(). If it's in the same folder (package) as your current class, then it's already in the classpath. To access it, just do:
public MyClass() {
URL url = getClass().getResource("filename.txt");
File file = new File(url.getPath());
InputStream input = new FileInputStream(file);
// ...
}
or
public MyClass() {
InputStream input = getClass().getResourceAsStream("filename.txt");
// ...
}

Try giving an absolute path to the filename.
Also, post the code so that we can see what exactly you're trying.

When you are opening a file with a relative file name in Java (and in general) it opens it relative to the working directory.
you can find the current working directory of your process using
String workindDir = new File(".").getAbsoultePath()
Make sure you are running your program from the correct directory (or change the file name so that it will be relative to where you are running it from).

If you're using Eclipse (or a similar IDE), the problem arises from the fact that your program is run from a few directories above where the actual source is located. Try moving your file up a level or two in the project tree.
Check out this question for more detail.

The simplest solution is to create a new file, then see where the output file is. That is the correct place to put your input file into.

If you put the file and the class working with it under same package can you use this:
Class A {
void readFile (String fileName) {
Url tmp = A.class.getResource (fileName);
// Or Url tmp = this.getClass().getResource (fileName);
File tmpFile = File (tmp);
if (tmpFile.exists())
System.out.print("I found the file.")
}
}
It will help if you read about classloaders.

say I have a text file input.txt which is located on the desktop
and input.txt has the following content
i came
i saw
i left
and below is the java code for reading that text file
public class ReadInputFromTextFile {
public static void main(String[] args) throws Exception
{
File file = new File(
"/Users/viveksingh/desktop/input.txt");
BufferedReader br
= new BufferedReader(new FileReader(file));
String st;
while ((st = br.readLine()) != null)
System.out.println(st);
}
}
output on the console:
i came
i saw
i left