Get file from project folder using java

Question

I want to get File from project folder by using "File class", How can I do that?

File file=new File("x1.txt");

Know

Yes. Something like that. – Keeshakeeshond

What error did you receive when you tried the code above? – Autobus

Answer 1

Well, there are many different ways to get a file in Java, but that's the general gist.

Don't forget that you'll need to wrap that up inside a try {} catch (Exception e){} at the very least, because File is part of java.io which means it must have try-catch block.

Not to step on Ericson's question, but if you are using actual packages, you'll have issues with locations of files, unless you explicitly use it's location. Relative pathing gets messed up with Packages.

ie,

src/
    main.java
    x.txt

In this example, using File f = new File("x.txt"); inside of main.java will throw a file-not-found exception.

However, using File f = new File("src/x.txt"); will work.

Hope that helps!

Speedwell

File itself does not throw any IOExceptions. Some of it's methods do, but simply creating a File object does not require any try-catch...It's also a bad idea to reference the src directory as it will not be available when the application is build and deployed. – Sawyor

Answer 2

This sounds like the file is embedded within your application.

You should be using getClass().getResource("/path/to/your/resource.txt"), which returns an URL or getClass().getResourceAsStream("/path/to/your/resource.txt");

If it's not an embedded resource, then you need to know the relative path from your application's execution context to where your file exists

Sawyor

You have to marvel at blind down votes - In what ways does this not answer the OP's question? In what ways could it be improved? Nothing mentioned, nothing changed – Sawyor

Would it be possible please to see an example of accessing not an embedded resource? Let's say I have a file {project directory}/somewhere/X.txt and I try to access that from {project directory}/src/java/GetX.java. How would that look like? – Esperanzaespial

@Esperanzaespial The problem you have is one of reference, where is the file in reference to the location in which the application was started. Since the application could be started from anywhere (aka the "working directory") it becomes difficult to reliably maintain a reference to the file. You could use a relative reference (ie ./somewhere/X.txt), but as I just stated, this breaks the moment the working directory changes. You could use an absolute path, but the problem becomes, where do you store the file. – Sawyor

@Esperanzaespial This problem isn't unique to Java, or developers have this problem. See this example for a general overview and this example for some details. Essentially, the answer is to follow the conventions of the OS where possible and store content in "well known" locations – Sawyor

Answer 3

If you don't specify any path and put just the file (Just like you did), the default directory is always the one of your project (It's not inside the "src" folder. It's just inside the folder of your project).

Crashaw

Answer 4

If you are trying to load a file which is not in the same directory as your Java class, you've got to use the runtime directory structure rather than the one which appears at design time.

To find out what the runtime directory structure is, check your {root project dir}/target/classes directory. This directory is accessible via the "." URL.

Based on user4284592's answer, the following worked for me:

ClassLoader cl = getClass().getClassLoader();
File file = new File(cl.getResource("./docs/doc.pdf").getFile());

with the following directory structure:

{root dir}/target/classes/docs/doc.pdf

Here's an explanation, so you don't just blindly copy and paste my code:

• java.lang.ClassLoader is an object that is responsible for loading classes. Every Class object contains a reference to the ClassLoader that defined it and can fetch it with getClassLoader() method.
• ClassLoader's getResource method finds the resource with the given name. A resource is some data that can be accessed by class code in a way that is independent of the location of the code.

Victor

Would this work for files outside the package? Just under the same project folder, for instance? – Scotsman

So, in my answer above, the "doc.pdf" file is completely outside the package. The package in question would have a location similar to this one: {root dir}/src/main/java/com/chris/n/whateverMyClassNameIs.java – Victor

Answer 5

String path = System.getProperty("user.dir")+"/config.xml";
File f=new File(path);

Mace

This is really useful in case your file is not in the resources folder, but just in the root folder. – Ulent

Answer 6

These lines worked in my case,

ClassLoader classLoader = getClass().getClassLoader();
File fi = new File(classLoader.getResource("test.txt").getFile());

provided test.txt file inside src

Eboni

Answer 7

Given a file application.yaml in test/resources

ll src/test/resources/
total 6
drwxrwx--- 1 root vboxsf 4096 Oct  6 12:23 ./
drwxrwx--- 1 root vboxsf    0 Sep 29 17:05 ../
-rwxrwx--- 1 root vboxsf  142 Sep 22 23:59 application.properties*
-rwxrwx--- 1 root vboxsf   78 Oct  6 12:23 application.yaml*
-rwxrwx--- 1 root vboxsf    0 Sep 22 17:31 db.properties*
-rwxrwx--- 1 root vboxsf  618 Sep 22 23:54 log4j2.json*

From the test context, I can get the file with

String file = getClass().getClassLoader().getResource("application.yaml").getPath(); 

which will actually point to the file in test-classes

ll target/test-classes/
total 10
drwxrwx--- 1 root vboxsf 4096 Oct  6 18:49 ./
drwxrwx--- 1 root vboxsf 4096 Oct  6 18:32 ../
-rwxrwx--- 1 root vboxsf  142 Oct  6 17:35 application.properties*
-rwxrwx--- 1 root vboxsf   78 Oct  6 17:35 application.yaml*
drwxrwx--- 1 root vboxsf    0 Oct  6 18:50 com/
-rwxrwx--- 1 root vboxsf    0 Oct  6 17:35 db.properties*
-rwxrwx--- 1 root vboxsf  618 Oct  6 17:35 log4j2.json*

Maltha

Answer 8

in javaFx 8

javafx.scene.image.Image img = new javafx.scene.image.Image("/myPic.jpg", true);

URL url = new URL(img.impl_getUrl());

Hautesalpes

Answer 9

Just did a quick google search and found that

System.getProperty("user.dir");

returns the current working directory as String. So to get a File out of this, just use

File projectDir = new File(System.getProperty("user.dir"));

Ephialtes

Answer 10

It is not trivial task (for any location for file request), unfortunately. Java is amazing but this simplest question places to confusion... Try start by this approach:

Path generatorPath;
generatorPath = Path.of(System.getProperty("user.dir") + GetPath(ConfigEnum.Debug));

Also, take into account Debug and Production variants

public String GetPath(ConfigEnum config){    
    if (config == ConfigEnum.Artifacts) {
        return "file:folder/file.txt";
    } else {
        return "/folder/file.txt";
    }
}

public enum ConfigEnum {
    Debug,
    Artifacts
}

Neoplasm