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Java can't get relative path for file

Image DescriptionTrying to access a test.txt file that is in the same location as my HelloController.java file but for some reason, it is showing that the file does not exist. I've tried moving the file around but it does not work.
Using the absolute path works, but this is a shared project so it will be ran on other computers. Any suggestions would be much appreciated.

Your best bet is to add it to the class path and reading it as a class path resource.

The relative path root is your "working directory". Which means if you try to access "." you will start at your working directory. This directory is only set once for your application and is normally the folder which was opened when you started it.
When working with IDEs (like in your case) the working directory will be the root folder of your project (So the folder in which the pom.xml and src folders are located.
If you want to access the file via the normal file API you are currently using, just put the file in that diretory and it should work (given you share it with the other people in the same location).
If you need the file to be inside your generated output jar-file, you will need to use the File as a resource (See duffymo's answer), as the file does not exist by itself on the file system, but as a file inside your jar-file.
If you want to know your current working directory, you can create a File refrence to "." and expand it to an absolute path (Which will replace refrences like "." and ".." and generate a file path from your root) and then write it to the console. This would look something like this:
// Get refrence to the current working directory
File workingDirectoryReference = new File(".");
// Convert it to an absolute path string
String absolutePath = workingDirectoryReference.getAbsolutePath();
// Output to console
System.out.println(absolutePath );

cannot locate properties file in java path

My code cannot locate the .properties file where i have stored login information.
I have put the file in the src folder to make sure it compiles, and it does correctly.
below is the current location of the file and how i am trying to access it.
I have tried various different paths but no luck.

Change your code;
ResourceBundle bundle = ResourceBundle.getBundle("Selenium/readme");
to
ResourceBundle bundle = ResourceBundle.getBundle("readme");

You don't compile a .properties file, you use it as is.
If you use a FileInputStream it will use the working directory which is set in your Run configuration (most likely the top directory)
But you are loading it as a resource which means it must be in your class path. The simplest thing to do is to create a sub-directory for your configuration and add this to your programs class path.

Read properties file like:
ResourceBundle.getBundle("src/properties/readme.properties"); //Or simply "properties/readme.properties"
Put readme.properties under src/properties directory.

you can use this.getClass().getResourceAsStream("readme.properties");
for more information read:

I see that the .properties file is not inside the src folder. Also check the build path of your project.It will show you the src folders and the output folders location. Once you build the project using eclipse build project option, make sure your properties file is now available in the output folder.

How to Specify Path for Properties file

I am using Config. properties file for passing parameters to my methods Now i am loading file from
Properties Config= new Properties();
Config.load(new FileInputStream("C:\\Config. properties "));
As i don't want to keep it hard coded how can i set it with package level. or within application.
Thanks in Advance.

Make use of ResourceBundle Class. You just need to specify the properties file name. It will take the file from any path,provided the path should be in the classpath.
Example:
// abc.properties is the properties file,which is placed in the class path.You just need to
// specify its name and the properties file gets loaded.
ResourceBundle s=ResourceBundle.getBundle("abc");
s.getString("key"); //any key from properties file...

I was also just going to suggest that but you can also pass in the full path to the config file via a command line argument for example:
java YourApp -config C:\\config.properties

A properties file packaged with the application should not be loaded using the file system, but using the class loader. Indeed, the properties file, once the application is packaged, will be embedded inside a jar file, with the .class files.
If the config.properties file is in the package com.foo.bar, then you should load it using
InputStream in = SomeClass.class.getResourceAsStream("/com/foo/bar/config.properties");
Or with
InputStream in = SomeClass.class.getClassLoader().getResourceAsStream("com/foo/bar/config.properties");
You may also load it with a relative path. If SomeClass is also in the package com.foo.bar, then you may load it with.
InputStream in = SomeClass.class.getResourceAsStream("config.properties");
Note that Java variables should always start with a lowercase letter: config and not Config.

If it's just the path you're worried about then you can use a relative path:
Config.load(new FileInputStream("Config.properties"));
This will look in the current working directory. The upsdie: dead simple. The downside: it's not that robust. If you start your application from somewhere else without changing the working directory before, the file won't be found.

Put the config file in the classpath (where your .class files are), and access it using
getClass().getClassLoader().getResourceAsStream(_path_to_config_file);

There are two ways to get the path of the config files at runtime.
a) Getting it from database.
b) Getting it from custom properties of JVM configured at server level
Best process is "b" , you can change the properties of JVM at any time if path is changed and just restart the server.

Absolute Path of Project's folder in Java

Lots of confusion in this topic. Several Questions have been asked. Things still seem unclear.
ClassLoader, Absolute File Paths etc etc
Suppose I have a project directory structure as,
MyProject--
--dist
--lib
--src
--test
I have a resource say "txtfile.txt" in "lib/txt" directory. I want to access it in a system independent way. I need the absolute path of the project.
So I can code the path as abspath+"/lib/Dictionary/txtfile.txt"
Suppose I do this
java.io.File file = new java.io.File(""); //Dummy file
String abspath=file.getAbsolutePath();
I get the current working directory which is not necessarily project root.
Suppose I execute the final 'prj.jar' from the 'dist' folder which also contains "lib/txt/txtfile.txt" directory structure and resource,It should work here too. I should absolute path of dist folder.
Hope the problem is clear.

You should really be using getResource() or getResourceAsStream() using your class loader for this sort of thing. In particular, these methods use your ClassLoader to determine the search context for resources within your project.
Specify something like getClass().getResource("lib/txtfile.txt") in order to pick up the text file.
To clarify: instead of thinking about how to get the path of the resource you ought to be thinking about getting the resource -- in this case a file in a directory somewhere (possibly inside your JAR). It's not necessary to know some absolute path in this case, only some URL to get at the file, and the ClassLoader will return this URL for you. If you want to open a stream to the file you can do this directly without messing around with a URL using getResourceAsStream.
The resources you're trying to access through the ClassLoader need to be on the Class-Path (configured in the Manifest of your JAR file). This is critical! The ClassLoader uses the Class-Path to find the resources, so if you don't provide enough context in the Class-Path it won't be able to find anything. If you add . the ClassLoader should resolve anything inside or outside of the JAR depending on how you refer to the resource, though you can certainly be more specific.
Referring to the resource prefixed with a . will cause the ClassLoader to also look for files outside of the JAR, while not prefixing the resource path with a period will direct the ClassLoader to look only inside the JAR file.
That means if you have some file inside the JAR in a directory lib with name foo.txt and you want to get the resource then you'd run getResource("lib/foo.txt");
If the same resource were outside the JAR you'd run getResource("./lib/foo.txt");

First, make sure the lib directory is in your classpath. You can do this by adding the command line parameter in your startup script:
$JAVA_HOME/bin/java -classpath .:lib com.example.MyMainClass
save this as MyProject/start.sh or any os dependent script.
Then you can access the textfile.txt (as rightly mentioned by Mark) as:
// if you want this as a File
URL res = getClass().getClassLoader().getResource("text/textfile.txt");
File f = new File(res.getFile());
// As InputStream
InputStream in = getClass().getClassLoader()
.getResourceAsStream("text/textfile.txt");

#Mark is correct. That is by far the simplest and most robust approach.
However, if you really have to have a File, then your best bet is to try the following:
turn the contents of the System property "java.class.path" into a list of pathnames,
identify the JAR pathname in the list based on its filename,
figure out what "../.." is relative to the JAR pathname to give you the "project" directory, and
build your target path relative to the project directory.
Another alternative is to embed the project directory name in a wrapper script and set it as a system property using a -D option. It is also possible to have a wrapper script figure out its own absolute pathname; e.g. using whence.

How to specify filepath in java?

I have created a java application for "Debian Linux." Now I want that that application reads a file placed in the directory where the jar file of that application is specified. So what to specify at the argument of the File Object?
File fileToBeReaded = new File(...);
What to specify as argument for the above statement to specify relative filepath representing the path where the jar file of the application has been placed?

If you know the name of the file, of course it's simply
new File("./myFileName")
If you don't know the name, you can use the File object's list() method to get a list of files in the current directory, and then pick the one you want.

Are you asking about escape character issues?
If that is the case then use forward slashes instead of backward slashes like
"C:/Users/You/Desktop/test.txt"
instead of
"C:\Users\You\Desktop\test.txt"

Using relative paths in java.io.File is fully dependent on the current working directory. This differs with the way you execute the JAR. If you're for example in /foo and you execute the JAR by java -jar /bar/jar/Bar.jar then the working directory is still /foo. But if you cd to /bar/jar and execute java -jar Bar.jar then the working directory is /bar/jar.
If you want the root path where the JAR is located, one of the ways would be:
File root = new File(Thread.currentThread().getContextClassLoader().getResource("").toURI());
This returns the root path of the JAR file (i.o.w. the classpath root). If you place your resource relative to the classpath root, you can access it as follows:
File resource = new File(root, "filename.ext");
Alternatively you can also just use:
File resource = new File(Thread.currentThread().getContextClassLoader().getResource("filename.ext").toURI());

I think this should do the trick:
File starting = new File(System.getProperty("user.dir"));
File fileToBeRead = new File(starting,"my_file.txt");
This way, the file will be searched in the user.dir property, which will be your app's working directory.

You could ask your classloader to give you the location of the jar:
getClass().getProtectionDomain().getCodeSource().getLocation().getPath();
...but I'd suggest to put the file you are looking for inside your jar file and read it as a resource (getClass().getResourceAsStream( "myFile.txt" )).

On IntelliJIDEA right click on the file then copy the absolute path, then in the double quotation paste the path as filePath.
for example it should be something like this:
"C:\\Users\\NameOfTheComputerUser\\IdeaProjects\\NameOfTheProject\\YourSubFolders\\name-of-the-file.example"