How to Create a File instance from the Jar/Classpath

Question

I have an executable JAR that has a file in it that i want to open as a java.io.File instance from code (not InputStream or anything else...just File).

Its a maven project and the file is at the root of "src/main/resources/file.xxx" The file is located at the root directory of the jar after packaging(verified that its there).

My first attempt: FileNotFoundException

java.io.File myFile = new java.io.File("file.xxx");
someMethodThatUsesTheFile(myFile); //I really need it to be a file!!!

Other attempts: FileNotFoundException

java.io.File myFile = new java.io.File("/file.xxx");
java.io.File myFile = new java.io.File("classpath:file.xxx");
java.io.File myFile = new java.io.File("classpath:/file.xxx");

I am not sure whats really going on. Web Apps can easily just load everything from the webapp root directory, Im confused as to why JAR apps behave differently.

Additional Info:

• Using Java8 as runtime/build
• command to run the JAR: "java -jar myjar.jar"
• Application Code and file are both located in the same jar

Answer 1

Short Answer: It is not a "File", so you just cant do it.

The JAR file is a File, but not its contents.

Alternatives would be:

• Try other overloaded versions of that method, InputStreams are usually your options, you can load it using this code:

  InputStream is = getClass().getResourceAsStream("file.xxx");

• Move that file out of the JAR and into the same directory of the JAR (consider the security risks)

Answer 2

Try this one may help you :

URL url = this.getClass().getResource("src/main/resources/file.xxx");
File file= new File(url.toURI());