尝试从资源文件夹中读取 txt 文件时出现 Java FileNotFoundException

Question

I'm trying to read a text file located in src/main/resources/test/file.txt . I'm trying to get the path of the file using String path = getClass().getResource("/text/file.txt").getFile();but when I try to read it I get a FileNotFoundException. I tried putting many different paths, all of which failed. How can I go about doing this?

Answer 1

Try giving complete path of the file from the disk.

C:\Users\MyUser\Desktop\file name with extension

Answer 2

The idea of putting something into the src/main/resources tree is that it will be copied into the JAR file that you build from your project. It will then be available to your application via the Class methods getResource(String) and getResourceAsStream(String) methods.

When you are running in your application in the development environment, it is certainly possible to use FileInputStream etcetera to access the resource. But this won't work in production. In production, the resources will then be inside your app's JAR file. FileInputStream cannot open a JAR file and its contents by name .

When you do this:

 getClass().getResource("/text/file.txt");

you get a URL for the resource, which will look something like this:

 jar:file:/path/to/your.jar!/text/file.txt"

It is not possible to turn that into a pathname the FileInputStream will understand. Whatever you try will give you a FileNotFoundException ... or something that is not the resource you want to read.

So what to do?

You have a few options, depending on your application's requirements.

1. You can use getResourceAsStream and use the resulting input stream directly.

2. You can copy the contents of getResourceAsStream to a temporary file, and then use the pathname of the temporary file.

3. You can create an application specific directory (eg in the user's home directory) and extract the file you need from the JAR into the directory. You might do this the first time the application runs.

4. You could open the JAR file as a JarFile and use that API to open an InputStream for the resource. But this assumes that that the resources are in a JAR... and on some platforms (eg Windows) you may encounter problems with file locking. (And it would be a bad idea to attempt to update the resource in the JAR.)