使用Java在Windows资源管理器中获取选定的文件路径

Question

I'm developping a Java application in which the user is able to change the background of a JFrame using a picture from his computer.

In order to do that I'm trying to get the path name of the picture he selects in windows explorer.

I use this code to open the explorer but I can't figure out what I got to do next to get the selected file path...

p = new ProcessBuilder("explorer.exe", "/select,C:\\directory\\selectedFile").start();

Any ideas ?

Thanks.

Answer 1

You should use a JFileChooser instead.

Then retrieve the selected file as a File object using the getSelectedFile() method.

Answer 2

更好的选择是JFileChooser，用于选择文件。

Answer 3

我建议使用JFileChooser 。

Answer 4

That may be possible, but it is much easier to use a JFileChooser . You can set the look and feel to mimic the current environment (Windows, in your case). I recommend googling JFileChooser as there are tons of examples (including in the API).