How to get short-filenames in Windows using Java?

问题

How to get the short-filename for a long-filename in Windows using Java?

I need to determine the short-filenames of files, stored on a Windows system, using Java(tm).

回答1:

Self Answer

There are related questions with related answers. I post this solution, however, because it uses Java(tm) code without the need for external libraries. Additional solutions for different versions of Java and/or Microsoft(R) Windows(tm) are welcome.

Main Concept

Main concept lies in calling CMD from Java(tm) by means of the runtime class:

cmd /c for %I in ("[long file name]") do @echo %~fsI

Solution

Tested on Java SE 7 running on Windows 7 system (Code has been reduced for brevity).

    public static String getMSDOSName(String fileName)
    throws IOException, InterruptedException {

    String path = getAbsolutePath(fileName);

    // changed "+ fileName.toUpperCase() +" to "path"
    Process process =
        Runtime.getRuntime().exec(
            "cmd /c for %I in (\"" + path + "\") do @echo %~fsI");

    process.waitFor();

    byte[] data = new byte[65536];
    int size = process.getInputStream().read(data);

    if (size <= 0)
        return null;

    return new String(data, 0, size).replaceAll("\\r\\n", "");
}

public static String getAbsolutePath(String fileName)
    throws IOException {
    File file = new File(fileName);
    String path = file.getAbsolutePath();

    if (file.exists() == false)
        file = new File(path);

    path = file.getCanonicalPath();

    if (file.isDirectory() && (path.endsWith(File.separator) == false))
        path += File.separator;

    return path;
}

来源：https://stackoverflow.com/questions/18893284/how-to-get-short-filenames-in-windows-using-java