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