Cách rename nhiều file [mp4,jpg,txt…] trong một thư mục code java

import java.io.File;
import java.io.IOException;

public class FileOps {
public static void main(String[] argv) throws IOException {

File folder = new File("C:\\Users\\iuiuadu9x\\Videos\\4K Video Downloader\\Test");//đường dẫn trong máy mình
File[] listOfFiles = folder.listFiles();
int j=1;

for (int i = 0; i < listOfFiles.length; i++) {

if (listOfFiles[i].isFile()) {

File f = new File("C:\\Users\\iuiuadu9x\\Videos\\4K Video Downloader\\Test\\"+listOfFiles[i].getName());
// f.renameTo(new File("C:\\Users\\iuiuadu9x\\Videos\\4K Video Downloader\\Test\\"+listOfFiles[i].getName().substring(5)));//lưu tên file mới

if(j<10){
f.renameTo(new File("C:\\Users\\iuiuadu9x\\Videos\\4K Video Downloader\\Test\\"+"0"+j+". "+listOfFiles[i].getName()));
j++;
System.out.println(new File("C:\\Users\\iuiuadu9x\\Videos\\4K Video Downloader\\Test\\"+"0"+j+".                 "+listOfFiles[i].getName()));
}else{
f.renameTo(new File("C:\\Users\\iuiuadu9x\\Videos\\4K Video Downloader\\Test\\"+j+". "+listOfFiles[i].getName()));
j++;
}
System.out.println(f);

}
}

System.out.println("conversion is done");
}
}

Addition

import java.io.File;
import java.io.IOException;

public class FileOps {
public static void main(String[] argv) throws IOException {

File folder = new File("C:\\test");// your destination
File[] listOfFiles = folder.listFiles();
//refer: http://regexr.com/
// text =© 2013 Anhdt, TA31M Soft™. 123 Chia Doi Loi Ve - Quang Ha.mp3
String regex = "^.*\\. ";

for (int i = 0; i < listOfFiles.length; i++) {

if (listOfFiles[i].isFile()) {

File f = new File("C:\\test\\"+ listOfFiles[i].getName());
String t = listOfFiles[i].getName().replaceAll(regex, "");
System.out.println(t);
f.renameTo(new File("C:\\test\\"+t));// file new name

}
}

System.out.println("conversion is done");
}
}