在Kotlin:很短的路

// fromPath : Path the file you want to copy 
// toPath :   The path where you want to save the file
// fileName : name of the file that you want to copy
// newFileName: New name for the copied file (you can put the fileName too instead of put a new name)    

val toPathF = File(toPath)
if (!toPathF.exists()) {
   path.mkdir()
}

File(fromPath, fileName).copyTo(File(toPath, fileName), replace)

这适用于任何文件，如图像和视频

在kotlin中你可以用

file1.copyTo(file2)

file1是原始文件的对象，而file2是要复制到的新文件的对象

Kotlin扩展它

fun File.copyTo(file: File) {
    inputStream().use { input ->
        file.outputStream().use { output ->
            input.copyTo(output)
        }
    }
}

这在Android O (API 26)上很简单，如你所见:

  @RequiresApi(api = Build.VERSION_CODES.O)
  public static void copy(File origin, File dest) throws IOException {
    Files.copy(origin.toPath(), dest.toPath());
  }

现在回答可能太迟了，但最方便的方法是使用

FileUtils's

文件srcFile，文件destFile

这就是我所做的

`

private String copy(String original, int copyNumber){
    String copy_path = path + "_copy" + copyNumber;
        try {
            FileUtils.copyFile(new File(path), new File(copy_path));
            return copy_path;
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }

`

在Kotlin:很短的路

// fromPath : Path the file you want to copy 
// toPath :   The path where you want to save the file
// fileName : name of the file that you want to copy
// newFileName: New name for the copied file (you can put the fileName too instead of put a new name)    

val toPathF = File(toPath)
if (!toPathF.exists()) {
   path.mkdir()
}

File(fromPath, fileName).copyTo(File(toPath, fileName), replace)

这适用于任何文件，如图像和视频