在我的应用程序中,我想用不同的名称保存某个文件的副本(这是我从用户那里得到的)

我真的需要打开文件的内容并将其写入另一个文件吗?

最好的方法是什么?


当前回答

下面是一个解决方案,如果在复制时发生错误,它实际上会关闭输入/输出流。这个解决方案利用apache Commons IO IOUtils方法来复制和处理流的关闭。

    public void copyFile(File src, File dst)  {
        InputStream in = null;
        OutputStream out = null;
        try {
            in = new FileInputStream(src);
            out = new FileOutputStream(dst);
            IOUtils.copy(in, out);
        } catch (IOException ioe) {
            Log.e(LOGTAG, "IOException occurred.", ioe);
        } finally {
            IOUtils.closeQuietly(out);
            IOUtils.closeQuietly(in);
        }
    }

其他回答

或者,您也可以使用FileChannel来复制文件。在复制大文件时,它可能比字节复制方法更快。如果你的文件大于2GB,你就不能使用它。

public void copy(File src, File dst) throws IOException {
    FileInputStream inStream = new FileInputStream(src);
    FileOutputStream outStream = new FileOutputStream(dst);
    FileChannel inChannel = inStream.getChannel();
    FileChannel outChannel = outStream.getChannel();
    inChannel.transferTo(0, inChannel.size(), outChannel);
    inStream.close();
    outStream.close();
}

这些对我来说很有效

public static void copyFileOrDirectory(String srcDir, String dstDir) {

    try {
        File src = new File(srcDir);
        File dst = new File(dstDir, src.getName());

        if (src.isDirectory()) {

            String files[] = src.list();
            int filesLength = files.length;
            for (int i = 0; i < filesLength; i++) {
                String src1 = (new File(src, files[i]).getPath());
                String dst1 = dst.getPath();
                copyFileOrDirectory(src1, dst1);

            }
        } else {
            copyFile(src, dst);
        }
    } catch (Exception e) {
        e.printStackTrace();
    }
}

public static void copyFile(File sourceFile, File destFile) throws IOException {
    if (!destFile.getParentFile().exists())
        destFile.getParentFile().mkdirs();

    if (!destFile.exists()) {
        destFile.createNewFile();
    }

    FileChannel source = null;
    FileChannel destination = null;

    try {
        source = new FileInputStream(sourceFile).getChannel();
        destination = new FileOutputStream(destFile).getChannel();
        destination.transferFrom(source, 0, source.size());
    } finally {
        if (source != null) {
            source.close();
        }
        if (destination != null) {
            destination.close();
        }
    }
}

在kotlin中,只需:

val fileSrc : File = File("srcPath")
val fileDest : File = File("destPath")

fileSrc.copyTo(fileDest)

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

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());
  }