如何在android中制作文件的副本？

Question

在我的应用中，我想用不同的名称（我从用户那里获得）保存某个文件的副本

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？

Answer 1

要复制文件并将其保存到目标路径，您可以使用以下方法。

public static void copy(File src, File dst) throws IOException {
    InputStream in = new FileInputStream(src);
    try {
        OutputStream out = new FileOutputStream(dst);
        try {
            // Transfer bytes from in to out
            byte[] buf = new byte[1024];
            int len;
            while ((len = in.read(buf)) > 0) {
                out.write(buf, 0, len);
            }
        } finally {
            out.close();
        }
    } finally {
        in.close();
    }
}

在 API 19+ 上，您可以使用 Java 自动资源管理：

public static void copy(File src, File dst) throws IOException {
    try (InputStream in = new FileInputStream(src)) {
        try (OutputStream out = new FileOutputStream(dst)) {
            // Transfer bytes from in to out
            byte[] buf = new byte[1024];
            int len;
            while ((len = in.read(buf)) > 0) {
                out.write(buf, 0, len);
            }
        }
    }
}

Answer 2

或者，您可以使用FileChannel 复制文件。在复制大文件时，它可能比字节复制方法快。 You can't use it if your file is bigger than 2GB though.

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();
}

Answer 3

它的 Kotlin 扩展

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

Answer 4

这在 Android O (API 26) 上很简单，如您所见：

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

Answer 5

这些对我很有效

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();
        }
    }
}

Answer 6

现在使用 Kotlin 变得更加简单：

 File("originalFileDir", "originalFile.name")
            .copyTo(File("newFileDir", "newFile.name"), true)

trueorfalse 用于覆盖目标文件

https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.io/java.io.-file/copy-to.html

Answer 7

答案可能为时已晚，但最方便的方法是使用

FileUtils的

static void copyFile(File srcFile, File destFile)

例如这就是我所做的

`

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;
    }

`

Answer 8

在 kotlin 中，只需：

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

fileSrc.copyTo(fileDest)

Answer 9

如果在复制时发生错误，这是一个实际关闭输入/输出流的解决方案。此解决方案利用 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);
        }
    }

Answer 10

在 Kotlin 中：一条捷径

// 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，例如图像和视频

Answer 11

现在在 kotlin 中你可以使用

file1.copyTo(file2)

其中 file1 是原始文件的对象，file2 是要复制到的新文件的对象