我在资产文件夹里有几个文件。我需要把它们都复制到一个文件夹,比如/sdcard/folder。我想从一个线程中做这件事。我该怎么做?


当前回答

您还可以使用Guava的ByteStream将文件从资产文件夹复制到SD卡。这是我最终得到的解决方案,将文件递归地从资产文件夹复制到SD卡:

/**
 * Copies all assets in an assets directory to the SD file system.
 */
public class CopyAssetsToSDHelper {

    public static void copyAssets(String assetDir, String targetDir, Context context) 
        throws IOException {
        AssetManager assets = context.getAssets();
        String[] list = assets.list(assetDir);
        for (String f : Objects.requireNonNull(list)) {
            if (f.indexOf(".") > 1) { // check, if this is a file
                File outFile = new File(context.getExternalFilesDir(null), 
                    String.format("%s/%s", targetDir, f));
                File parentFile = outFile.getParentFile();
                if (!Objects.requireNonNull(parentFile).exists()) {
                    if (!parentFile.mkdirs()) {
                        throw new IOException(String.format("Could not create directory %s.", 
                            parentFile));
                    }
                }
                try (InputStream fin = assets.open(String.format("%s/%s", assetDir, f));
                     OutputStream fout = new FileOutputStream(outFile)) {
                    ByteStreams.copy(fin, fout);
                }
            } else { // This is a directory
                copyAssets(String.format("%s/%s", assetDir, f), String.format("%s/%s", targetDir, f), 
                    context);
            }
        }
    }

}

其他回答

复制所有文件和目录从资产到您的文件夹!

为了更好地复制,使用apache Commons IO

public void doCopyAssets() throws IOException {
    File externalFilesDir = context.getExternalFilesDir(null);

    doCopy("", externalFilesDir.getPath());

}

//这是复制的主要方法

private void doCopy(String dirName, String outPath) throws IOException {

    String[] srcFiles = assets.list(dirName);//for directory
    for (String srcFileName : srcFiles) {
        String outFileName = outPath + File.separator + srcFileName;
        String inFileName = dirName + File.separator + srcFileName;
        if (dirName.equals("")) {// for first time
            inFileName = srcFileName;
        }
        try {
            InputStream inputStream = assets.open(inFileName);
            copyAndClose(inputStream, new FileOutputStream(outFileName));
        } catch (IOException e) {//if directory fails exception
            new File(outFileName).mkdir();
            doCopy(inFileName, outFileName);
        }

    }
}

public static void closeQuietly(AutoCloseable autoCloseable) {
    try {
        if(autoCloseable != null) {
            autoCloseable.close();
        }
    } catch(IOException ioe) {
        //skip
    }
}

public static void copyAndClose(InputStream input, OutputStream output) throws IOException {
    copy(input, output);
    closeQuietly(input);
    closeQuietly(output);
}

public static void copy(InputStream input, OutputStream output) throws IOException {
    byte[] buffer = new byte[1024];
    int n = 0;
    while(-1 != (n = input.read(buffer))) {
        output.write(buffer, 0, n);
    }
}

使用这个问题答案中的一些概念,我编写了一个名为assetcopyer的类来简化复制/assets/。它在github上可用,可以通过jitpack.io访问:

new AssetCopier(MainActivity.this)
        .withFileScanning()
        .copy("tocopy", destDir);

详情见https://github.com/flipagram/android-assetcopier。

如果其他人也有同样的问题,我就是这么做的

private void copyAssets() {
    AssetManager assetManager = getAssets();
    String[] files = null;
    try {
        files = assetManager.list("");
    } catch (IOException e) {
        Log.e("tag", "Failed to get asset file list.", e);
    }
    if (files != null) for (String filename : files) {
        InputStream in = null;
        OutputStream out = null;
        try {
          in = assetManager.open(filename);
          File outFile = new File(getExternalFilesDir(null), filename);
          out = new FileOutputStream(outFile);
          copyFile(in, out);
        } catch(IOException e) {
            Log.e("tag", "Failed to copy asset file: " + filename, e);
        }     
        finally {
            if (in != null) {
                try {
                    in.close();
                } catch (IOException e) {
                    // NOOP
                }
            }
            if (out != null) {
                try {
                    out.close();
                } catch (IOException e) {
                    // NOOP
                }
            }
        }  
    }
}
private void copyFile(InputStream in, OutputStream out) throws IOException {
    byte[] buffer = new byte[1024];
    int read;
    while((read = in.read(buffer)) != -1){
      out.write(buffer, 0, read);
    }
}

参考:使用Java移动文件

使用AssetManager,它允许读取资产中的文件。然后使用常规的Java IO将文件写入sdcard。

谷歌是你的朋友,搜索一个例子。

您还可以使用Guava的ByteStream将文件从资产文件夹复制到SD卡。这是我最终得到的解决方案,将文件递归地从资产文件夹复制到SD卡:

/**
 * Copies all assets in an assets directory to the SD file system.
 */
public class CopyAssetsToSDHelper {

    public static void copyAssets(String assetDir, String targetDir, Context context) 
        throws IOException {
        AssetManager assets = context.getAssets();
        String[] list = assets.list(assetDir);
        for (String f : Objects.requireNonNull(list)) {
            if (f.indexOf(".") > 1) { // check, if this is a file
                File outFile = new File(context.getExternalFilesDir(null), 
                    String.format("%s/%s", targetDir, f));
                File parentFile = outFile.getParentFile();
                if (!Objects.requireNonNull(parentFile).exists()) {
                    if (!parentFile.mkdirs()) {
                        throw new IOException(String.format("Could not create directory %s.", 
                            parentFile));
                    }
                }
                try (InputStream fin = assets.open(String.format("%s/%s", assetDir, f));
                     OutputStream fout = new FileOutputStream(outFile)) {
                    ByteStreams.copy(fin, fout);
                }
            } else { // This is a directory
                copyAssets(String.format("%s/%s", assetDir, f), String.format("%s/%s", targetDir, f), 
                    context);
            }
        }
    }

}