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

基于Yoram Cohen的回答,这里有一个支持非静态目标目录的版本。

使用copyFileOrDir(getDataDir(), "")来写入应用程序内部存储文件夹/data/data/pkg_name/

Supports subfolders. Supports custom and non-static target directory Avoids copying "images" etc fake asset folders like private void copyFileOrDir(String TARGET_BASE_PATH, String path) { AssetManager assetManager = this.getAssets(); String assets[] = null; try { Log.i("tag", "copyFileOrDir() "+path); assets = assetManager.list(path); if (assets.length == 0) { copyFile(TARGET_BASE_PATH, path); } else { String fullPath = TARGET_BASE_PATH + "/" + path; Log.i("tag", "path="+fullPath); File dir = new File(fullPath); if (!dir.exists() && !path.startsWith("images") && !path.startsWith("sounds") && !path.startsWith("webkit")) if (!dir.mkdirs()) Log.i("tag", "could not create dir "+fullPath); for (int i = 0; i < assets.length; ++i) { String p; if (path.equals("")) p = ""; else p = path + "/"; if (!path.startsWith("images") && !path.startsWith("sounds") && !path.startsWith("webkit")) copyFileOrDir(TARGET_BASE_PATH, p + assets[i]); } } } catch (IOException ex) { Log.e("tag", "I/O Exception", ex); } } private void copyFile(String TARGET_BASE_PATH, String filename) { AssetManager assetManager = this.getAssets(); InputStream in = null; OutputStream out = null; String newFileName = null; try { Log.i("tag", "copyFile() "+filename); in = assetManager.open(filename); if (filename.endsWith(".jpg")) // extension was added to avoid compression on APK file newFileName = TARGET_BASE_PATH + "/" + filename.substring(0, filename.length()-4); else newFileName = TARGET_BASE_PATH + "/" + filename; out = new FileOutputStream(newFileName); byte[] buffer = new byte[1024]; int read; while ((read = in.read(buffer)) != -1) { out.write(buffer, 0, read); } in.close(); in = null; out.flush(); out.close(); out = null; } catch (Exception e) { Log.e("tag", "Exception in copyFile() of "+newFileName); Log.e("tag", "Exception in copyFile() "+e.toString()); } }

其他回答

基于您的解决方案,我自己做了一些事情来允许子文件夹。有些人可能会觉得这很有用:

...

copyFileOrDir("myrootdir");

...

private void copyFileOrDir(String path) {
    AssetManager assetManager = this.getAssets();
    String assets[] = null;
    try {
        assets = assetManager.list(path);
        if (assets.length == 0) {
            copyFile(path);
        } else {
            String fullPath = "/data/data/" + this.getPackageName() + "/" + path;
            File dir = new File(fullPath);
            if (!dir.exists())
                dir.mkdir();
            for (int i = 0; i < assets.length; ++i) {
                copyFileOrDir(path + "/" + assets[i]);
            }
        }
    } catch (IOException ex) {
        Log.e("tag", "I/O Exception", ex);
    }
}

private void copyFile(String filename) {
    AssetManager assetManager = this.getAssets();

    InputStream in = null;
    OutputStream out = null;
    try {
        in = assetManager.open(filename);
        String newFileName = "/data/data/" + this.getPackageName() + "/" + filename;
        out = new FileOutputStream(newFileName);

        byte[] buffer = new byte[1024];
        int read;
        while ((read = in.read(buffer)) != -1) {
            out.write(buffer, 0, read);
        }
        in.close();
        in = null;
        out.flush();
        out.close();
        out = null;
    } catch (Exception e) {
        Log.e("tag", e.getMessage());
    }

}

试试这个,它更简单,这将帮助你:

// Open your local db as the input stream
    InputStream myInput = _context.getAssets().open(YOUR FILE NAME);

    // Path to the just created empty db
    String outFileName =SDCARD PATH + YOUR FILE NAME;

    // Open the empty db as the output stream
    OutputStream myOutput = new FileOutputStream(outFileName);

    // transfer bytes from the inputfile to the outputfile
    byte[] buffer = new byte[1024];
    int length;
    while ((length = myInput.read(buffer)) > 0) {
        myOutput.write(buffer, 0, length);
    }
    // Close the streams
    myOutput.flush();
    myOutput.close();
    myInput.close();

对于正在更新到Kotlin的用户:

按照以下步骤避免FileUriExposedExceptions, 假设用户已授予WRITE_EXTERNAL_STORAGE权限,并且您的文件位于assets/pdfs/mypdf.pdf。

private fun openFile() {
    var inputStream: InputStream? = null
    var outputStream: OutputStream? = null
    try {
        val file = File("${activity.getExternalFilesDir(null)}/$PDF_FILE_NAME")
        if (!file.exists()) {
            inputStream = activity.assets.open("$PDF_ASSETS_PATH/$PDF_FILE_NAME")
            outputStream = FileOutputStream(file)
            copyFile(inputStream, outputStream)
        }

        val uri = FileProvider.getUriForFile(
            activity,
            "${BuildConfig.APPLICATION_ID}.provider.GenericFileProvider",
            file
        )
        val intent = Intent(Intent.ACTION_VIEW).apply {
            setDataAndType(uri, "application/pdf")
            addFlags(Intent.FLAG_GRANT_READ_URI_PERMISSION)
            addFlags(Intent.FLAG_ACTIVITY_NO_HISTORY)
        }
        activity.startActivity(intent)
    } catch (ex: IOException) {
        ex.printStackTrace()
    } catch (ex: ActivityNotFoundException) {
        ex.printStackTrace()
    } finally {
        inputStream?.close()
        outputStream?.flush()
        outputStream?.close()
    }
}

@Throws(IOException::class)
private fun copyFile(input: InputStream, output: OutputStream) {
    val buffer = ByteArray(1024)
    var read: Int = input.read(buffer)
    while (read != -1) {
        output.write(buffer, 0, read)
        read = input.read(buffer)
    }
}

companion object {
    private const val PDF_ASSETS_PATH = "pdfs"
    private const val PDF_FILE_NAME = "mypdf.pdf"
}

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。

在Kotlin中,这可以用一行完成!

为InputStream添加扩展乐趣

fun InputStream.toFile(to: File){
    this.use { input->
        to.outputStream().use { out->
            input.copyTo(out)
        }
    }
}

然后使用它

MainActivity.kt

assets.open("test.zip").toFile(File(filesDir,"test.zip"))