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());
    }

}

import android.app.Activity;
import android.content.Intent;
import android.content.res.AssetManager;
import android.net.Uri;
import android.os.Environment;
import android.os.Bundle;
import android.util.Log;
import java.io.BufferedOutputStream;
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;


public class MainActivity extends Activity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        copyReadAssets();
    }


    private void copyReadAssets()
    {
        AssetManager assetManager = getAssets();

        InputStream in = null;
        OutputStream out = null;

        String strDir = Environment.getExternalStoragePublicDirectory(Environment.DIRECTORY_DOWNLOADS)+ File.separator + "Pdfs";
        File fileDir = new File(strDir);
        fileDir.mkdirs();   // crear la ruta si no existe
        File file = new File(fileDir, "example2.pdf");



        try
        {

            in = assetManager.open("example.pdf");  //leer el archivo de assets
            out = new BufferedOutputStream(new FileOutputStream(file)); //crear el archivo


            copyFile(in, out);
            in.close();
            in = null;
            out.flush();
            out.close();
            out = null;
        } catch (Exception e)
        {
            Log.e("tag", e.getMessage());
        }

        Intent intent = new Intent(Intent.ACTION_VIEW);
        intent.setDataAndType(Uri.parse("file://" + Environment.getExternalStoragePublicDirectory(Environment.DIRECTORY_DOWNLOADS) + File.separator + "Pdfs" + "/example2.pdf"), "application/pdf");
        startActivity(intent);
    }

    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);
        }
    }
}

修改部分代码，如下所示:

out = new BufferedOutputStream(new FileOutputStream(file));

前面的例子是针对pdf文件的，如果是。txt

FileOutputStream fos = new FileOutputStream(file);

这是目前为止我能在网上找到的最好的解决方案。 我使用了以下链接https://gist.github.com/mhasby/026f02b33fcc4207b302a60645f6e217，但它有一个错误，我修复了，然后它的工作就像一个魅力。 这是我的代码。您可以轻松使用它，因为它是一个独立的java类。

public class CopyAssets {
public static void copyAssets(Context context) {
    AssetManager assetManager = context.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);

            out = new FileOutputStream(Environment.getExternalStorageDirectory()+"/www/resources/" + filename);
            copyFile(in, out);
        } catch(IOException e) {
            Log.e("tag", "Failed to copy asset file: " + filename, e);
        }
        finally {
            if (in != null) {
                try {
                    in.close();
                    in = null;
                } catch (IOException e) {

                }
            }
            if (out != null) {
                try {
                    out.flush();
                    out.close();
                    out = null;
                } catch (IOException e) {

                }
            }
        }
    }
}

public static 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);
    }
}}

As you can see, just create an instance of CopyAssets in your java class which has an activity. Now this part is important, as far as my testing and researching on the internet, You cannot use AssetManager if the class has no activity . It has something to do with the context of the java class. Now, the c.copyAssets(getApplicationContext()) is an easy way to access the method, where c is and instance of CopyAssets class. As per my requirement, I allowed the program to copy all my resource files inside the asset folder to the /www/resources/ of my internal directory. You can easily find out the part where you need to make changes to the directory as per your use. Feel free to ping me if you need any help.

你可以用Kotlin在几个步骤中做到这一点，在这里我只复制几个文件，而不是所有从资产到我的应用程序文件目录。

private fun copyRelatedAssets() {
    val assets = arrayOf("myhome.html", "support.css", "myscript.js", "style.css")
    assets.forEach {
        val inputStream = requireContext().assets.open(it)
        val nameSplit = it.split(".")
        val name = nameSplit[0]
        val extension = nameSplit[1]
        val path = inputStream.getFilePath(requireContext().filesDir, name, extension)
        Log.v(TAG, path)
    }
}

这是扩展函数，

fun InputStream.getFilePath(dir: File, name: String, extension: String): String {
    val file = File(dir, "$name.$extension")
    val outputStream = FileOutputStream(file)
    this.copyTo(outputStream, 4096)
    return file.absolutePath
}

洛格猫

/data/user/0/com.***.***/files/myhome.html
/data/user/0/com.***.***/files/support.css
/data/user/0/com.***.***/files/myscript.js
/data/user/0/com.***.***/files/style.css

由于一些错误，上述解决方案无法工作:

目录创建失败 Android返回的资产也包含三个文件夹:图像，声音和webkit 增加了处理大文件的方法:在项目的资产文件夹中添加扩展名.mp3到文件，在复制目标文件时将没有.mp3扩展名

下面是代码(我留下了Log语句，但你现在可以删除它们):

final static String TARGET_BASE_PATH = "/sdcard/appname/voices/";

private void copyFilesToSdCard() {
    copyFileOrDir(""); // copy all files in assets folder in my project
}

private void copyFileOrDir(String path) {
    AssetManager assetManager = this.getAssets();
    String assets[] = null;
    try {
        Log.i("tag", "copyFileOrDir() "+path);
        assets = assetManager.list(path);
        if (assets.length == 0) {
            copyFile(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( p + 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;
    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());
    }

}

编辑:更正了一个错位的“;”，这是抛出一个系统的“无法创建目录”错误。

我知道这已经回答，但我有一个稍微更优雅的方式从资产目录复制到sdcard上的文件。它不需要“for”循环，而是使用文件流和通道来完成工作。

(注)如果使用任何类型的压缩文件，APK, PDF，…你可能想重命名文件扩展名之前插入到资产，然后重命名一旦你复制到SDcard)

AssetManager am = context.getAssets();
AssetFileDescriptor afd = null;
try {
    afd = am.openFd( "MyFile.dat");

    // Create new file to copy into.
    File file = new File(Environment.getExternalStorageDirectory() + java.io.File.separator + "NewFile.dat");
    file.createNewFile();

    copyFdToFile(afd.getFileDescriptor(), file);

} catch (IOException e) {
    e.printStackTrace();
}

一种复制文件而不必循环遍历它的方法。

public static void copyFdToFile(FileDescriptor src, File dst) throws IOException {
    FileChannel inChannel = new FileInputStream(src).getChannel();
    FileChannel outChannel = new FileOutputStream(dst).getChannel();
    try {
        inChannel.transferTo(0, inChannel.size(), outChannel);
    } finally {
        if (inChannel != null)
            inChannel.close();
        if (outChannel != null)
            outChannel.close();
    }
}