assets文件夹及其子文件夹中的文件列表

Question

我的 Android 项目的“assets”文件夹中有一些包含 HTML 文件的文件夹。我需要在列表中显示资产子文件夹中的这些 HTML 文件。我已经编写了一些关于制作此列表的代码。

lv1 = (ListView) findViewById(R.id.listView);
// Insert array in ListView

// In the next row I need to insert an array of strings of file names
// so please, tell me, how to get this array

lv1.setAdapter(new ArrayAdapter<String>(this, android.R.layout.simple_list_item_1, filel));
lv1.setTextFilterEnabled(true);
// onclick items in ListView:
lv1.setOnItemClickListener(new OnItemClickListener() {
    public void onItemClick(AdapterView<?> a, View v, int position, long id) {
        //Clicked item position
        String itemname = new Integer(position).toString();  
        Intent intent = new Intent();
        intent.setClass(DrugList.this, Web.class);
        Bundle b = new Bundle();
        //I don't know what it's doing here
        b.putString("defStrID", itemname); 
        intent.putExtras(b);
        //start Intent
        startActivity(intent);
    }
});

Answer 1

private boolean listAssetFiles(String path) {

    String [] list;
    try {
        list = getAssets().list(path);
        if (list.length > 0) {
            // This is a folder
            for (String file : list) {
                if (!listAssetFiles(path + "/" + file))
                    return false;
                else {
                    // This is a file
                    // TODO: add file name to an array list
                }
            }
        } 
    } catch (IOException e) {
        return false;
    }

    return true; 
} 

使用资源文件夹的根文件夹名称调用 listAssetFiles。

    listAssetFiles("root_folder_name_in_assets");

如果根文件夹是资产文件夹，则使用

调用它

    listAssetFiles("");    

Answer 2

尝试一下，它会适合你的情况

f = getAssets().list("");
for(String f1 : f){
    Log.v("names",f1);
}

上面的代码片段将显示资产根目录的内容。

例如...如果下面是资产结构..

assets
 |__Dir1
 |__Dir2
 |__File1

片段的输出将是...... 目录 1 目录 2 文件 1

如果您需要目录Dir1的内容

在列表函数中传递目录名称。

  f = getAssets().list("Dir1");

Answer 3

希望有帮助：

以下代码将复制所有文件夹及其内容和子文件夹的内容到SD卡位置：

 private void getAssetAppFolder(String dir) throws Exception{

    {
        File f = new File(sdcardLocation + "/" + dir);
        if (!f.exists() || !f.isDirectory())
            f.mkdirs();
    }
     AssetManager am=getAssets();

     String [] aplist=am.list(dir);

     for(String strf:aplist){
        try{
             InputStream is=am.open(dir+"/"+strf);
             copyToDisk(dir,strf,is);
         }catch(Exception ex){


            getAssetAppFolder(dir+"/"+strf);
         }
     }



 }


 public void copyToDisk(String dir,String name,InputStream is) throws IOException{
     int size;
        byte[] buffer = new byte[2048];

        FileOutputStream fout = new FileOutputStream(sdcardLocation +"/"+dir+"/" +name);
        BufferedOutputStream bufferOut = new BufferedOutputStream(fout, buffer.length);

        while ((size = is.read(buffer, 0, buffer.length)) != -1) {
            bufferOut.write(buffer, 0, size);
        }
        bufferOut.flush();
        bufferOut.close();
        is.close();
        fout.close();
 }

Answer 4

这是我的问题的解决方案，我发现它可以 100% 列出所有目录和文件，甚至子目录和子目录中的文件。

注意：就我而言

1. 文件名有一个 .在他们中。即 .htm .txt 等

2. 目录名称没有任何 .在他们里面。

   listAssetFiles2(path); // <<-- Call function where required
   
   
   //function to list files and directories
   public void listAssetFiles2 (String path){
   String [] list;
   
   try {
       list = getAssets().list(path);
       if(list.length > 0){
           for(String file : list){
               System.out.println("File path = "+file);
   
               if(file.indexOf(".") < 0) { // <<-- check if filename has a . then it is a file - hopefully directory names dont have . 
                   System.out.println("This is a folder = "+path+"/"+file);
                   listAssetFiles2(file); // <<-- To get subdirectory files and directories list and check 
               }else{
                   System.out.println("This is a file = "+path+"/"+file);
               }
           }
   
       }else{
           System.out.println("Failed Path = "+path);
           System.out.println("Check path again.");
       }
   }catch(IOException e){
       e.printStackTrace();
   }
   }//now completed
   

谢谢

Answer 5

我认为最好检查文件是否为目录，或者尝试，捕获！

 public static  List<String> listAssetFiles(Context c,String rootPath) {
    List<String> files =new ArrayList<>();
    try {
        String [] Paths = c.getAssets().list(rootPath);
        if (Paths.length > 0) {
            // This is a folder
            for (String file : Paths) {
                String path = rootPath + "/" + file;
                if (new File(path).isDirectory())
                    files.addAll(listAssetFiles(c,path));
                else files.add(path);
            }
        }
    } catch (IOException e) {
        e.printStackTrace();
    }
   return files;
}

Answer 6

基于@Kammaar 的回答。此 kotlin 代码扫描文件树中的叶子：

private fun listAssetFiles(path: String, context: Context): List<String> {
    val result = ArrayList<String>()
    context.assets.list(path).forEach { file ->
        val innerFiles = listAssetFiles("$path/$file", context)
        if (!innerFiles.isEmpty()) {
            result.addAll(innerFiles)
        } else {
            // it can be an empty folder or file you don't like, you can check it here
            result.add("$path/$file")
        }
    }
    return result
}

Answer 7

此方法返回 Assets 文件夹中目录中的文件名

 private fun getListOfFilesFromAsset(path: String, context: Context): ArrayList<String> {
            val listOfAudioFiles = ArrayList<String>()
            context.assets.list(path)?.forEach { file ->
                val innerFiles = getListOfFilesFromAsset("$path/$file", context)
                if (innerFiles.isNotEmpty()) {
                    listOfAudioFiles.addAll(innerFiles)
                } else {
                    // it can be an empty folder or file you don't like, you can check it here
                    listOfAudioFiles.add("$path/$file")
                }
            }
            return listOfAudioFiles
        }

例如您想从声音文件夹加载音乐文件路径

您可以像这样获取所有声音：

  private const val SOUND_DIRECTORY = "sound"


   fun fetchSongsFromAssets(context: Context): ArrayList<String> {
        return getListOfFilesFromAsset(SOUND_DIRECTORY, context)
    }

Answer 8

public static String[] getDirectoryFilesRecursive(String path)
{
    ArrayList<String> result  = new ArrayList<String>();
    try
    {
        String[] files = Storage.AssetMgr.list(path);
        for(String file : files)
        {
            String filename = path + (path.isEmpty() ? "" : "/") + file;
            String[] tmp = Storage.AssetMgr.list(filename);
            if(tmp.length!=0) {
                result.addAll(Arrays.asList(getDirectoryFilesRecursive(filename)));
            }
            else {
                result.add(filename);
            }
        }
    }
    catch (IOException e)
    {
        Native.err("Failed to get asset file list: " + e);
    }
    Object[] objectList = result.toArray();
    return Arrays.copyOf(objectList,objectList.length,String[].class);
}

Answer 9

@Kammaar 的答案的改进版本，在 Kotlin [递归]

fun listAssetFiles(
    context: Context,
    path: String,
    dirCallback: ((dirPath: String) -> Unit)? = null,
    fileCallback: (filePath: String) -> Unit,
): Boolean {
    try {
        context.assets.list(path)?.also { files ->
            if (files.isNotEmpty()) {
                for (file in files) {
                    val relativePath = if (path.isEmpty()) file else "$path${File.separatorChar}$file"
                    if (!listAssetFiles(context, relativePath, dirCallback,fileCallback))
                        fileCallback.invoke(relativePath) else dirCallback?.invoke( relativePath)
                }
            } else return false
        }
    } catch (e: IOException) {return false}
    return true
}

使用方法：

    listAssetFiles(getApplication(), ""){Log.e("TAG", "File found ->  $it")}

另一个示例，用于检索单独的文件和文件夹列表：

    val fileList = mutableListOf<String>()
    val folderList = mutableListOf<String>()
    listAssetFiles(
        context = getApplication(),
        path = "sample_folder",
        dirCallback = {folderList.add(it)},
        fileCallback = {fileList.add(it)})