如何从 Android Oreo (8.1) 或更高版本的 URI 获取文件路径

Question

预期行为

当我select读取存储在"Download"中的文件时，它应该能够检索其文件名和路径

实际行为

当我 select 访问存储在 "Download" 中的文件时，它 returns 为空。

重现问题的步骤

调用picking file方法时，显示Android
去下载，select一个文件
从 URI 获取真实路径returns null

Here is the code what i implemented

public static String getPath(final Context context, final Uri uri) {
   String id = DocumentsContract.getDocumentId(uri);
   if (!TextUtils.isEmpty(id)) {
      if (id.startsWith("raw:")) {
          return id.replaceFirst("raw:", "");
      }
      try {
         final boolean isOreo = Build.VERSION.SDK_INT >= Build.VERSION_CODES.O;
         String stringContentURI;
         Uri contentUri;
         if(isOreo){ 
            stringContentURI = "content://downloads/my_downloads";
         }else{
            stringContentURI = "content://downloads/public_downloads";
         }
         contentUri = ContentUris.withAppendedId(
                    Uri.parse(stringContentURI), Long.valueOf(id));
         return getDataColumn(context, contentUri, null, null);
      } catch (NumberFormatException e) {
         return null;
      }
   }
}

public static String getDataColumn(Context context, Uri uri, String selection,
                                   String[] selectionArgs) {
    Cursor cursor = null;
    final String column = "_data";
    final String[] projection = {   column};
    try {
        cursor = context.getContentResolver().query(uri, projection, selection, selectionArgs,
                null);
        if (cursor != null && cursor.moveToFirst()) {
            final int index = cursor.getColumnIndexOrThrow(column);
            return cursor.getString(index);
        }
    } finally {
        if (cursor != null)
            cursor.close();
    }
    return null;
}

但是，当 select 从 Android 设备中的其他文件夹中读取文件时，它可以正常工作

请指教。谢谢大家:)

Answer 1

目前，获取路径的最佳方法是：

从 URI 获取物理文件作为 InputStream， ContentResolver.openInputStream() 允许您在不知道其真实路径的情况下访问文件的内容

String id = DocumentsContract.getDocumentId(uri);
InputStream inputStream = getContentResolver().openInputStream(uri);

然后将其作为临时文件写入缓存中

File file = new File(getCacheDir().getAbsolutePath()+"/"+id);
writeFile(inputStream, file);
String filePath = file.getAbsolutePath();

Here is the method to write temporary file into cached storage

void writeFile(InputStream in, File file) {
     OutputStream out = null;
     try {
          out = new FileOutputStream(file);
          byte[] buf = new byte[1024];
          int len;
          while((len=in.read(buf))>0){
            out.write(buf,0,len);
          }
     } catch (Exception e) {
          e.printStackTrace();
     }
     finally {
          try {
            if ( out != null ) {
                 out.close();
            }
            in.close();
          } catch ( IOException e ) {
               e.printStackTrace();
          }
     }
}

不确定这是否是最好的方法，但代码运行正常:D

Answer 2

这是我从这个要点中找到的另一个解决方案。 FileUtils.java

这里是使用方法。

@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    super.onActivityResult(requestCode, resultCode, data);
    Uri uri = data.getData();
    File originalFile = new File(FileUtils.getRealPath(this,uri));
}

Answer 3

这是我刚刚想出的（它在 Android Pie 上测试并假设有 SD 卡）：

private String getParentDirectory(@NonNull Uri uri) {
    String uriPath = uri.getPath();
    String filePath = Environment.getExternalStorageDirectory().getAbsolutePath();
    if(uriPath != null) {filePath = new File(filePath.concat("/" + uriPath.split(":")[1])).getParent();}
    return filePath;
}

private String getAbsolutePath(@NonNull Uri uri) {
    String uriPath = uri.getPath();
    String filePath = Environment.getExternalStorageDirectory().getAbsolutePath();
    if(uriPath != null) {filePath = filePath.concat("/" + uriPath.split(":")[1]);}
    return filePath;
}

Answer 4

this is activity instance variable

*************************************************************************
*************************************************************************  

    Uri filePath;
    String strAttachmentFileName = "",//attached file name
            strAttachmentCoded = "";//attached file in byte code Base64
    int PICK_REQUEST =1;
*************************************************************************
*************************************************************************

this is in activity method

*************************************************************************
*************************************************************************  

Button buttonChoose.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                Intent intent = new Intent();
                intent.setType("file/*");
                intent.setAction(Intent.ACTION_GET_CONTENT);
                startActivityForResult(Intent.createChooser(intent, "Select any file"), PICK_REQUEST );
            }
        });
*************************************************************************
*************************************************************************

this is overrride activity method

*************************************************************************
*************************************************************************



    @Override
    protected void onActivityResult(int requestCode, int resultCode, @Nullable Intent data) {
        super.onActivityResult(requestCode, resultCode, data);
        if (requestCode == PICK_REQUEST && resultCode == Activity.RESULT_OK && data != null && data.getData() != null) {
            filePath = data.getData();
            File uploadFile = new File(FileUtils.getRealPath(activity.this, filePath));
            try {
                if (uploadFile != null) {
                    strAttachmentFileName = uploadFile.getName();
                    FileInputStream objFileIS = new FileInputStream(uploadFile);
                    ByteArrayOutputStream objByteArrayOS = new ByteArrayOutputStream();
                    byte[] byteBufferString = new byte[1024];
                    int readNum;
                    readNum = objFileIS.read(byteBufferString);
                    while (readNum != -1) {
                        objByteArrayOS.write(byteBufferString, 0, readNum);
                        readNum = objFileIS.read(byteBufferString);
                    }
                    strAttachmentCoded  = Base64.encodeToString(objByteArrayOS.toByteArray(), Base64.DEFAULT);

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

*************************************************************************
*************************************************************************

Please create file FileUtils.java as below

link HBiSoft

对于 .Net 中的任何类型的文件 ---

byte[] p = Convert.FromBase64String("byte string");

MemoryStream ms = new MemoryStream(p);
FileStream fs = new FileStream
        (System.Web.Hosting.HostingEnvironment.MapPath("~/ComplaintDetailsFile/") +
                item.FileName, FileMode.Create);
ms.WriteTo(fs);
ms.Close();
fs.Close();
fs.Dispose();

Answer 5

下面 API 将 useful.We 需要在我们的应用程序缓存中创建文件，并且 return 相同的路径因为限制，即 Scoped 存储。

fun createFileAndGetPathFromCacheFolder(context: Context, uri: Uri): String? {
    var inputStream: FileInputStream? = null
    var outputStream: FileOutputStream? = null
    try {
        val parcelFileDescriptor = context.contentResolver.openFileDescriptor(uri, "r", null)
        inputStream = FileInputStream(parcelFileDescriptor?.fileDescriptor)
        val file = File(context.cacheDir, context.contentResolver.getFileName(uri))
        outputStream = FileOutputStream(file)
        val buffer = ByteArray(1024)
        var len: Int
        while (inputStream.read(buffer).also { len = it } != -1) {
            outputStream.write(buffer, 0, len)
        }
        return file.absolutePath
    } catch (e: Exception) {

    } finally {
        inputStream?.close()
        outputStream?.close()
    }
    return " "
}

如何从 Android Oreo (8.1) 或更高版本的 URI 获取文件路径

How to Get File Path from URI in Android Oreo (8.1) or above

android

uri

android-contentresolver

android-fileprovider

android-8.1-oreo