如何从视频文件的内容 URI 获取文件路径？

Question

我想附加来自 Android 存储的图像或视频文件。当我 select 来自图库的视频时，它 returns 内容 uri 路径 。

如何从内容 URI 中获取带有扩展名的 文件路径？

我尝试了以下代码，但它 returns 在 lollipop 中为空：

void pickVideo() {
    Intent videoIntent = new Intent(Intent.ACTION_GET_CONTENT);
    videoIntent.setType("video/*");
    startActivityForResult(videoIntent, PICK_VIDEO_FILE);
}


@Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
    super.onActivityResult(requestCode, resultCode, data);
    try {
        if (resultCode != Activity.RESULT_OK)
            return;

        switch (requestCode) {
            case PICK_VIDEO_FILE:
                Uri videoUri = data.getData();
                String path = getRealPathFromURI(getContext(), videoUri);
                Log.d("Video uri path", "path" + path);
                if (mChooseFileDialogListener != null) {
                    mChooseFileDialogListener.onVideoClick(videoUri, HealthRecordViewModel.FILE_TYPE_VIDEO);
                }
                break;
        }
    }
}


public String getRealPathFromURI(Context context, Uri contentUri) {
    Cursor cursor = null;
    try {
        String[] proj = { MediaStore.Video.Media.DATA };
        cursor = context.getContentResolver().query(contentUri,  proj, null, null, null);
        int column_index = cursor.getColumnIndexOrThrow(MediaStore.Video.Media.DATA);
        cursor.moveToFirst();
        return cursor.getString(column_index);
    } finally {
        if (cursor != null) {
            cursor.close();
        }
    }
}

Answer 1

How to get the file path with extension?

你不知道。不要求用户 select 一段内容作为文件保存在您具有读取权限的位置。并且没有要求 ContentProvider 提供一些将 Uri 转换为文件路径的方法。

使用ContentResolver和openInputStream()读入Uri标识的内容。

Answer 2

这可能有帮助

import android.content.ContentResolver
import android.content.Context
import android.net.Uri
import android.webkit.MimeTypeMap
import java.io.File
import java.io.FileOutputStream
import java.io.IOException
import java.io.InputStream
import java.io.OutputStream
import java.text.SimpleDateFormat
import java.util.Date
import java.util.Locale.getDefault

/**
 * This class will create a temporary file in the cache if need.
 *
 * When the uri already have `file://` schema we don't need to create a new file.
 * The temporary file will always override a previous one, saving memory.
 * Using the cache memory(context.cacheDir) we guarantee to not leak memory
 *
 * @param context used to access Android APIs, like content resolve, it is your activity/fragment.
 * @param uri the URI to load the image from.
 * @param uniqueName If true, make each image cropped have a different file name, this could cause
 * memory issues, use wisely.
 *
 * @return string value of the File path.
 */
 fun getFilePathFromUri(context: Context, uri: Uri, uniqueName: Boolean): String =
    if (uri.path?.contains("file://") == true) uri.path!!
    else getFileFromContentUri(context, uri, uniqueName).path

private fun getFileFromContentUri(context: Context, contentUri: Uri, uniqueName: Boolean): File {
    // Preparing Temp file name
    val fileExtension = getFileExtension(context, contentUri) ?: ""
    val timeStamp = SimpleDateFormat("yyyyMMdd_HHmmss", getDefault()).format(Date())
    val fileName = ("temp_file_" + if (uniqueName) timeStamp else "") + ".$fileExtension"
    // Creating Temp file
    val tempFile = File(context.cacheDir, fileName)
    tempFile.createNewFile()
    // Initialize streams
    var oStream: FileOutputStream? = null
    var inputStream: InputStream? = null

    try {
        oStream = FileOutputStream(tempFile)
        inputStream = context.contentResolver.openInputStream(contentUri)

        inputStream?.let { copy(inputStream, oStream) }
        oStream.flush()
    } catch (e: Exception) {
        e.printStackTrace()
    } finally {
        // Close streams
        inputStream?.close()
        oStream?.close()
    }

    return tempFile
}

private fun getFileExtension(context: Context, uri: Uri): String? =
    if (uri.scheme == ContentResolver.SCHEME_CONTENT)
        MimeTypeMap.getSingleton().getExtensionFromMimeType(context.contentResolver.getType(uri))
    else uri.path?.let { MimeTypeMap.getFileExtensionFromUrl(Uri.fromFile(File(it)).toString()) }

@Throws(IOException::class)
private fun copy(source: InputStream, target: OutputStream) {
    val buf = ByteArray(8192)
    var length: Int
    while (source.read(buf).also { length = it } > 0) {
        target.write(buf, 0, length)
    }
}

如何从视频文件的内容 URI 获取文件路径？

How do I get the file path from the content URI for video files?

android

uri

android-contentprovider