什么是最简单的方法从android.net.Uri对象持有一个文件:类型转换为java.io.File对象在Android?

我尝试了下面的方法,但不管用:

File file = new File(Environment.getExternalStorageDirectory(), "read.me");
Uri uri = Uri.fromFile(file);
File auxFile = new File(uri.toString());
assertEquals(file.getAbsolutePath(), auxFile.getAbsolutePath());

当前回答

要正确地使用context uri获取文件, 感谢来自@Mohsents, @Bogdan Kornev, @CommonsWare, @Juan Camilo Rodriguez的回答Durán;

我从uri中创建了一个inputStream,并使用这个iStream创建了一个临时文件,最后我能够从这个文件中提取uri和路径。

fun createFileFromContentUri(fileUri : Uri) : File{

    var fileName : String = ""

    fileUri.let { returnUri ->
        requireActivity().contentResolver.query(returnUri,null,null,null)
    }?.use { cursor ->
        val nameIndex = cursor.getColumnIndex(OpenableColumns.DISPLAY_NAME)
        cursor.moveToFirst()
        fileName = cursor.getString(nameIndex)
    }
    
    //  For extract file mimeType
    val fileType: String? = fileUri.let { returnUri ->
        requireActivity().contentResolver.getType(returnUri)
    }

    val iStream : InputStream = requireActivity().contentResolver.openInputStream(fileUri)!!
    val outputDir : File = context?.cacheDir!!
    val outputFile : File = File(outputDir,fileName)
    copyStreamToFile(iStream, outputFile)
    iStream.close()
    return  outputFile
}

fun copyStreamToFile(inputStream: InputStream, outputFile: File) {
    inputStream.use { input ->
        val outputStream = FileOutputStream(outputFile)
        outputStream.use { output ->
            val buffer = ByteArray(4 * 1024) // buffer size
            while (true) {
                val byteCount = input.read(buffer)
                if (byteCount < 0) break
                output.write(buffer, 0, byteCount)
            }
            output.flush()
        }
    }
}

其他回答

你可以使用这个函数从uri中获取文件在新的android和旧的

fun getFileFromUri(context: Context, uri: Uri?): File? {
    uri ?: return null
    uri.path ?: return null

    var newUriString = uri.toString()
    newUriString = newUriString.replace(
        "content://com.android.providers.downloads.documents/",
        "content://com.android.providers.media.documents/"
    )
    newUriString = newUriString.replace(
        "/msf%3A", "/image%3A"
    )
    val newUri = Uri.parse(newUriString)

    var realPath = String()
    val databaseUri: Uri
    val selection: String?
    val selectionArgs: Array<String>?
    if (newUri.path?.contains("/document/image:") == true) {
        databaseUri = MediaStore.Images.Media.EXTERNAL_CONTENT_URI
        selection = "_id=?"
        selectionArgs = arrayOf(DocumentsContract.getDocumentId(newUri).split(":")[1])
    } else {
        databaseUri = newUri
        selection = null
        selectionArgs = null
    }
    try {
        val column = "_data"
        val projection = arrayOf(column)
        val cursor = context.contentResolver.query(
            databaseUri,
            projection,
            selection,
            selectionArgs,
            null
        )
        cursor?.let {
            if (it.moveToFirst()) {
                val columnIndex = cursor.getColumnIndexOrThrow(column)
                realPath = cursor.getString(columnIndex)
            }
            cursor.close()
        }
    } catch (e: Exception) {
        Log.i("GetFileUri Exception:", e.message ?: "")
    }
    val path = realPath.ifEmpty {
        when {
            newUri.path?.contains("/document/raw:") == true -> newUri.path?.replace(
                "/document/raw:",
                ""
            )
            newUri.path?.contains("/document/primary:") == true -> newUri.path?.replace(
                "/document/primary:",
                "/storage/emulated/0/"
            )
            else -> return null
        }
    }
    return if (path.isNullOrEmpty()) null else File(path)
}

文件imageToUpload =新文件(新URI(androidURI.toString()));如果这是你在外部存储中创建的文件,则有效。

例如file:///storage/emulated/0/(一些目录和文件名)

对于那些在这里寻找图像解决方案的人,特别是在这里。

private Bitmap getBitmapFromUri(Uri contentUri) {
        String path = null;
        String[] projection = { MediaStore.Images.Media.DATA };
        Cursor cursor = getContentResolver().query(contentUri, projection, null, null, null);
        if (cursor.moveToFirst()) {
            int columnIndex = cursor.getColumnIndexOrThrow(MediaStore.Images.Media.DATA);
            path = cursor.getString(columnIndex);
        }
        cursor.close();
        Bitmap bitmap = BitmapFactory.decodeFile(path);
        return bitmap;
    }

在寻找了很长一段时间后,这对我来说是有效的:

File file = new File(getPath(uri));


public String getPath(Uri uri) 
    {
        String[] projection = { MediaStore.Images.Media.DATA };
        Cursor cursor = getContentResolver().query(uri, projection, null, null, null);
        if (cursor == null) return null;
        int column_index =             cursor.getColumnIndexOrThrow(MediaStore.Images.Media.DATA);
        cursor.moveToFirst();
        String s=cursor.getString(column_index);
        cursor.close();
        return s;
    }

科特林 2022

suspend fun Context.createFileFromAsset(assetName: String, fileName: String): File? {
    return withContext(Dispatchers.IO) {
        runCatching {
            val stream = assets.open(assetName)
            val file = File(cacheDir.absolutePath, fileName)
            org.apache.commons.io.FileUtils.copyInputStreamToFile(stream, file)
            file
        }.onFailure { Timber.e(it) }.getOrNull()
    }
}

处理完文件后,请确保对其调用.delete()。向@Mohsent致敬