Android使用contentResolver從內容URI獲取文件路徑

Question

我正在嘗試獲取內容 URI 的文件路徑。 URL 如下所示：content://com.android.providers.downloads.documents/document/31

游標對象不為空，但 cursor.getString(column_index) 返回空。
列索引始終為 0。

   public static String getPath(Context context, Uri uri) throws URISyntaxException {
    if ("content".equalsIgnoreCase(uri.getScheme())) {
       String[] projection = { "data"};
        Cursor cursor = null;

        try {
            cursor = context.getContentResolver().query(uri, projection, null, null, null);
            int column_index = cursor.getColumnIndexOrThrow( "_data");
            if (cursor.moveToFirst()) {
                // Method returns here with null value
                return cursor.getString(column_index);
            }
        } catch (Exception e) {
            // Eat it
        }
    }
    else if ("file".equalsIgnoreCase(uri.getScheme())) {
        return uri.getPath();
    }

    return null;
}

編輯：內容 URI 是從文件管理器返回的，因此它應該代表一個實際的文件。

public void filePicker(View view) {
    Intent intent = new Intent(Intent.ACTION_GET_CONTENT);
    intent.setType("*/*");
    intent.addCategory(Intent.CATEGORY_OPENABLE);

    try {
        startActivityForResult(
                Intent.createChooser(intent, "Select a File to Upload"),
                FILE_SELECT_CODE);
    } catch (android.content.ActivityNotFoundException ex) {
        // Potentially direct the user to the Market with a Dialog
        Toast.makeText(this, "Please install a File Manager.",
                Toast.LENGTH_SHORT).show();
    }
}

Answer 1

我正在嘗試獲取內容URI的文件路徑

不需要content:// Uri指向文件，更不用說您可以訪問的文件了。 如果要訪問文件中的數據，請使用ContentResolver和openInputStream()或openOutputStream() 。

Answer 2

這個助手類可以提供幫助，我在 github 庫中在線找到

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)
    }
}