Android getting file path from content URI using contentResolver

Question

I'm trying to get the file path for a content URI. URL looks like this: content://com.android.providers.downloads.documents/document/31

The cursor object is not null but the cursor.getString(column_index) returns null.
Column index is always 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;
}

EDIT: The content URI is returned from the file manager so it should represent an actual file.

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();
    }
}

score 2 · Answer 1 · answered Jun 11 '14 at 01:11

2

I'm trying to get the file path for a content URI

There is no requirement that a content:// Uri point to a file, let alone one that you can access. If you want to access the data in the file, use a ContentResolver and openInputStream() or openOutputStream().

answered Jun 11 '14 at 01:11

CommonsWare

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The content Uri is returned from the filemanager (see edited question for actual code) – Merch Jun 11 '14 at 01:16
2

@Merch: There are thousands of apps for Android -- on devices and on the Play Store -- that can respond to `ACTION_GET_CONTENT` for `*/*`. There is no requirement that what they return point to a file, let alone one that you can access. And it is the *user*, not you, who chooses what app will respond to your `startActivity()` call. – CommonsWare Jun 11 '14 at 10:50
1

Users will be internal to my company not randoms so they are expected to pick a file manager. The goal is to upload the selected file to an internal web server. – Merch Jun 12 '14 at 03:11
1

@Merch: If you want a path to a file, you cannot use `ACTION_GET_CONTENT`. Use one of the "File/Directory Chooser" libraries from [the Android Arsenal](http://android-arsenal.com/) as a library. – CommonsWare Jun 12 '14 at 09:33
If you're just reading a file you probably want `getContentResolver().openInputStream(uri)`. This will work no matter *where* the data comes from. – Edward Falk Mar 31 '16 at 01:39

score 1 · Answer 2 · answered Jan 23 '22 at 05:11

This helper class can help, i found online in a github library

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)
    }
}

Android getting file path from content URI using contentResolver

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