How to extract the file name from URI returned from Intent.ACTION_GET_CONTENT?

Question

I am using 3rd party file manager to pick a file (PDF in my case) from the file system.

This is how I launch the activity:

Intent intent = new Intent(Intent.ACTION_GET_CONTENT);
intent.setType(getString(R.string.app_pdf_mime_type));
intent.addCategory(Intent.CATEGORY_OPENABLE);

String chooserName = getString(R.string.Browse);
Intent chooser = Intent.createChooser(intent, chooserName);

startActivityForResult(chooser, ActivityRequests.BROWSE);

This is what I have in onActivityResult:

Uri uri = data.getData();
if (uri != null) {
    if (uri.toString().startsWith("file:")) {
        fileName = uri.getPath();
    } else { // uri.startsWith("content:")

        Cursor c = getContentResolver().query(uri, null, null, null, null);

        if (c != null && c.moveToFirst()) {

            int id = c.getColumnIndex(Images.Media.DATA);
            if (id != -1) {
                fileName = c.getString(id);
            }
        }
    }
}

Code snippet is borrowed from Open Intents File Manager instructions available here:
http://www.openintents.org/en/node/829

The purpose of if-else is backwards compatibility. I wonder if this is a best way to get the file name as I have found that other file managers return all kind of things.

For example, Documents ToGo return something like the following:

content://com.dataviz.dxtg.documentprovider/document/file%3A%2F%2F%2Fsdcard%2Fdropbox%2FTransfer%2Fconsent.pdf

on which getContentResolver().query() returns null.

To make things more interesting, unnamed file manager (I got this URI from client log) returned something like:

/./sdcard/downloads/.bin

Is there a preferred way of extracting the file name from URI or one should resort to string parsing?

Same question with maybe better answer: http://stackoverflow.com/questions/8646246/uri-from-intent-action-get-content-into-file — Rémi, Jul 26 '12 at 10:04

score 226 · Answer 1 · answered Jul 28 '14 at 22:17

226

developer.android.com has nice example code for this: https://developer.android.com/guide/topics/providers/document-provider.html

A condensed version to just extract the file name (assuming "this" is an Activity):

public String getFileName(Uri uri) {
  String result = null;
  if (uri.getScheme().equals("content")) {
    Cursor cursor = getContentResolver().query(uri, null, null, null, null);
    try {
      if (cursor != null && cursor.moveToFirst()) {
        result = cursor.getString(cursor.getColumnIndex(OpenableColumns.DISPLAY_NAME));
      }
    } finally {
      cursor.close();
    }
  }
  if (result == null) {
    result = uri.getPath();
    int cut = result.lastIndexOf('/');
    if (cut != -1) {
      result = result.substring(cut + 1);
    }
  }
  return result;
}

answered Jul 28 '14 at 22:17

Stefan Haustein

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in case when you attempt read mimeType or fileName of file from camera, firstly you have to notify MediaScanner that will convert **URI** of your just created file from `file://` to `content://` in `onScanCompleted(String path, Uri uri)` method https://stackoverflow.com/a/5815005/2163045 – murt Oct 04 '17 at 09:11
17

Adding `new String[]{OpenableColumns.DISPLAY_NAME}` as a second parameter to query will filter the columns for a more efficient request. – JM Lord Jan 19 '18 at 16:34
`OpenableColumns.DISPLAY_NAME` didn't work for me, i used `MediaStore.Files.FileColumns.TITLE` instead. – Dmitry Kopytov Feb 20 '20 at 20:48
This will crash for videos when creating the cursor with a `java.lang.IllegalArgumentException: Invalid column latitude` unfortunately. Works perfectly for photos though! – Lucas P. Feb 28 '20 at 12:47
5

In your fallback, you could simply call `uri.getLastPathSegment()` instead of parsing it yourself. – Grishka Sep 16 '20 at 18:15
this work only for some apps, not at Files at Android 8 – Josef Vancura Nov 28 '20 at 14:11
@JosefVancura have you find any solution? OpenableColumns.DISPLAY_NAME is working on a few phones. – iamkdblue Dec 04 '20 at 13:06
Does the result contains name including file extension or excluding it? – Sourav Kannantha B May 12 '21 at 07:14
Much thanks for this old post. It's 2022 and this still works. ``` – Anonymous User Jun 14 '22 at 18:47

score 56 · Answer 2 · answered Jul 11 '16 at 10:01

56

Taken from Retrieving File information | Android developers

Retrieving a File's name.

private String queryName(ContentResolver resolver, Uri uri) {
    Cursor returnCursor =
            resolver.query(uri, null, null, null, null);
    assert returnCursor != null;
    int nameIndex = returnCursor.getColumnIndex(OpenableColumns.DISPLAY_NAME);
    returnCursor.moveToFirst();
    String name = returnCursor.getString(nameIndex);
    returnCursor.close();
    return name;
}

answered Jul 11 '16 at 10:01

Bhargav

8,118
6
40
63

1

I've been looking for this for way too long! – dasfima Oct 13 '18 at 21:20
16

Thank you. Good lord why do they have to make such a trivial thing so annoying? – TylerJames Nov 02 '18 at 15:50
3

It only works with content files. See https://stackoverflow.com/a/25005243/6325722 for full method – Johnny Five Mar 29 '19 at 09:09
may simplify calling -- queryName( getContentResolver(), uri) ; in MainActivity – Noor Hossain Nov 20 '20 at 21:34

score 49 · Answer 3 · edited Jun 29 '14 at 11:29

49

I'm using something like this:

String scheme = uri.getScheme();
if (scheme.equals("file")) {
    fileName = uri.getLastPathSegment();
}
else if (scheme.equals("content")) {
    String[] proj = { MediaStore.Images.Media.TITLE };
    Cursor cursor = context.getContentResolver().query(contentUri, proj, null, null, null);
    if (cursor != null && cursor.getCount() != 0) {
        int columnIndex = cursor.getColumnIndexOrThrow(MediaStore.Images.Media.TITLE);
        cursor.moveToFirst();
        fileName = cursor.getString(columnIndex);
    }
    if (cursor != null) {
        cursor.close();
    }
}

edited Jun 29 '14 at 11:29

flx

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answered Apr 06 '11 at 16:02

Ken Fehling

2,051
23
32

4

I have run some tests on your code. On URI returned by OI File Manager IllegalArgumentException is thrown as column 'title' does not exist. On URI returened by Documents ToGo cursor is null. On URI returned by unknown file manager scheme is (obviously) null. – Viktor Brešan Apr 07 '11 at 10:38
2

Hmm, interesting. Yeah, testing for scheme != null is a good idea. Actually I don't think TITLE is what you want. I was using that because certain media types in Android (like songs chosen through the music picker) have URIs like content://media/external/audio/media/78 and I wanted to display something more relevant than the ID number. You can simply use uri.getLastPathSegment() like my code uses for file:// URIs if you have a URI like content://...somefile.pdf – Ken Fehling Apr 17 '11 at 06:15
1

uri.getLastPathSegment(); - you have saved my day :) – Lumis May 05 '11 at 21:49
@Ken Fehling: This didn't work for me. It works when I click a file in a file browser, but when I click on an email attachment, I still get the content://... thing. I tried all the suggestions here without luck. Any idea why? – Luis A. Florit May 04 '13 at 21:41
1

Hi, why the `cursor` is not `close()`d? – fikr4n Aug 27 '13 at 06:10
Get the uri and use the function getLastPathSegment() to get the file name. `uri.getLastPathSegment()` – V_J Aug 15 '15 at 00:45
I think that only MediaStore.MediaColumns.DISPLAY_NAME and MediaStore.MediaColumns.SIZE are guaranteed to exist for content:// Uris. See [OpenableColumns](https://developer.android.com/reference/android/provider/OpenableColumns.html). If there's a reliable way to know what columns are available for any given Content:// Uri, I haven't found it. – Edward Falk Apr 01 '16 at 23:34

Metu · Answer 4 · 2018-05-28T06:36:23.720

Easiest ways to get file name:

val fileName = File(uri.path).name
// or
val fileName = uri.pathSegments.last()

If they don't give you the right name you should use:

fun Uri.getName(context: Context): String {
    val returnCursor = context.contentResolver.query(this, null, null, null, null)
    val nameIndex = returnCursor.getColumnIndex(OpenableColumns.DISPLAY_NAME)
    returnCursor.moveToFirst()
    val fileName = returnCursor.getString(nameIndex)
    returnCursor.close()
    return fileName
}

Tamim Attafi · Answer 5 · 2021-05-25T12:12:10.120

30

For Kotlin you can use something like this:

fun Context.getFileName(uri: Uri): String? = when(uri.scheme) {
    ContentResolver.SCHEME_CONTENT -> getContentFileName(uri)
    else -> uri.path?.let(::File)?.name
}

private fun Context.getContentFileName(uri: Uri): String? = runCatching {
    contentResolver.query(uri, null, null, null, null)?.use { cursor ->
        cursor.moveToFirst()
        return@use cursor.getColumnIndexOrThrow(OpenableColumns.DISPLAY_NAME).let(cursor::getString)
    }
}.getOrNull()

edited May 25 '21 at 12:12

answered May 21 '19 at 08:42

Tamim Attafi

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Good answer, but on my opinion it's better to insert these funs (Context.getFileName and Context.getCursorContent) in a file with name 'Context', delete words object FileUtils and use it as extention, for example: val txtFileName = context.getFileName(uri) – Alexey Simchenko Jun 17 '20 at 20:28
@LumisD you can import these extensions in any file, your suggestion is correct but I just hate when methods are global. I want to see certain methods only if I import them and not always. And sometimes I prefer having same method names and different sources so I import the ones I need in the right place:) – Tamim Attafi Jul 06 '20 at 13:41
You can import it this way: import com.example.FileUtils.getFileName or just write context.getFileName(uri) and hit Alt+Enter and your IDE will do it for you:) – Tamim Attafi Jul 06 '20 at 13:44

score 16 · Answer 6 · answered Sep 29 '16 at 09:00

16

If you want it short this should work.

Uri uri= data.getData();
File file= new File(uri.getPath());
file.getName();

answered Sep 29 '16 at 09:00

Samuel

9,883
5
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1

I am getting a name with file.getName() but its not actual name – Tushar Thakur Jul 12 '17 at 09:24
1

My fileName is user.jpg but I am getting "6550" in response – Tushar Thakur Jul 12 '17 at 09:24
1

It doesn't return actual file name. Actually it return last part of your resource url. – Emdadul Sawon Sep 25 '17 at 08:48
This does not work as a file! Mostly URIs have to be decoded by the MediaStore today. Reason why I and others are getting the numbers. Because those are the IDs of those files. – sud007 Apr 18 '20 at 12:27
1

This was a response given 4 years ago. Android is obviously changing every release. – Samuel Apr 18 '20 at 13:16
then what should I do for getting the actual name of the file? – EAS Jan 27 '22 at 04:57

score 14 · Answer 7 · edited May 05 '20 at 06:52

14

The most condensed version:

public String getNameFromURI(Uri uri) {
    Cursor c = getContentResolver().query(uri, null, null, null, null);
    c.moveToFirst();
    return c.getString(c.getColumnIndex(OpenableColumns.DISPLAY_NAME));
}

edited May 05 '20 at 06:52

סטנלי גרונן

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answered May 05 '20 at 06:22

artdeell

161
1
5

3

do not forget to close cursor :) – BekaBot Jun 12 '20 at 12:25

score 11 · Answer 8 · answered Aug 24 '16 at 09:53

I use below code to get File Name & File Size from Uri in my project.

/**
 * Used to get file detail from uri.
 * <p>
 * 1. Used to get file detail (name & size) from uri.
 * 2. Getting file details from uri is different for different uri scheme,
 * 2.a. For "File Uri Scheme" - We will get file from uri & then get its details.
 * 2.b. For "Content Uri Scheme" - We will get the file details by querying content resolver.
 *
 * @param uri Uri.
 * @return file detail.
 */
public static FileDetail getFileDetailFromUri(final Context context, final Uri uri) {
    FileDetail fileDetail = null;
    if (uri != null) {
        fileDetail = new FileDetail();
        // File Scheme.
        if (ContentResolver.SCHEME_FILE.equals(uri.getScheme())) {
            File file = new File(uri.getPath());
            fileDetail.fileName = file.getName();
            fileDetail.fileSize = file.length();
        }
        // Content Scheme.
        else if (ContentResolver.SCHEME_CONTENT.equals(uri.getScheme())) {
            Cursor returnCursor =
                    context.getContentResolver().query(uri, null, null, null, null);
            if (returnCursor != null && returnCursor.moveToFirst()) {
                int nameIndex = returnCursor.getColumnIndex(OpenableColumns.DISPLAY_NAME);
                int sizeIndex = returnCursor.getColumnIndex(OpenableColumns.SIZE);
                fileDetail.fileName = returnCursor.getString(nameIndex);
                fileDetail.fileSize = returnCursor.getLong(sizeIndex);
                returnCursor.close();
            }
        }
    }
    return fileDetail;
}

/**
 * File Detail.
 * <p>
 * 1. Model used to hold file details.
 */
public static class FileDetail {

    // fileSize.
    public String fileName;

    // fileSize in bytes.
    public long fileSize;

    /**
     * Constructor.
     */
    public FileDetail() {

    }
}

Helped solve my problem with loading file names from external or internal data! Very useful thank you! — Brandon, Oct 24 '18 at 02:28

score 7 · Answer 9 · answered Nov 06 '18 at 10:33

If you want to have the filename with extension I use this function to get it. It also works with google drive file picks

public static String getFileName(Uri uri) {
    String result;

    //if uri is content
    if (uri.getScheme() != null && uri.getScheme().equals("content")) {
        Cursor cursor = global.getInstance().context.getContentResolver().query(uri, null, null, null, null);
        try {
            if (cursor != null && cursor.moveToFirst()) {
                //local filesystem
                int index = cursor.getColumnIndex("_data");
                if(index == -1)
                    //google drive
                    index = cursor.getColumnIndex("_display_name");
                result = cursor.getString(index);
                if(result != null)
                    uri = Uri.parse(result);
                else
                    return null;
            }
        } finally {
            cursor.close();
        }
    }

    result = uri.getPath();

    //get filename + ext of path
    int cut = result.lastIndexOf('/');
    if (cut != -1)
        result = result.substring(cut + 1);
    return result;
}

Thanks for your answer, this helped me. – Sakhawat Hossain Sep 15 '20 at 06:18 — Sakhawat Hossain, Sep 15 '20 at 06:18
For me it was that leading underscore on the column name. – David U Oct 12 '22 at 08:34 — David U, Oct 12 '22 at 08:34

score 6 · Answer 10 · answered Mar 20 '22 at 11:33

My answer would be an overkill, but here is how you can get filename from 4 different uri types in android.

Content provider uri [content://com.example.app/sample.png]
File uri [file://data/user/0/com.example.app/cache/sample.png]
Resource uri [android.resource://com.example.app/1234567890] or [android.resource://com.example.app/raw/sample]
Http uri [https://example.com/sample.png]

fun Uri.name(context: Context): String {
    when (scheme) {
        ContentResolver.SCHEME_FILE -> {
            return toFile().nameWithoutExtension
        }
        ContentResolver.SCHEME_CONTENT -> {
            val cursor = context.contentResolver.query(
                this,
                arrayOf(OpenableColumns.DISPLAY_NAME),
                null,
                null,
                null
            ) ?: throw Exception("Failed to obtain cursor from the content resolver")
            cursor.moveToFirst()
            if (cursor.count == 0) {
                throw Exception("The given Uri doesn't represent any file")
            }
            val displayNameColumnIndex = cursor.getColumnIndex(OpenableColumns.DISPLAY_NAME)
            val displayName = cursor.getString(displayNameColumnIndex)
            cursor.close()
            return displayName.substringBeforeLast(".")
        }
        ContentResolver.SCHEME_ANDROID_RESOURCE -> {
            // for uris like [android.resource://com.example.app/1234567890]
            var resourceId = lastPathSegment?.toIntOrNull()
            if (resourceId != null) {
                return context.resources.getResourceName(resourceId)
            }
            // for uris like [android.resource://com.example.app/raw/sample]
            val packageName = authority
            val resourceType = if (pathSegments.size >= 1) {
                pathSegments[0]
            } else {
                throw Exception("Resource type could not be found")
            }
            val resourceEntryName = if (pathSegments.size >= 2) {
                pathSegments[1]
            } else {
                throw Exception("Resource entry name could not be found")
            }
            resourceId = context.resources.getIdentifier(
                resourceEntryName,
                resourceType,
                packageName
            )
            return context.resources.getResourceName(resourceId)
        }
        else -> {
            // probably a http uri
            return toString().substringBeforeLast(".").substringAfterLast("/")
        }
    }
}

score 4 · Answer 11 · answered Nov 01 '13 at 08:12

public String getFilename() 
{
/*  Intent intent = getIntent();
    String name = intent.getData().getLastPathSegment();
    return name;*/
    Uri uri=getIntent().getData();
    String fileName = null;
    Context context=getApplicationContext();
    String scheme = uri.getScheme();
    if (scheme.equals("file")) {
        fileName = uri.getLastPathSegment();
    }
    else if (scheme.equals("content")) {
        String[] proj = { MediaStore.Video.Media.TITLE };
        Uri contentUri = null;
        Cursor cursor = context.getContentResolver().query(uri, proj, null, null, null);
        if (cursor != null && cursor.getCount() != 0) {
            int columnIndex = cursor.getColumnIndexOrThrow(MediaStore.Video.Media.TITLE);
            cursor.moveToFirst();
            fileName = cursor.getString(columnIndex);
        }
    }
    return fileName;
}

this didn't work in my case , but Vasanth answer did. – Nabzi Sep 18 '18 at 14:07 — Nabzi, Sep 18 '18 at 14:07

score 2 · Answer 12 · answered Nov 11 '18 at 01:30

This actually worked for me:

private String uri2filename() {

    String ret;
    String scheme = uri.getScheme();

    if (scheme.equals("file")) {
        ret = uri.getLastPathSegment();
    }
    else if (scheme.equals("content")) {
        Cursor cursor = getContentResolver().query(uri, null, null, null, null);
        if (cursor != null && cursor.moveToFirst()) {
            ret = cursor.getString(cursor.getColumnIndex(OpenableColumns.DISPLAY_NAME));
        }
   }
   return ret;
}

score 2 · Answer 13 · edited Nov 19 '19 at 04:14

Please try this :

  private String displayName(Uri uri) {

             Cursor mCursor =
                     getApplicationContext().getContentResolver().query(uri, null, null, null, null);
             int indexedname = mCursor.getColumnIndex(OpenableColumns.DISPLAY_NAME);
             mCursor.moveToFirst();
             String filename = mCursor.getString(indexedname);
             mCursor.close();
             return filename;
 }

OhhhThatVarun · Answer 14 · 2021-08-13T10:50:51.567

2

If anyone is looking for a Kotlin answer especially an extension function, here is the way to go.

fun Uri.getOriginalFileName(context: Context): String? {
    return context.contentResolver.query(this, null, null, null, null)?.use {
        val nameColumnIndex = it.getColumnIndex(OpenableColumns.DISPLAY_NAME)
        it.moveToFirst()
        it.getString(nameColumnIndex)
    }
}

edited Aug 13 '21 at 10:50

answered Aug 13 '21 at 10:34

OhhhThatVarun

3,981
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score 2 · Answer 15 · answered Nov 01 '21 at 14:24

Here is my utils method to achieve this. You can copy/paste and use it from anywhere.

public class FileUtils {

    /**
     * Return file name from Uri given.
     * @param context the context, cannot be null.
     * @param uri uri request for file name, cannot be null
     * @return the corresponding display name for file defined in uri or null if error occurs.
     */
    public String getNameFromURI(@NonNull Context context,  @NonNull Uri uri) {
        String result = null;
        Cursor c = null;
        try {
            c = context.getContentResolver().query(uri, null, null, null, null);
            c.moveToFirst();
            result = c.getString(c.getColumnIndex(OpenableColumns.DISPLAY_NAME));
        }
        catch (Exception e){
            // error occurs
        }
        finally {
            if(c != null){
                c.close();
            }
        }
        return result;
    }
...
}

And usage.

String fileName = FileUtils.getNameFromContentUri(context, myuri);
if(fileName != null){
    // do stuff
}

Regards.

score 2 · Answer 16 · answered Dec 28 '21 at 13:52

2

How about this?

 Uri uri = result.getData().getClipData().getItemAt(i).getUri();
 uri = Uri.parse(uri.getLastPathSegment());
 System.out.println(uri.getLastPathSegment());

This prints the file name with extension

answered Dec 28 '21 at 13:52

WhiteVulpes

29
1

This will not work with all uris. For example, if I use a uri I received from Google Photos `content://com.google.../image%2Fjpeg/484274255` this will not return a valid file name with an extension. – Lifes Apr 08 '22 at 07:53

score 1 · Answer 17 · answered Aug 16 '15 at 20:14

1

String Fpath = getPath(this, uri) ;
File file = new File(Fpath);
String filename = file.getName();

answered Aug 16 '15 at 20:14

user5233139

19
1

6

Could you explain a bit what your code does? This way your answer will be more useful for other users stumbling over this issue. Thanks – wmk Aug 16 '15 at 20:17
for some reason this doesn't work on marshmellow. It just outputs "audio and some number" – Alex Apr 20 '16 at 21:03

score 1 · Answer 18 · answered Aug 19 '15 at 09:24

My version of the answer is actually very similar to the @Stefan Haustein. I found the answer on Android Developer page Retrieving File Information; the information here is even more condensed on this specific topic than on Storage Access Framework guide site. In the result from the query the column index containing file name is OpenableColumns.DISPLAY_NAME. None of other answers/solutions for column indexes worked for me. Below is the sample function:

 /**
 * @param uri uri of file.
 * @param contentResolver access to server app.
 * @return the name of the file.
 */
def extractFileName(uri: Uri, contentResolver: ContentResolver): Option[String] = {

    var fileName: Option[String] = None
    if (uri.getScheme.equals("file")) {

        fileName = Option(uri.getLastPathSegment)
    } else if (uri.getScheme.equals("content")) {

        var cursor: Cursor = null
        try {

            // Query the server app to get the file's display name and size.
            cursor = contentResolver.query(uri, null, null, null, null)

            // Get the column indexes of the data in the Cursor,
            // move to the first row in the Cursor, get the data.
            if (cursor != null && cursor.moveToFirst()) {

                val nameIndex = cursor.getColumnIndex(OpenableColumns.DISPLAY_NAME)
                fileName = Option(cursor.getString(nameIndex))
            }

        } finally {

            if (cursor != null) {
                cursor.close()
            }

        }

    }

    fileName
}

Ayaz Alifov · Answer 19 · 2016-01-10T13:15:01.710

1

First, you need to convert your URI object to URL object, and then use File object to retrieve a file name:

try
    {
        URL videoUrl = uri.toURL();
        File tempFile = new File(videoUrl.getFile());
        String fileName = tempFile.getName();
    }
    catch (Exception e)
    {

    }

That's it, very easy.

edited Jan 10 '16 at 13:15

answered Jan 10 '16 at 13:09

Ayaz Alifov

8,334
4
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score 1 · Answer 20 · answered Oct 30 '18 at 08:29

Stefan Haustein function for xamarin/c#:

public string GetFilenameFromURI(Android.Net.Uri uri)
        {
            string result = null;
            if (uri.Scheme == "content")
            {
                using (var cursor = Application.Context.ContentResolver.Query(uri, null, null, null, null))
                {
                    try
                    {
                        if (cursor != null && cursor.MoveToFirst())
                        {
                            result = cursor.GetString(cursor.GetColumnIndex(OpenableColumns.DisplayName));
                        }
                    }
                    finally
                    {
                        cursor.Close();
                    }
                }
            }
            if (result == null)
            {
                result = uri.Path;
                int cut = result.LastIndexOf('/');
                if (cut != -1)
                {
                    result = result.Substring(cut + 1);
                }
            }
            return result;
        }

score 1 · Answer 21 · answered Jul 13 '20 at 18:59

Combination of all the answers

Here is what I have arrived at after a read of all the answers presented here as well what some Airgram has done in their SDKs - A utility that I have open sourced on Github:

https://github.com/mankum93/UriUtilsAndroid/tree/master/app/src/main/java/com/androiduriutils

Usage

As simple as calling, UriUtils.getDisplayNameSize(). It provides both the name and size of the content.

Note: Only works with a content:// Uri

Here is a glimpse on the code:

/**
 * References:
 * - https://www.programcreek.com/java-api-examples/?code=MLNO/airgram/airgram-master/TMessagesProj/src/main/java/ir/hamzad/telegram/MediaController.java
 * - https://stackoverflow.com/questions/5568874/how-to-extract-the-file-name-from-uri-returned-from-intent-action-get-content
 *
 * @author Manish@bit.ly/2HjxA0C
 * Created on: 03-07-2020
 */
public final class UriUtils {


    public static final int CONTENT_SIZE_INVALID = -1;

    /**
     * @param context context
     * @param contentUri content Uri, i.e, of the scheme <code>content://</code>
     * @return The Display name and size for content. In case of non-determination, display name
     * would be null and content size would be {@link #CONTENT_SIZE_INVALID}
     */
    @NonNull
    public static DisplayNameAndSize getDisplayNameSize(@NonNull Context context, @NonNull Uri contentUri){

        final String scheme = contentUri.getScheme();
        if(scheme == null || !scheme.equals(ContentResolver.SCHEME_CONTENT)){
            throw new RuntimeException("Only scheme content:// is accepted");
        }

        final DisplayNameAndSize displayNameAndSize = new DisplayNameAndSize();
        displayNameAndSize.size = CONTENT_SIZE_INVALID;

        String[] projection = new String[]{MediaStore.Images.Media.DATA, OpenableColumns.DISPLAY_NAME, OpenableColumns.SIZE};
        Cursor cursor = context.getContentResolver().query(contentUri, projection, null, null, null);
        try {
            if (cursor != null && cursor.moveToFirst()) {

                // Try extracting content size

                int sizeIndex = cursor.getColumnIndex(OpenableColumns.SIZE);
                if (sizeIndex != -1) {
                    displayNameAndSize.size = cursor.getLong(sizeIndex);
                }

                // Try extracting display name
                String name = null;

                // Strategy: The column name is NOT guaranteed to be indexed by DISPLAY_NAME
                // so, we try two methods
                int nameIndex = cursor.getColumnIndex(OpenableColumns.DISPLAY_NAME);
                if (nameIndex != -1) {
                    name = cursor.getString(nameIndex);
                }

                if (nameIndex == -1 || name == null) {
                    nameIndex = cursor.getColumnIndex(MediaStore.Images.Media.DATA);
                    if (nameIndex != -1) {
                        name = cursor.getString(nameIndex);
                    }
                }
                displayNameAndSize.displayName = name;
            }
        }
        finally {
            if(cursor != null){
                cursor.close();
            }
        }

        // We tried querying the ContentResolver...didn't work out
        // Try extracting the last path segment
        if(displayNameAndSize.displayName == null){
            displayNameAndSize.displayName = contentUri.getLastPathSegment();
        }

        return displayNameAndSize;
    }
}

score 0 · Answer 22 · answered Feb 09 '21 at 06:53

我从开发者官网找到一些信息

I got some information from the developer's website

取得游标

val cursor = context.contentResolver.query(fileUri, null, null, null, null)

接着就可以获取名称和文件大小

val nameIndex = cursor.getColumnIndex(OpenableColumns.DISPLAY_NAME)
val sizeIndex = cursor.getColumnIndex(OpenableColumns.SIZE)
cursor.moveToFirst()
val fileName = cursor.getString(nameIndex)
val size = cursor.getLong(sizeIndex)

别忘记关闭资源

Don't forget to close resources

Retrieving file information

kulikovman · Answer 23 · 2021-05-15T05:11:19.867

0

This will return the filename from Uri without file extension.

fun Uri.getFileName(): String? {
    return this.path?.let { path -> File(path).name }
}

Here I described a way to get the filename with the extension.

edited May 15 '21 at 05:11

answered May 13 '21 at 05:08

kulikovman

333
4
8

1

Please add an explanation to improve the insight and give the OP some more information – ucMedia May 13 '21 at 17:27

score 0 · Answer 24 · answered Aug 30 '22 at 17:47

Try this,

Intent data = result.getData();
// check condition
if (data != null) {
    Uri sUri = data.getData();
    @SuppressLint("Recycle")
    Cursor returnCursor = getContentResolver().query(sUri, null, null, null, null);
    int nameIndex = returnCursor.getColumnIndex(OpenableColumns.DISPLAY_NAME);
    returnCursor.moveToFirst();
    String file_name = returnCursor.getString(nameIndex);
}

How to extract the file name from URI returned from Intent.ACTION_GET_CONTENT?

24 Answers24

Retrieving a File's name.

Combination of all the answers

Usage

Linked