如何从Intent.ACTION_GET_CONTENT返回的URI中提取文件名？

Question

我正在使用第三方文件管理器从文件系统中选择一个文件（在我的情况下为PDF）。

这是我启动活动的方式：

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);

这就是我在onActivityResult中所拥有的：

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);
            }
        }
    }
}

代码片段借用于此处提供的 Open Intents文件管理器说明：
http://www.openintents.org/en/node/829

if-else的目的是向后兼容。我想知道这是否是获取文件名的最佳方式，因为我发现其他文件管理器会返回所有类型的东西
例如， Documents ToGo 返回如下内容：

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

getContentResolver().query()返回null
为了使事情更有趣，未命名的文件管理器（我从客户端日志中获取此URI）返回类似于：

/./sdcard/downloads/.bin

是否有从URI中提取文件名的首选方法，或者应该求助于字符串解析？

Answer 1

developer.android.com有很好的示例代码： https://developer.android.com/guide/topics/providers/document-provider.html

只提取文件名的精简版（假设＆＃34;此＆＃34;是一个活动）：

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;
}

Answer 2

我正在使用这样的东西：

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();
    }
}

Answer 3

取自Retrieving File information | Android developers

检索文件名。

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;
}

Answer 4

如果你想要它很短，这应该有用。

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

Answer 5

最简洁的版本：

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

Answer 6

获取文件名的最简单方法：

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

如果他们没有给你正确的名字，你应该使用：

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
}

Answer 7

我使用下面的代码来获取文件名称＆amp;我项目中Uri的文件大小。

/**
 * 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() {

    }
}

Answer 8

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;
}

Answer 9

- (CGSize)collectionView:(UICollectionView *)collectionView layout:(UICollectionViewLayout*)collectionViewLayout sizeForItemAtIndexPath:(NSIndexPath *)indexPath{

    return CGSizeMake(collectionView.frame.size.width/2.0, collectionView.frame.size.height/2.0);
}

Answer 10

首先，您需要将URI对象转换为URL对象，然后使用File对象检索文件名：

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

    }

就是这样，很容易。

Answer 11

我的答案版本与@Stefan Haustein非常相似。我在Android Developer页面Retrieving File Information找到了答案;此处的信息比Storage Access Framework指南网站上的特定主题更加浓缩。在查询结果中，包含文件名的列索引为OpenableColumns.DISPLAY_NAME。列索引的其他答案/解决方案都不适用于我。以下是示例函数：

 /**
 * @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
}

Answer 12

xamarin / c＃的Stefan Haustein函数：

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;
        }

Answer 13

如果要使用扩展名的文件名，可以使用此功能获取。 它还适用于Google驱动器文件选择

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;
}

Answer 14

这实际上对我有用：

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;
}

Answer 15

对于Kotlin，您可以使用类似这样的内容：

object FileUtils {

fun getFileName(context: Context, uri : Uri) : String {
    var fileName = "Unknown"
    when (uri.scheme) {
        ContentResolver.SCHEME_FILE -> {
            fileName = File(uri.path).name
        }
        ContentResolver.SCHEME_CONTENT -> {
            try {
                context.contentResolver.query(
                    uri,
                    null,
                    null,
                    null,
                    null
                )?.apply {
                    if (moveToFirst()) {
                        fileName = getString(getColumnIndex(OpenableColumns.DISPLAY_NAME))
                    }
                    close()
                }
            } catch (e : Exception) {
                return fileName
            }
        }
    }
    return fileName
}
}

Answer 16

请尝试：

  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;
 }

Answer 17

所有答案的组合

在阅读完此处给出的所有答案以及一些Airgram在其SDK中所做的工作后，我得出的结论是-我在Github上开源了该实用程序：

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

用法

就像调用UriUtils.getDisplayNameSize()一样简单。它提供了内容的名称和大小。

注意：仅适用于内容：// Uri

这里是代码的一瞥：

/**
 * 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;
    }
}

Answer 18

这将从不带文件扩展名的 Uri 返回文件名。

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

Here 我描述了一种获取带有扩展名的文件名的方法。

Answer 19

我从开发者官网找到了一些信息

我从开发者的网站上得到了一些信息

取得游标

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

然后就可以获取姓名和文件大小

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

不要忘记关闭资源

Retrieving file information