Android 4.4 (KitKat)上的Android Gallery为Intent返回不同的URI。ACTION_GET_CONTENT

在奇巧(或新画廊)之前，意图。ACTION_GET_CONTENT返回一个这样的URI

内容:/ /媒体/外部/图片/媒体/ 3951。

使用ContentResolver并查询 media . data返回文件URL。

然而，在奇巧，画廊返回一个URI(通过“Last”)像这样:

内容:/ / com.android.providers.media.documents /文档/图片:3951

我该怎么处理呢?

当前回答

问题

如何从URI中获得实际的文件路径

回答

据我所知，我们不需要从URI获取文件路径，因为对于大多数情况，我们可以直接使用URI来完成我们的工作(如1。获取位图2。向服务器发送文件，等等)

1. 发送到服务器

我们可以直接使用URI将文件发送到服务器。

使用URI，我们可以获得InputStream，我们可以使用MultiPartEntity直接将其发送到服务器。

例子

/**
 * Used to form Multi Entity for a URI (URI pointing to some file, which we got from other application).
 *
 * @param uri     URI.
 * @param context Context.
 * @return Multi Part Entity.
 */
public MultipartEntity formMultiPartEntityForUri(final Uri uri, final Context context) {
    MultipartEntity multipartEntity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE, null, Charset.forName("UTF-8"));
    try {
        InputStream inputStream = mContext.getContentResolver().openInputStream(uri);
        if (inputStream != null) {
            ContentBody contentBody = new InputStreamBody(inputStream, getFileNameFromUri(uri, context));
            multipartEntity.addPart("[YOUR_KEY]", contentBody);
        }
    }
    catch (Exception exp) {
        Log.e("TAG", exp.getMessage());
    }
    return multipartEntity;
}

/**
 * Used to get a file name from a URI.
 *
 * @param uri     URI.
 * @param context Context.
 * @return File name from URI.
 */
public String getFileNameFromUri(final Uri uri, final Context context) {

    String fileName = null;
    if (uri != null) {
        // Get file name.
        // File Scheme.
        if (ContentResolver.SCHEME_FILE.equals(uri.getScheme())) {
            File file = new File(uri.getPath());
            fileName = file.getName();
        }
        // 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);
                fileName = returnCursor.getString(nameIndex);
                returnCursor.close();
            }
        }
    }
    return fileName;
}

2. 从URI中获取位图

如果URI指向图像，我们将得到位图，否则为null:

/**
 * Used to create bitmap for the given URI.
 * <p>
 * 1. Convert the given URI to bitmap.
 * 2. Calculate ratio (depending on bitmap size) on how much we need to subSample the original bitmap.
 * 3. Create bitmap bitmap depending on the ration from URI.
 * 4. Reference - http://stackoverflow.com/questions/3879992/how-to-get-bitmap-from-an-uri
 *
 * @param context       Context.
 * @param uri           URI to the file.
 * @param bitmapSize Bitmap size required in PX.
 * @return Bitmap bitmap created for the given URI.
 * @throws IOException
 */
public static Bitmap createBitmapFromUri(final Context context, Uri uri, final int bitmapSize) throws IOException {

    // 1. Convert the given URI to bitmap.
    InputStream input = context.getContentResolver().openInputStream(uri);
    BitmapFactory.Options onlyBoundsOptions = new BitmapFactory.Options();
    onlyBoundsOptions.inJustDecodeBounds = true;
    onlyBoundsOptions.inDither = true;//optional
    onlyBoundsOptions.inPreferredConfig = Bitmap.Config.ARGB_8888;//optional
    BitmapFactory.decodeStream(input, null, onlyBoundsOptions);
    input.close();
    if ((onlyBoundsOptions.outWidth == -1) || (onlyBoundsOptions.outHeight == -1)) {
        return null;
    }

    // 2. Calculate ratio.
    int originalSize = (onlyBoundsOptions.outHeight > onlyBoundsOptions.outWidth) ? onlyBoundsOptions.outHeight : onlyBoundsOptions.outWidth;
    double ratio = (originalSize > bitmapSize) ? (originalSize / bitmapSize) : 1.0;

    // 3. Create bitmap.
    BitmapFactory.Options bitmapOptions = new BitmapFactory.Options();
    bitmapOptions.inSampleSize = getPowerOfTwoForSampleRatio(ratio);
    bitmapOptions.inDither = true;//optional
    bitmapOptions.inPreferredConfig = Bitmap.Config.ARGB_8888;//optional
    input = context.getContentResolver().openInputStream(uri);
    Bitmap bitmap = BitmapFactory.decodeStream(input, null, bitmapOptions);
    input.close();

    return bitmap;
}

/**
 * For Bitmap option inSampleSize - We need to give value in power of two.
 *
 * @param ratio Ratio to be rounded of to power of two.
 * @return Ratio rounded of to nearest power of two.
 */
private static int getPowerOfTwoForSampleRatio(final double ratio) {
    int k = Integer.highestOneBit((int) Math.floor(ratio));
    if (k == 0) return 1;
    else return k;
}

Android没有提供任何从URI获取文件路径的方法，在上面的大多数答案中，我们都硬编码了一些常量，这可能会在功能发布中中断(对不起，我可能错了)。在直接从URI获取文件路径的解决方案之前，尝试是否可以用URI和Android默认方法解决您的用例。

参考

https://developer.android.com/guide/topics/providers/content-provider-basics.html https://developer.android.com/reference/android/content/ContentResolver.html https://hc.apache.org/httpcomponents-client-ga/httpmime/apidocs/org/apache/http/entity/mime/content/InputStreamBody.html

2016-08-25 12:40:49

其他回答

这是一个完全的hack，但这是我所做的…

因此，在设置DocumentsProvider时，我注意到样例代码(在getDocIdForFile中，大约在第450行)根据文件相对于指定根的路径(即在第96行中设置mBaseDir的路径)为所选文档生成了唯一的id。

所以URI最终看起来像这样:

内容:/ / com.example.provider /文档/根:路径/ / /文件

正如文档所说，它假设只有一个根(在我的情况下是Environment.getExternalStorageDirectory()，但你可以在其他地方使用…然后它获取文件路径，从根目录开始，并使其成为唯一的ID，前置“root:”。所以我可以通过消除uri.getPath()中的“/document/root:”部分来确定路径，通过这样做来创建一个实际的文件路径:

public void onActivityResult(int requestCode, int resultCode, Intent data) {
// check resultcodes and such, then...
uri = data.getData();
if (uri.getAuthority().equals("com.example.provider"))  {
    String path = Environment.getExternalStorageDirectory(0.toString()
                 .concat("/")
                 .concat(uri.getPath().substring("/document/root:".length())));
    doSomethingWithThePath(path); }
else {
    // another provider (maybe a cloud-based service such as GDrive)
    // created this uri.  So handle it, or don't.  You can allow specific
    // local filesystem providers, filter non-filesystem path results, etc.
}

我知道。这很可耻，但它奏效了。同样，这依赖于您在应用程序中使用自己的文档提供程序来生成文档ID。

(此外，还有一种更好的方法来构建路径，不假设“/”是路径分隔符，等等。但你懂的。)

2013-11-26 06:16:32

试试这个:

if (Build.VERSION.SDK_INT <19){
    Intent intent = new Intent(); 
    intent.setType("image/jpeg");
    intent.setAction(Intent.ACTION_GET_CONTENT);
    startActivityForResult(Intent.createChooser(intent, getResources().getString(R.string.select_picture)),GALLERY_INTENT_CALLED);
} else {
    Intent intent = new Intent(Intent.ACTION_OPEN_DOCUMENT);
    intent.addCategory(Intent.CATEGORY_OPENABLE);
    intent.setType("image/jpeg");
    startActivityForResult(intent, GALLERY_KITKAT_INTENT_CALLED);
}

@Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
    super.onActivityResult(requestCode, resultCode, data);
    if (resultCode != Activity.RESULT_OK) return;
    if (null == data) return;
    Uri originalUri = null;
    if (requestCode == GALLERY_INTENT_CALLED) {
        originalUri = data.getData();
    } else if (requestCode == GALLERY_KITKAT_INTENT_CALLED) {
        originalUri = data.getData();
        final int takeFlags = data.getFlags()
                & (Intent.FLAG_GRANT_READ_URI_PERMISSION
                | Intent.FLAG_GRANT_WRITE_URI_PERMISSION);
        // Check for the freshest data.
        getContentResolver().takePersistableUriPermission(originalUri, takeFlags);
    }

    loadSomeStreamAsynkTask(originalUri);

}

可能需要

@SuppressLint（“NewApi”）

for

takePersistableUriPermission

2013-11-09 09:57:11

问题

如何从URI中获得实际的文件路径

回答

据我所知，我们不需要从URI获取文件路径，因为对于大多数情况，我们可以直接使用URI来完成我们的工作(如1。获取位图2。向服务器发送文件，等等)

1. 发送到服务器

我们可以直接使用URI将文件发送到服务器。

使用URI，我们可以获得InputStream，我们可以使用MultiPartEntity直接将其发送到服务器。

例子

/**
 * Used to form Multi Entity for a URI (URI pointing to some file, which we got from other application).
 *
 * @param uri     URI.
 * @param context Context.
 * @return Multi Part Entity.
 */
public MultipartEntity formMultiPartEntityForUri(final Uri uri, final Context context) {
    MultipartEntity multipartEntity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE, null, Charset.forName("UTF-8"));
    try {
        InputStream inputStream = mContext.getContentResolver().openInputStream(uri);
        if (inputStream != null) {
            ContentBody contentBody = new InputStreamBody(inputStream, getFileNameFromUri(uri, context));
            multipartEntity.addPart("[YOUR_KEY]", contentBody);
        }
    }
    catch (Exception exp) {
        Log.e("TAG", exp.getMessage());
    }
    return multipartEntity;
}

/**
 * Used to get a file name from a URI.
 *
 * @param uri     URI.
 * @param context Context.
 * @return File name from URI.
 */
public String getFileNameFromUri(final Uri uri, final Context context) {

    String fileName = null;
    if (uri != null) {
        // Get file name.
        // File Scheme.
        if (ContentResolver.SCHEME_FILE.equals(uri.getScheme())) {
            File file = new File(uri.getPath());
            fileName = file.getName();
        }
        // 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);
                fileName = returnCursor.getString(nameIndex);
                returnCursor.close();
            }
        }
    }
    return fileName;
}

2. 从URI中获取位图

如果URI指向图像，我们将得到位图，否则为null:

/**
 * Used to create bitmap for the given URI.
 * <p>
 * 1. Convert the given URI to bitmap.
 * 2. Calculate ratio (depending on bitmap size) on how much we need to subSample the original bitmap.
 * 3. Create bitmap bitmap depending on the ration from URI.
 * 4. Reference - http://stackoverflow.com/questions/3879992/how-to-get-bitmap-from-an-uri
 *
 * @param context       Context.
 * @param uri           URI to the file.
 * @param bitmapSize Bitmap size required in PX.
 * @return Bitmap bitmap created for the given URI.
 * @throws IOException
 */
public static Bitmap createBitmapFromUri(final Context context, Uri uri, final int bitmapSize) throws IOException {

    // 1. Convert the given URI to bitmap.
    InputStream input = context.getContentResolver().openInputStream(uri);
    BitmapFactory.Options onlyBoundsOptions = new BitmapFactory.Options();
    onlyBoundsOptions.inJustDecodeBounds = true;
    onlyBoundsOptions.inDither = true;//optional
    onlyBoundsOptions.inPreferredConfig = Bitmap.Config.ARGB_8888;//optional
    BitmapFactory.decodeStream(input, null, onlyBoundsOptions);
    input.close();
    if ((onlyBoundsOptions.outWidth == -1) || (onlyBoundsOptions.outHeight == -1)) {
        return null;
    }

    // 2. Calculate ratio.
    int originalSize = (onlyBoundsOptions.outHeight > onlyBoundsOptions.outWidth) ? onlyBoundsOptions.outHeight : onlyBoundsOptions.outWidth;
    double ratio = (originalSize > bitmapSize) ? (originalSize / bitmapSize) : 1.0;

    // 3. Create bitmap.
    BitmapFactory.Options bitmapOptions = new BitmapFactory.Options();
    bitmapOptions.inSampleSize = getPowerOfTwoForSampleRatio(ratio);
    bitmapOptions.inDither = true;//optional
    bitmapOptions.inPreferredConfig = Bitmap.Config.ARGB_8888;//optional
    input = context.getContentResolver().openInputStream(uri);
    Bitmap bitmap = BitmapFactory.decodeStream(input, null, bitmapOptions);
    input.close();

    return bitmap;
}

/**
 * For Bitmap option inSampleSize - We need to give value in power of two.
 *
 * @param ratio Ratio to be rounded of to power of two.
 * @return Ratio rounded of to nearest power of two.
 */
private static int getPowerOfTwoForSampleRatio(final double ratio) {
    int k = Integer.highestOneBit((int) Math.floor(ratio));
    if (k == 0) return 1;
    else return k;
}

参考

2016-08-25 12:40:49

如果有人感兴趣，我为ACTION_GET_CONTENT做了一个工作的Kotlin版本:

var path: String = uri.path // uri = any content Uri
val databaseUri: Uri
val selection: String?
val selectionArgs: Array<String>?
if ("/document/image:" in path || "/document/image%3A" in path) {
    // files selected from "Documents"
    databaseUri = MediaStore.Images.Media.EXTERNAL_CONTENT_URI
    selection = "_id=?"
    selectionArgs = arrayOf(DocumentsContract.getDocumentId(uri).split(":")[1])
} else { // files selected from all other sources, especially on Samsung devices
    databaseUri = uri
    selection = null
    selectionArgs = null
}
try {
    val projection = arrayOf(MediaStore.Images.Media.DATA,
        MediaStore.Images.Media._ID,
        MediaStore.Images.Media.ORIENTATION,
        MediaStore.Images.Media.DATE_TAKEN) // some example data you can query
    val cursor = context.contentResolver.query(databaseUri,
        projection, selection, selectionArgs, null)
    if (cursor.moveToFirst()) {
        // do whatever you like with the data
    }
    cursor.close()
} catch (e: Exception) {
    Log.e(TAG, e.message, e)
}

2017-03-08 06:33:12

This is what I do: Uri selectedImageURI = data.getData(); imageFile = new File(getRealPathFromURI(selectedImageURI)); private String getRealPathFromURI(Uri contentURI) { Cursor cursor = getContentResolver().query(contentURI, null, null, null, null); if (cursor == null) { // Source is Dropbox or other similar local file path return contentURI.getPath(); } else { cursor.moveToFirst(); int idx = cursor.getColumnIndex(MediaStore.Images.ImageColumns.DATA); return cursor.getString(idx); } } NOTE: managedQuery() method is deprecated, so I am not using it.

这个答案是来自m3n0R的问题安卓得到真正的路径Uri.getPath()和我声称没有信用。我只是想那些还没有解决这个问题的人可以使用这个。

2014-01-17 20:02:42

Android 4.4 (KitKat)上的Android Gallery为Intent返回不同的URI。ACTION_GET_CONTENT

推荐文章

最新文章

标签