根据Paul Burke的回答，我在解决外部SD卡的URI路径时遇到了许多问题，因为大多数建议的“内置”函数返回的路径都不能解析为文件。

然而，这是我的方法 // TODO处理非主卷。

String resolvedPath = "";
File[] possibleExtSdComposites = context.getExternalFilesDirs(null);
for (File f : possibleExtSdComposites) {
    // Reset final path
    resolvedPath = "";

    // Construct list of folders
    ArrayList<String> extSdSplit = new ArrayList<>(Arrays.asList(f.getPath().split("/")));

    // Look for folder "<your_application_id>"
    int idx = extSdSplit.indexOf(BuildConfig.APPLICATION_ID);

    // ASSUMPTION: Expected to be found at depth 2 (in this case ExtSdCard's root is /storage/0000-0000/) - e.g. /storage/0000-0000/Android/data/<your_application_id>/files
    ArrayList<String> hierarchyList = new ArrayList<>(extSdSplit.subList(0, idx - 2));

    // Construct list containing full possible path to the file
    hierarchyList.add(tail);
    String possibleFilePath = TextUtils.join("/", hierarchyList);

    // If file is found --> success
    if (idx != -1 && new File(possibleFilePath).exists()) {
        resolvedPath = possibleFilePath;
        break;
    }
}

if (!resolvedPath.equals("")) {
    return resolvedPath;
} else {
    return null;
}

注意，它取决于层次结构，可能在每个手机制造商上都是不同的-我没有全部测试过(到目前为止，它在Xperia Z3 API 23和三星Galaxy A3 API 23上运行良好)。

请确认其他地方是否表现不佳

@paul burke的答案适用于API级别19及以上的相机和画廊图片，但如果你的Android项目的最低SDK设置为19以下，它就不起作用，上面提到的一些答案并不适用于画廊和相机。好吧，我已经修改了@paul burke的代码，它适用于API级别低于19的代码。下面是代码。

public static String getPath(final Context context, final Uri uri) {

    final boolean isKitKat = Build.VERSION.SDK_INT >=
                             Build.VERSION_CODES.KITKAT;
    Log.i("URI",uri+"");
    String result = uri+"";

    // DocumentProvider
    // if (isKitKat && DocumentsContract.isDocumentUri(context, uri)) {
    if (isKitKat && (result.contains("media.documents"))) {

        String[] ary = result.split("/");
        int length = ary.length;
        String imgary = ary[length-1];
        final String[] dat = imgary.split("%3A");

        final String docId = dat[1];
        final String type = dat[0];

        Uri contentUri = null;
        if ("image".equals(type)) {
            contentUri = MediaStore.Images.Media.EXTERNAL_CONTENT_URI;
        }
        else if ("video".equals(type)) {
        }
        else if ("audio".equals(type)) {
        }

        final String selection = "_id=?";
        final String[] selectionArgs = new String[] {
            dat[1]
        };

        return getDataColumn(context, contentUri, selection, selectionArgs);
    }
    else
    if ("content".equalsIgnoreCase(uri.getScheme())) {
        return getDataColumn(context, uri, null, null);
    }
    // File
    else if ("file".equalsIgnoreCase(uri.getScheme())) {
        return uri.getPath();
    }

    return null;
}

public static String getDataColumn(Context context, Uri uri, String selection,
                                   String[] selectionArgs) {
    Cursor cursor = null;
    final String column = "_data";
    final String[] projection = {
            column
    };

    try {
        cursor = context.getContentResolver().query(uri, projection, selection, selectionArgs,
                null);
        if (cursor != null && cursor.moveToFirst()) {
            final int column_index = cursor.getColumnIndexOrThrow(column);
            return cursor.getString(column_index);
        }
    }
    finally {
        if (cursor != null)
            cursor.close();
    }
    return null;
}

对于这种类型的uri 内容:/ / % 3 a19298 com.android.providers.media.documents /文件/文档 或者uri.getAuthority()是这些中的任何一个

"com.google.android.apps.docs.storage".equals(uri.getAuthority()) || "com.google.android.apps.docs.storage.legacy".equals(uri.getAuthority());

使用这个函数

private static String getDriveFilePath(Uri uri, Context context) {
        Uri returnUri = uri;
        Cursor returnCursor = context.getContentResolver().query(returnUri, null, null, null, null);

        int nameIndex = returnCursor.getColumnIndex(OpenableColumns.DISPLAY_NAME);
        int sizeIndex = returnCursor.getColumnIndex(OpenableColumns.SIZE);
        returnCursor.moveToFirst();
        String name = (returnCursor.getString(nameIndex));
        String size = (Long.toString(returnCursor.getLong(sizeIndex)));
        File file = new File(context.getCacheDir(), name);
        try {
            InputStream inputStream = context.getContentResolver().openInputStream(uri);
            FileOutputStream outputStream = new FileOutputStream(file);
            int read = 0;
            int maxBufferSize = 1 * 1024 * 1024;
            int bytesAvailable = inputStream.available();

            //int bufferSize = 1024;
            int bufferSize = Math.min(bytesAvailable, maxBufferSize);

            final byte[] buffers = new byte[bufferSize];
            while ((read = inputStream.read(buffers)) != -1) {
                outputStream.write(buffers, 0, read);
            }
            Log.e("File Size", "Size " + file.length());
            inputStream.close();
            outputStream.close();
            Log.e("File Path", "Path " + file.getPath());
            Log.e("File Size", "Size " + file.length());
        } catch (Exception e) {
            Log.e("Exception", e.getMessage());
        }
        return file.getPath();
    }

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。

正如commonware提到的，你不应该假设，你通过ContentResolver得到的流可以转换成文件。

你真正应该做的是从ContentProvider打开InputStream，然后从中创建一个位图。而且它在4.4和更早的版本上也可以工作，不需要反射。

    //cxt -> current context

    InputStream input;
    Bitmap bmp;
    try {
        input = cxt.getContentResolver().openInputStream(fileUri);
        bmp = BitmapFactory.decodeStream(input);
    } catch (FileNotFoundException e1) {

    }

当然，如果你处理大图片，你应该用适当的inSampleSize: http://developer.android.com/training/displaying-bitmaps/load-bitmap.html加载它们。但那是另一个话题了。

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;
}

评论

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