我在这里尝试了几个答案，我认为我有一个解决方案，每次都可以工作，并且还可以管理权限。

这是基于LEO的巧妙解决方案。这篇文章应该包含所有的代码，你需要使这个工作，它应该工作在任何手机和Android版本;)

为了能够从SD卡中选择文件，你需要在你的清单中这样做:

<uses-permission android:name="android.permission.READ_EXTERNAL_STORAGE" />

常量:

private static final int PICK_IMAGE = 456; // Whatever number you like
public static final int MY_PERMISSIONS_REQUEST_READ_EXTERNAL = 28528; // Whatever number you like
public static final String FILE_TEMP_NAME = "temp_image"; // Whatever file name you like

检查权限和launchImagePick如果可能的话

if (ContextCompat.checkSelfPermission(getThis(),
        Manifest.permission.READ_EXTERNAL_STORAGE)
        != PackageManager.PERMISSION_GRANTED) {

    ActivityCompat.requestPermissions(getThis(),
            new String[]{Manifest.permission.READ_EXTERNAL_STORAGE},
            MY_PERMISSIONS_REQUEST_READ_EXTERNAL);
}
else {
    launchImagePick();
}

允许响应

@Override
public void onRequestPermissionsResult(int requestCode,
                                       @NonNull
                                         String permissions[],
                                       @NonNull
                                         int[] grantResults) {

    if (manageReadExternalPermissionResponse(this, requestCode, grantResults)) {
        launchImagePick();
    }
}

管理权限响应

public static boolean manageReadExternalPermissionResponse(final Activity activity, int requestCode, int[] grantResults) {

    if (requestCode == MY_PERMISSIONS_REQUEST_READ_EXTERNAL) {

        // If request is cancelled, the result arrays are empty.

        if (grantResults.length > 0
                && grantResults[0] == PackageManager.PERMISSION_GRANTED) {

            // Permission was granted, yay! Do the
            // contacts-related task you need to do.
            return true;

        } else if (grantResults.length > 0 && grantResults[0] == PackageManager.PERMISSION_DENIED) {

            boolean showRationale = ActivityCompat.shouldShowRequestPermissionRationale(activity,
                    Manifest.permission.READ_EXTERNAL_STORAGE);

            if (!showRationale) {
                // The user also CHECKED "never ask again".
                // You can either enable some fall back,
                // disable features of your app
                // or open another dialog explaining
                // again the permission and directing to
                // the app setting.

            } else {
                // The user did NOT check "never ask again".
                // This is a good place to explain the user
                // why you need the permission and ask if he/she wants
                // to accept it (the rationale).
            }
        } else {
            // Permission denied, boo! Disable the
            // functionality that depends on this permission.
        }
    }
    return false;
}

发射图像选择

private void launchImagePick() {

    Intent intent = new Intent(Intent.ACTION_GET_CONTENT);
    intent.setType("image/*");
    intent.addCategory(Intent.CATEGORY_OPENABLE);
    startActivityForResult(intent, PICK_IMAGE);

    // see onActivityResult
}

管理图像选择响应

@Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
    super.onActivityResult(requestCode, resultCode, data);
    if (requestCode == PICK_IMAGE) {

        if (resultCode == Activity.RESULT_OK) {
            if (data != null && data.getData() != null) {

                try {
                     InputStream inputStream = getContentResolver().openInputStream(data.getData())
                     if (inputStream != null) {

                        // No special persmission needed to store the file like that
                        FileOutputStream fos = openFileOutput(FILE_TEMP_NAME, Context.MODE_PRIVATE);

                        final int BUFFER_SIZE = 1 << 10 << 3; // 8 KiB buffer
                        byte[] buffer = new byte[BUFFER_SIZE];
                        int bytesRead = -1;
                        while ((bytesRead = inputStream.read(buffer)) > -1) {
                            fos.write(buffer, 0, bytesRead);
                        }
                        inputStream.close();
                        fos.close();

                        File tempImageFile = new File(getFilesDir()+"/"+FILE_TEMP_NAME);

                        // Do whatever you want with the File

                        // Delete when not needed anymore
                        deleteFile(FILE_TEMP_NAME);
                    }
                }
                catch (Exception e) {
                    e.printStackTrace();
                }
            } else {
                // Error display
            }
        } else {
            // The user did not select any image
        }
    }
}

这就是所有的人;这在我所有的电话上都适用。

这对我来说很有效:

else if(requestCode == GALLERY_ACTIVITY_NEW && resultCode == Activity.RESULT_OK)
{
    Uri uri = data.getData();
    Log.i(TAG, "old uri =  " + uri);
    dumpImageMetaData(uri);

    try {
        ParcelFileDescriptor parcelFileDescriptor =
                getContentResolver().openFileDescriptor(uri, "r");
        FileDescriptor fileDescriptor = parcelFileDescriptor.getFileDescriptor();
        Log.i(TAG, "File descriptor " + fileDescriptor.toString());

        final BitmapFactory.Options options = new BitmapFactory.Options();
        options.inJustDecodeBounds = true;
        BitmapFactory.decodeFileDescriptor(fileDescriptor, null, options);

        options.inSampleSize =
           BitmapHelper.calculateInSampleSize(options,
                                              User.PICTURE_MAX_WIDTH_IN_PIXELS,
                                              User.PICTURE_MAX_HEIGHT_IN_PIXELS);
        options.inJustDecodeBounds = false;

        Bitmap bitmap = BitmapFactory.decodeFileDescriptor(fileDescriptor, null, options);
        imageViewPic.setImageBitmap(bitmap);

        ByteArrayOutputStream stream = new ByteArrayOutputStream();
        bitmap.compress(Bitmap.CompressFormat.JPEG, 100, stream);
        // get byte array here
        byte[] picData = stream.toByteArray();
        ParseFile picFile = new ParseFile(picData);
        user.setProfilePicture(picFile);
    }
    catch(FileNotFoundException exc)
    {
        Log.i(TAG, "File not found: " + exc.toString());
    }
}

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

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

Just wanted to say that this answer is brilliant and I'm using it for a long time without problems. But some time ago I've stumbled upon a problem that DownloadsProvider returns URIs in format content://com.android.providers.downloads.documents/document/raw%3A%2Fstorage%2Femulated%2F0%2FDownload%2Fdoc.pdf and hence app is crashed with NumberFormatException as it's impossible to parse its uri segments as long. But raw: segment contains direct uri which can be used to retrieve a referenced file. So I've fixed it by replacing isDownloadsDocument(uri) if content with following:

final String id = DocumentsContract.getDocumentId(uri);
if (!TextUtils.isEmpty(id)) {
if (id.startsWith("raw:")) {
    return id.replaceFirst("raw:", "");
}
try {
    final Uri contentUri = ContentUris.withAppendedId(
            Uri.parse("content://downloads/public_downloads"), Long.valueOf(id));
    return getDataColumn(context, contentUri, null, null);
} catch (NumberFormatException e) {
    Log.e("FileUtils", "Downloads provider returned unexpected uri " + uri.toString(), e);
    return null;
}
}

我相信已经发布的回复应该会让人们朝着正确的方向前进。然而，以下是我所做的对我正在更新的遗留代码有意义的事情。遗留代码使用图库中的URI来更改和保存图像。

在4.4之前(和谷歌驱动器)，uri看起来是这样的: 内容:/ /媒体/外部/图片/媒体/ 41

正如问题中所述，它们通常是这样的: 内容:/ / com.android.providers.media.documents /文档/图片:3951

因为我需要保存图像的能力，而不打扰已经存在的代码，我只是从图库中复制URI到应用程序的数据文件夹中。然后从数据文件夹中保存的图像文件中产生一个新的URI。

这个想法是这样的:

Intent intent = new Intent(Intent.ACTION_GET_CONTENT);
intent.setType("image/*");
startActivityForResult(intent), CHOOSE_IMAGE_REQUEST);

public void onActivityResult(int requestCode, int resultCode, Intent data) {

    super.onActivityResult(requestCode, resultCode, data);

    File tempFile = new File(this.getFilesDir().getAbsolutePath(), "temp_image");

    //Copy URI contents into temporary file.
    try {
        tempFile.createNewFile();
        copyAndClose(this.getContentResolver().openInputStream(data.getData()),new FileOutputStream(tempFile));
    }
    catch (IOException e) {
        //Log Error
    }

    //Now fetch the new URI
    Uri newUri = Uri.fromFile(tempFile);

    /* Use new URI object just like you used to */
 }

注意- copyAndClose()只是执行文件I/O，将InputStream复制到FileOutputStream。代码没有被发布。