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?
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
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。
其他回答
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
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)
}
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
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。