I am using CameraX API to take pictures in my android app, save them and then display them from their path. With the previous version alpha-09 I was able to do so with onImageSaved(File file). However with the alpha-10 I have to use onImageSaved(OutputFileResults outputFileResults) and then get the path from the uri retrieved by the outputFileResults. But the Uri I get is always wrong. For instance when my image is saved at: "/external/images/media/1581680878237.jpg" I get the uri's path: "/external/images/media/113758".
Here is my code:
ContentValues contentValues = new ContentValues();
contentValues.put(MediaStore.MediaColumns.DISPLAY_NAME, "NEW_IMAGE");
contentValues.put(MediaStore.MediaColumns.MIME_TYPE, "image/jpg");
ImageCapture.OutputFileOptions outputFileOptions = new ImageCapture.OutputFileOptions.Builder(
activity.getContentResolver(),
MediaStore.Images.Media.EXTERNAL_CONTENT_URI,
contentValues).build();
imageCapture.takePicture(outputFileOptions, Runnable::run, new ImageCapture.OnImageSavedCallback() {
#Override
public void onImageSaved(#NonNull ImageCapture.OutputFileResults outputFileResults) {
Uri uri = outputFileResults.getSavedUri();
if(uri != null){
System.out.println("URI PATH" + uri.getPath());
System.out.println("URI PATH" + uri.toString());
activity.runOnUiThread(cameraProvider::unbindAll);
galleryAddPic(uri);
Bundle params = new Bundle();
params.putString("FILE_PATH", uri.getPath());
Navigation.findNavController(root).navigate(R.id.navigation_edit_image, params);
}
}
#Override
public void onError(#NonNull ImageCaptureException exception) {
exception.printStackTrace();
}
});
So I finally managed to save the image taken by ImageCapture by using an other method (especially an other ImageCapture.OutputFileOptions.Builde). I didn't use an Uri object to save the image but a File object.
File mImageDir = new File(Environment.getExternalStoragePublicDirectory(Environment.DIRECTORY_PICTURES), "YOUR_DIRECTORY");
boolean isDirectoryCreated = mImageDir.exists() || mImageDir.mkdirs();
if(isDirectoryCreated){
File file = new File(Environment.getExternalStoragePublicDirectory(Environment.DIRECTORY_PICTURES) + "/YOUR_DIRECTORY", "YOUR_IMAGE.jpg");
ImageCapture.OutputFileOptions.Builder outputFileOptionsBuilder =
new ImageCapture.OutputFileOptions.Builder(file);
imageCapture.takePicture(outputFileOptionsBuilder.build(), Runnable::run, new ImageCapture.OnImageSavedCallback() {
#Override
public void onImageSaved(#NonNull ImageCapture.OutputFileResults outputFileResults) {
Bundle params = new Bundle();
params.putString("FILE_PATH", file.getPath());
Navigation.findNavController(root).navigate(R.id.navigation_edit_image, params);
}
#Override
public void onError(#NonNull ImageCaptureException exception) {
exception.printStackTrace();
}
});
}
Be aware that if you use outputFileResults.getSavedUri() with this method you will always have a null uri.
As of CameraX alpha 10, ImageCapture supports 3 types of save location: File, MediaStore URI and OutputStream, depending on which OutputFileOptions.Builder() is used.
The Uri field in OutputFileResults is only populated if the OutputFileOptions is MediaStore URI type. For File type, the caller should have the save location already, there is no need to return the info; for OutputStream type, the save location is unknown to CameraX. See the JavaDoc:
public Uri getSavedUri ()
Returns the Uri of the saved file.
This field is only returned if the ImageCapture.OutputFileOptions is
backed by MediaStore constructed with #Builder(ContentResolver, Uri,
ContentValues).
For more info, please checkout the developer doc.
I am writing a App for a Open Source Conference.
Originally each attendees will receive different link via email or SMS like
https://example.com/?token=fccfc8bfa07643a1ca8015cbe74f5f17
then use this link to open app, we can know the user is which attendee by the token.
Firebase release a new feature Dynamic Links in I/O 2016, it provide better experience for users.
I had try that, but I can't find any way to pass the custom parameters (the token) in dynamic links, how to use the same link with different parameters to my users?
Thanks.
I don't think you can use the short url:
https://<my app>.app.goo.gl/Gk3m
unless you create one for each user, but you can use the long url:
https://<my app>.app.goo.gl/?link=https://example.com/?token=fccfc8bfa07643a1ca8015cbe74f5f17 ...(add other parameters as needed)
and set new token for each user.
I assume you generate the tokens automatically. In that case you can use this to shorten the links.
1) From https://console.firebase.google.com/ (no need for custom parameter here.)
2) Create link somewhere, f.e. on your confluence page (here we add our parameter):
https://PROJECTNAME.page.link/?link=https://PROJECTNAME.page.link/LINKNAME?PARAMETER=1&ofl=https://www.PROJECTNAME.com/
PARAMETER is your custom parameter.
ofl is a link where to go if click the link from another platform (PC, Mac).
3) Getting link data from android project Kotlin code:
Firebase.dynamicLinks
.getDynamicLink(intent)
.addOnSuccessListener { pendingDynamicLinkData ->
val parameter: String =
pendingDynamicLinkData?.link?.getQueryParameter("PARAMETER").orEmpty()
}
If you want to use dynamic links with custom arguments with REST, here is an example of a payload:
{
"dynamicLinkInfo": {
"dynamicLinkDomain": "example.app.goo.gl",
"link": "http://someurl.com?my_first_param=test&my_second_param=test2"
},
"suffix": {
"option":"UNGUESSABLE"
}
}
Make sure your remove 'https://' from your dynamicLinkDomain
Julien
Case A. If you want the short link to expand to a link with multiple parameters:
In the part where you setup a dynamic link, any parameter you append to the deep link URL will go on all platforms (web, iOS, android)
Case B. If you want to use dynamic parameters, you should use the api to create a short link
see documentation
Now you can create short links using the Firebase SDK through the FirebaseDynamicLinks.getInstance().createDynamicLink():
https://firebase.google.com/docs/dynamic-links/android/create
Sample code:
Task shortLinkTask = FirebaseDynamicLinks.getInstance().createDynamicLink()
.setLink(Uri.parse("https://example.com/"))
.setDynamicLinkDomain("abc123.app.goo.gl")
// Set parameters
// ...
.buildShortDynamicLink()
.addOnCompleteListener(this, new OnCompleteListener() {
#Override
public void onComplete(#NonNull Task task) {
if (task.isSuccessful()) {
// Short link created
Uri shortLink = task.getResult().getShortLink();
Uri flowchartLink = task.getResult().getPreviewLink();
} else {
// Error
}
}
});
Create a dynamic link
Go to link details
Copy the long dynamic link and add your parameter in the link parameter of the URL, e.g., PARAMETER=132323
https://link.projectname.com/?link=https://link.projectname.com/LINK?PARAMETER=132323&apn=com.projectname.app&afl=https://link.projectname.com/LINK
1 First Change your Dynamic Link in firebase console from http://exampleandroid/test to http://exampleandroid/test?data 2. You send the query paramter data with this
DynamicLink dynamicLink = FirebaseDynamicLinks.getInstance().createDynamicLink()
// .setLink(dynamicLinkUri)
.setLink(Uri.parse("http://exampleandroid/test?data=dsads"))
.setDomainUriPrefix("https://App_Name.page.link")
// Open links with this app on Android
.setAndroidParameters(new DynamicLink.AndroidParameters.Builder().build())
// Open links with com.example.ios on iOS
.setIosParameters(new DynamicLink.IosParameters.Builder("com.appinventiv.ios").build())
.buildDynamicLink();
dynamicLinkUri = dynamicLink.getUri();
You can add extra parameter to your link to generate Short URL from Firebase.
Here I given example of Short URL generation using Firebase API.
Here ServiceRequestHelper(this).PostCall is my generic class to make API request
String url = "https://firebasedynamiclinks.googleapis.com/v1/shortLinks?key=YOUR_KEY";
try {
PackageManager manager = this.getPackageManager();
PackageInfo info = manager.getPackageInfo(this.getPackageName(), 0);
JSONObject jsonObject = new JSONObject();
JSONObject androidInfoObject = new JSONObject();
androidInfoObject.put("androidPackageName", getApplicationContext().getPackageName());
androidInfoObject.put("androidMinPackageVersionCode",String.valueOf(info.versionCode));
JSONObject iosInfoObject = new JSONObject();
iosInfoObject.put("iosBundleId", "tettares.Test_ID");
JSONObject dynamicLinkInfoObj = new JSONObject();
dynamicLinkInfoObj.put("dynamicLinkDomain", "wgv3v.app.goo.gl");
dynamicLinkInfoObj.put("link", "https://test.in/?UserId=14&UserName=Naveen"); // Pass your extra paramters to here
dynamicLinkInfoObj.put("androidInfo", androidInfoObject);
dynamicLinkInfoObj.put("iosInfo", iosInfoObject);
JSONObject suffixObject = new JSONObject();
suffixObject.put("option" , "SHORT");
jsonObject.put("dynamicLinkInfo", dynamicLinkInfoObj);
jsonObject.put("suffix", suffixObject);
Log.d("JSON Object : " , jsonObject.toString());
new ServiceRequestHelper(this).PostCall(url, jsonObject, false, new CallBackJson() {
#Override
public void done(JSONObject result) throws JSONException {
try {
if (result.has("shortLink")) {
DEEP_LINK_URL = result.getString("shortLink"); }
} catch(Exception e)
{
e.printStackTrace();
}
}
});
} catch (JSONException | PackageManager.NameNotFoundException e) {
e.printStackTrace();
}
In Your Receiving Activity:
boolean autoLaunchDeepLink = false;
AppInvite.AppInviteApi.getInvitation(mGoogleApiClient, this, autoLaunchDeepLink)
.setResultCallback(
new ResultCallback<AppInviteInvitationResult>() {
#Override
public void onResult(#NonNull AppInviteInvitationResult result) {
if (result.getStatus().isSuccess()) {
// Extract deep link from Intent
Intent intent = result.getInvitationIntent();
String deepLink = AppInviteReferral.getDeepLink(intent);
// Handle the deep link. For example, open the linked
// content, or apply promotional credit to the user's
// account.
// [START_EXCLUDE]
// Display deep link in the UI
Log.d(TAG, "deeplink URL: " + deeplink); // Here you get https://test.in/?UserId=14&UserName=Naveen
// [END_EXCLUDE]
} else {
Log.d(TAG, "getInvitation: no deep link found.");
}
}
});
I tried all above but none work. So I think you should change the link http://example.com/?userid=123tohttp://example.com/userid/123
No need of all the hustle
Don't shorten the url if you want to pass parameters
Write the links like this.
//APP_CODE is firebase link
String link = "https://APP_CODE.app.goo.gl/?refferer=" + userId;
Intent intent = new AppInviteInvitation.IntentBuilder(getString(R.string.invitation_title))
.setMessage(getString(R.string.invitation_custom_message)))
.setDeepLink(Uri.parse(link))
.setCustomImage(Uri.parse(getString(R.string.invitation_custom_image)))
.setCallToActionText(getString(R.string.invitation_cta))
.build();
startActivityForResult(intent, REQUEST_INVITE);
For web apps, which generating dynamic links..
const data = {
dynamicLinkInfo: {
domainUriPrefix: 'subdomain.page.link',
link: url,
},
suffix: {
option: 'SHORT'
}
};
return fetch(`https://firebasedynamiclinks.googleapis.com/v1/shortLinks?key=${environment.firebaseConfig.apiKey}`, {
method: 'POST',
headers: {
'Content-Type': 'application/json'
},
body: JSON.stringify(data),
}).then(res => res.json()).then(response => response.shortLink).catch(console.error);
In a web api method I am generating a file and then streaming it to the response like so
public async Task<HttpResponseMessage> GetFile() {
FileInfo file = generateFile();
var msg = Request.CreateResponse(HttpStatusCode.OK);
msg.Content = new StreamContent(file.OpenRead());
msg.Content.Headers.ContentType = new MediaTypeHeaderValue("application/pdf");
msg.Content.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment") {FileName = file.Name};
return msg;
}
because this a generated file I want to delete it after the response has finished streaming but I can't seem to find a hook in the pipeline for this.
I suppose that I can put a reference to the file in a static and set up a custom MessageHandler that pulls values out of this same static variable and deletes. However, this seems like it can't possibly be right both because of the use of a static (when this should all be per-request) and because I'd have to register a separate route.
I've seen this question but it seems to not really have much of a useful response.
Nice scenario!...the problem with using message handlers is that response writing happens at the host layers and below message handlers layer, so they are not ideal...
Following is an example of how you could do it:
msg.Content = new CustomStreamContent(generatedFilePath);
public class CustomStreamContent : StreamContent
{
string filePath;
public CustomStreamContent(string filePath)
: this(File.OpenRead(filePath))
{
this.filePath = filePath;
}
private CustomStreamContent(Stream fileStream)
: base(content: fileStream)
{
}
protected override void Dispose(bool disposing)
{
//close the file stream
base.Dispose(disposing);
try
{
File.Delete(this.filePath);
}
catch (Exception ex)
{
//log this exception somewhere so that you know something bad happened
}
}
}
By the way, are you generating this file because you are converting some data into PDF. If yes, then I think you could use PushStreamContent for this purpose by directly writing the converted data into the response stream. This way you need not generate a file first and then worry about deleting it later.
We performed same action in WebAPI. I needed to delete file just after it downloaded form server.
We can create custom response message class. It takes file path as parameter and delete it once its transmitted.
public class FileResponseMessage : HttpResponseMessage
{
private readonly string _filePath;
public FileHttpResponseMessage(string filePath)
{
this._filePath= filePath;
}
protected override void Dispose(bool disposing)
{
base.Dispose(disposing);
Content.Dispose();
File.Delete(_filePath);
}
}
Use this class as below code and it will delete your file once it will be written on response stream.
var response = new FileResponseMessage(filePath);
response.StatusCode = HttpStatusCode.OK;
response.Content = new StreamContent(new FileStream(filePath, FileMode.Open, FileAccess.Read));
response.Content.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
{
FileName = "MyReport.pdf"
};
return response;
I've implemented a custom exception filter to my Web API. It is working as intended, except for one small detail...
In the following code sample, SaveToErrorLog saves exception details and tries to get the request url from context.Request.RawUrl. But context.Request does not contain the url that the API tried to serve when the exception happened. Is there a way to get the url when using an exception filter like this?
public class APIExceptionFilter : ExceptionFilterAttribute
{
private HttpContextBase context;
public APIExceptionFilter()
{
context = new HttpContextWrapper(HttpContext.Current);
}
public override void OnException(HttpActionExecutedContext actionContext)
{
actionContext.Response = new HttpResponseMessage(HttpStatusCode.InternalServerError);
if (actionContext != null && context != null)
{
facade.SaveToErrorLog(actionContext.Exception, context.Request);
}
throw new HttpResponseException(new HttpResponseMessage(HttpStatusCode.InternalServerError)
{
Content = new StringContent(actionContext.Exception.Message),
ReasonPhrase = "APIException"
});
}
}
As per the comment above by #emre_nevayeshirazi, you need to use the HttpActionExecutedContext. This gives you access to the request and then the required Uri.
public override void OnException(HttpActionExecutedContext actionExecutedContext)
{
var requestedUri = actionExecutedContext.Request.RequestUri;
//Do something
}
public class Sampleontroller:apicontroller
{
public void PostBodyMethod() {
HttpRequestMessage request=this.request;
//How to read the multi part data in the method
}
}
I am sending a multi part data to webapi controller.
How to read the contents in the method?
An 'async' example:
public async Task<HttpResponseMessage> PostSurveys()
{
// Verify that this is an HTML Form file upload request
if (!Request.Content.IsMimeMultipartContent("form-data"))
{
return Request.CreateResponse(HttpStatusCode.BadRequest);
}
//Destination folder
string uploadFolder = "mydestinationfolder";
// Create a stream provider for setting up output streams that saves the output under -uploadFolder-
// If you want full control over how the stream is saved then derive from MultipartFormDataStreamProvider and override what you need.
MultipartFormDataStreamProvider streamProvider = new MultipartFormDataStreamProvider(uploadFolder );
MultipartFileStreamProvider multipartFileStreamProvider = await Request.Content.ReadAsMultipartAsync(streamProvider);
// Get the file names.
foreach (MultipartFileData file in streamProvider.FileData)
{
//Do something awesome with the files..
}
}
Have a look at the article by Mike Wasson:
http://www.asp.net/web-api/overview/working-with-http/sending-html-form-data,-part-2
Or if you are doing file uploads, here: www.strathweb.com/2012/08/a-guide-to-asynchronous-file-uploads-in-asp-net-web-api-rtm/