J'essaie d'ajouter l'URL d'Instagram à mon application dans iOS9, mais je reçois l'avertissement suivant:
-canOpenURL: failed for URL: "instragram://media?id=MEDIA_ID" - error: "This app is not allowed to query for scheme instragram"
Cependant, j'ai ajouté ce qui suit au LSApplicationQueriesSchemes
dans mon info.plist
;
<key>LSApplicationQueriesSchemes</key>
<array>
<string>instagram</string>
<string>instagram://media?id=MEDIA_ID</string>//this one seems to be the issue
</array>
Toute aide est grandement appréciée?
MODIFIER 1
Voici le code que j'utilise pour ouvrir instagram:
NSURL * instagramURL = [NSURL URLWithString:@"instragram://media?id=MEDIA_ID"];//edit: note, to anyone copy pasting this code, please notice the typo OP has in the url, that being "instragram" instead of "instagram". This typo was discovered after this StackOverflow question was posted.
if ([[UIApplication sharedApplication] canOpenURL:instagramURL]) {
//do stuff
}
else{
NSLog(@"NO instgram found");
}
basé sur cet exemple.