canOpenUrl – 此应用程序不允许查询schemeinstragram

我试图添加到我的应用程序在iOS9的Instagram的url，但我收到以下警告：

-canOpenURL: failed for URL: "instragram://media?id=MEDIA_ID" - error: "This app is not allowed to query for scheme instragram" 

但是，我在info.plist的LSApplicationQueriesSchemes中添加了以下内容;

 <key>LSApplicationQueriesSchemes</key> <array> <string>instagram</string> <string>instagram://media?id=MEDIA_ID</string>//this one seems to be the issue </array> 

任何帮助是极大的赞赏？

编辑1

这是我用来打开instagram的代码：

  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"); } 

基于这个例子。

Answer 1

1. 您的LSApplicationQueriesSchemes条目应该只有计划。 第二项没有意义。

    <key>LSApplicationQueriesSchemes</key> <array> <string>instagram</string> </array> 

2. 阅读错误。 您正尝试在计划中打开带有拼写错误的url。 修复您在调用canOpenURL:对canOpenURL: 。

Answer 2

只放置<string>instagram</string> 。 这是没有必要的完整path，但计划url的基础。

Answer 3

对于需要Facebook的人：

 <key>LSApplicationQueriesSchemes</key> <array> <string>fbauth</string> <string>fbauth2</string> <string>fb-messenger-api20140430</string> <string>fbapi20130214</string> <string>fbapi20130410</string> <string>fbapi20130702</string> <string>fbapi20131010</string> <string>fbapi20131219</string> <string>fbapi20140410</string> <string>fbapi20140116</string> <string>fbapi20150313</string> <string>fbapi20150629</string> <string>fbshareextension</string> </array>