canOpenUrl - 此应用程序不允许查询方案instragram

Question

我试图将iOS网址添加到我在iOS9中的应用程序但是我收到以下警告:

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

但是,我LSApplicationQueriesSchemes在我的内容中添加了以下内容info.plist;

<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参赛作品应该只有计划.第二个条目没有意义.

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

2. 阅读错误.您正尝试在方案中使用拼写错误打开URL.instragram在您的通话中修复您的引用canOpenURL:.

Answer 2

只放<string>instagram</string>。不一定是完整路径，而是方案url的基础。

Answer 3

对于需要帮助的Facebook：

<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>