canOpenUrl-不允许该应用查询方案图

我正在尝试将Instagram网址添加到iOS9中的应用中，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");
}

基于此示例。

1. 您的LSApplicationQueriesSchemes条目应仅包含方案。第二个条目没有意义。

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

2. 读取错误。您正在尝试在方案中输入错误的URL。修复您参考，以instragram在您的来电canOpenURL:。

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

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>