canOpenUrl - 此应用不允许查询方案 instagram

Question

我试图在 iOS9 中将 Instagram 网址添加到我的应用中，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>

非常感谢任何帮助？

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。

Answer 1

1. 您的LSApplicationQueriesSchemes 条目应该只有方案。第二个条目没有意义。

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

2. 读取错误。您正在尝试打开方案中存在拼写错误的 URL。修正您对canOpenURL: 的调用中对instragram 的引用。

Answer 2

对于有需要的 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>

Answer 3

只输入<string>instagram</string>。不需要完整路径，而是方案 url 的基础。