iOS 9 failed for URL: "XXX://@" - error: "This app is not allowed to query for scheme XXX" iOS 从APP里
2016-09-14 19:38
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iOS 9 failed for URL: "XXX://@" - error: "This app is not allowed to
query for scheme XXX" iOS 从APP里启动另一APP
iOS 从C APP里启动 D APP
首先在D APP里设置 URL Schemes
在info.plist 文件里添加URL Schemes
URL Types -->item0 --> URL Schemes --> TestD
然后再回到C APP 找到info.plist 文件
添加 LSApplicationQueriesSchemes --> item0 TestD
添加代码
NSURL *urlT = [NSURL URLWithString:@"TestD://@lksdjflksdl"]; //注意“://”后面可以任意传参数。这些参数传过去后当跳到B时会在-(BOOL)application:(UIApplication *)application handleOpenURL:(NSURL *)url 这个方法里实现。 if ([[UIApplication sharedApplication] canOpenURL:urlT]) { NSLog(@"xxxx"); [[UIApplication sharedApplication] openURL:urlT]; }
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