App Link Request phone

1, use OpenURL execution: [[uiapplication sharedapplication] openurl:[nsurl urlwithstring:@ "tel:07551111"]; Jumps are not returned to the app and are not prompted.

Note: use @ "telprompt:" The URL of the link is can be done in the box and return the fact that prompt has been told as a warning box, is understandable to show the popup returned

2, use UIWebView or Wkwebkit class------can frame and can return to the app interface

[Web loadrequest:[nsurlrequest requestwithurl:[nsurl urlwithstring:@ "tel:18612126979"]];

[Self.view Addsubview:web]; Be sure to add this web-to-interface display for working with the frame interface.

App Link Request phone