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openURL(string) Error

When I use openURL(string) on a buttion, it returns this error.

Screen Shot 2018-10-15 at 11.45.49 AM

Is there any way for it to automatically use safari as the defult browser rather than constantly choosing the application to open, I have to do this every time and it wont register it as safari :( .

1 reply

null
    • Mconneen
    • 5 yrs ago
    • Reported - view

    Make sure to prefix your string starts with http:// or https:// ...  I was able to repeat this error by dropping the prefix. 

    So  openURL("https://www.google.com")  

    works.

Content aside

  • 5 yrs agoLast active
  • 1Replies
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