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Question by nickeeromo · Oct 19, 2017 at 04:20 PM · c#iosoptionsopenurl

Launch an iOS app with options C#

I usually don't have to ask questions here, but all I can find is the code using swift, objective-c or some Android examples, and they set key names and values with specific commands, not in one string.


As I understand it, I should be able to use:


Application.OpenURL("targetName://");


to open an iOS app with the URL schema to accept targetName. However, would it read options if I were to put


?key=value


at the end of this, is there a separate command or set of commands I should be using here, and if so, can I pass an object or would variables have to be separate, and would I separate with question marks or something else if I can just use a straight string here?


I appreciate any and all help. Thank you.

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avatar image nickeeromo · Oct 25, 2017 at 06:45 PM 0
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So it launched the other application when I sent it:

Application.OpenURL("targetName://key=value");

However, the options are not read in the DidFinishLaunchingWithOptions function. I'll have to return to it later to see what is going on, as I know zero xcode stuff, and that maybe is where my solution will lie.

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