How to implement URL Schema in my application?

Question

I have implement all the required things in my application to enable url schema, and its working fine. I want to open my application from safari. I have pass one URL with parameter like : https://www.myURL.com/signup/callback?signupCode=123?id=20. I want to open my application whenever I press on this link. But its not working.

Please suggest to me what I can do to achieve this. I already added the required tags in the .plist file (see below screenshot for reference).

plist infomation

Did you implement "- application:openURL:options:" ? ref: https://developer.apple.com/library/ios/documentation/UIKit/Reference/UIApplicationDelegate_Protocol/index.html#//apple_ref/occ/intfm/UIApplicationDelegate/application:openURL:options: — firstinq, Aug 29 '16 at 07:06
Yes, I do it. I also find it that I have to mplementing Custom URL Schemes, Do you know how to do this — dipen gtl, Aug 29 '16 at 07:18

score 0 · Answer 1 · answered Aug 29 '16 at 06:50

0

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"https://www.myURL.com/signup/callback?signupCode=123?id=20"]];

This is actual code. But hear your link (https://www.myURL.com/signup/callback?signupCode=123?id=20) is not working. So problem is accure.

answered Aug 29 '16 at 06:50

Keyur Hirani

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I want to open this from safari, Not from my application. – dipen gtl Aug 29 '16 at 06:53
Yes i know. this link open in safari. – Keyur Hirani Aug 29 '16 at 06:54
Yes I know that, this is dummy URL just for you information. But actual URL also not ask for open my application – dipen gtl Aug 29 '16 at 06:54
You can take reference from http://stackoverflow.com/questions/31361655/how-to-open-url-schemes-from-safari-in-ios9 – Keyur Hirani Aug 29 '16 at 06:57

How to implement URL Schema in my application?

1 Answers1