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JS / React Native - open link to a website (and trigger a popup window)

Not sure if it's a javascript question or something related to Expo's Linking in React Native. I'm using:

Linking.openURL('https://www.example.com');

To direct a user to a website from the app. Is there a way to pass an attribute in the url that would trigger a popup available on the website. I've got a class of the button that triggers the popup on the website.

Thanks

about 4 years ago · Juan Pablo Isaza
1 answers
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You need to pass the query string with the URL and in the website, you need to check the query string, and based on the value you have to show the popup. Like this:

Linking.openURL('https://www.example.com?showModel=true');
about 4 years ago · Juan Pablo Isaza Report
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