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Open a URL from a buttonsAsync input

Topic Labels: Scripting extentions
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Hello all,
Forgive me for what seems like a silly question, I’m new to this and can’t figure this out. I would like to launch a website when I click a button in the scripting block. Is this possible?

let purpose = await input.buttonsAsync(
'Are you Creating a Record?',
[
    {label: 'Create', value: 'creating'}
],
);
if (purpose === 'creating') {
//LAUNCH URL HERE
}
1 Solution

Accepted Solutions

Scripting block itself cannot launch a webpage. Scripting block can display a link (in markdown format) to a web page, and when the user clicks the link, the web page will open. Note that the web page will open the browser window, no in the Scripting block window.

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1 Reply 1

Scripting block itself cannot launch a webpage. Scripting block can display a link (in markdown format) to a web page, and when the user clicks the link, the web page will open. Note that the web page will open the browser window, no in the Scripting block window.