Hi,
I'm launching a survey PowerApp from an Admin PowerApp. The first time it worked great and passed the parameter that I have included in the launch URL.
Launch("https://web.powerapps.com/apps/76836bd7-f6fb-4362-b41b-8b3ab43c9f12?personId="&thisPatid);
Now the launch doesn't pass the parameter correctly from the browser (the target PowerApp no longer see's it) and when I launch from a published power app the launched app gets the message: "This app isn't opening correctly. This app didn't install correctly. Please try again." Session ID: 178061cf-2fdc-9baf-e6b3-6171be39c0ce
In the OnStart event of the target PowerApp:
Set(thisPatId,Text(Param("personId")));
Please help...Steve
BTW: I logged out of PowerApps on my tablet and logged back in - same symptoms
Solved! Go to Solution.
If you want to launch a PowerApp within another app, then you can use the app id directly in the Launch function:
Launch("76836bd7-f6fb-4362-b41b-8b3ab43c9f12", "personId", thisPatid)
Hi Carlos,
As you pointed out, I wasn't using the Launch function correctly to pass parameters, however, after I fixed the parameter passing it didn't work with your syntax below. I couldn't launch another PowerApp from within an active PowerApp using just the AppId. What is launching the 2nd PowerApp is:
Launch("https://ms.web.powerapps.com/apps/76836bd7-f6fb-4362-b41b-8b3ab43c9f12","personId",thisPatid);
And, my launched PowerApp is happily getting the personId parameter onStart.
Thanks...Steve
If you want to launch a PowerApp within another app, then you can use the app id directly in the Launch function:
Launch("76836bd7-f6fb-4362-b41b-8b3ab43c9f12", "personId", thisPatid)
Hi Carlos,
As you pointed out, I wasn't using the Launch function correctly to pass parameters, however, after I fixed the parameter passing it didn't work with your syntax below. I couldn't launch another PowerApp from within an active PowerApp using just the AppId. What is launching the 2nd PowerApp is:
Launch("https://ms.web.powerapps.com/apps/76836bd7-f6fb-4362-b41b-8b3ab43c9f12","personId",thisPatid);
And, my launched PowerApp is happily getting the personId parameter onStart.
Thanks...Steve
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