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Launch(fileurl) giving error

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Posted on by SM-20092000-0

I have the following code that calls a Power Automate flow get create a .csv file and then return the URL. The powerautomate generates the uRL and sends it back to Powerapps, but i can't get the Launch(fileurl) to work

ClearCollect(colfilteredResults,galResults.AllItems);
Set(
varJSON,
JSON(
colfilteredResults,
JSONFormat.IndentFour & JSONFormat.IgnoreUnsupportedTypes & JSONFormat.IgnoreBinaryData
)
);
Set(fileurl,
'ITS-TO_Onboarding_Checklist_Export_to_CSV'.Run(varJSON));
Launch(fileurl);
 
I get this message:

Categories:

Power Apps Pro Dev & ISV

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Verified answer

Michael E. Gernaey 53,433 Super User 2025 Season 2 on at

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Hi

the issue is you are doing Run(varJSon) and on the other side you are probably using a Respond to Canvas App / Flow

The issue is the control passes back an Object of values (record), and you aren't grabbing the actual VALUE of your output

So if you response is NewUrl (parameter)

you need to say

Set(fileUrl, Flow.Run(parameters).NewUrl)

or with what you have say launch(fileUrl.NewUrl)

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Verified answer

Hack-7 226 Moderator on at

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What I have always done is something like this. It helps me visualize and breakdown the steps in the process.

Set(fileurl, nameOfFlow.Run(varJSON, User().FullName)); // Opens New File Launch(fileurl.filelink);

I hope this helps.

Cheers

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SM-20092000-0 16 on at

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So i was finally able to get it working with the following:

Launch(fileurl.fileurl)

Now it launches that url in a new tab. How can i get it to download to my downloads folder and not open in a new tab?

thanks for your help.

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