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Chandu
Helper III
Helper III

How to pass Datacardvalue along with URL in Launch() in Powerapps

In Powerapps, I am having a Button(View Report). In Button Onselect : Launch("http://Server/reports/report/folder/reportname"). With this formula I am getting all the records in the report.

powerapp.PNG

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

But, I should display records based on TEST_ID. Now along with that URL, I need to pass Test_ID Datacardvalue. Based on that Test_ID, the SSRS report must be filter in browser. In below, SSRS Report, the table must be filter according to the Test_ID passing from Powerapps.

SSRS.PNG

For this, In Button Onselect : Launch("http://Server/reports/report/folder/MFTestTaskMap" && Test_IDDataCardValue). This formula is not working.

 

Can anyne help me how to solve this.

1 ACCEPTED SOLUTION

Accepted Solutions
v-siky-msft
Community Support
Community Support

Hi @Chandu

 

 

Refer to this article, to set a report parameter within a URL, use the following syntax: 

parameter=value

Hence, define 'Test_ID' as a parameter in the report, then use the following URL:

Launch("http://Server/reports/report/folder/reportname&Test_ID="&Test_IDDataCardValue.Text)

 

Hope this helps.

Sik 

View solution in original post

2 REPLIES 2
timl
Super User
Super User

Hi @Chandu 

I would try this...

Launch("http://Server/reports/report/folder/MFTestTaskMap" & Test_IDDataCardValue.Text)
v-siky-msft
Community Support
Community Support

Hi @Chandu

 

 

Refer to this article, to set a report parameter within a URL, use the following syntax: 

parameter=value

Hence, define 'Test_ID' as a parameter in the report, then use the following URL:

Launch("http://Server/reports/report/folder/reportname&Test_ID="&Test_IDDataCardValue.Text)

 

Hope this helps.

Sik 

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