PowerApps Launch a SharePoint link

[SumanthDundi7]

Hi,

I am using a Gallery to fetch and display SharePoint list items on a screen. In the background, I am creating folders in a document library based on current list item field values like Employee Name, Employee ID, Created. I am using Power Automate to create folders in a SharePoint Library with attachments moved from current SharePoint list to SharePoint Library.

There is a need to launch a link using a HTML Text field or a button on PowerApps form with

https://test.sharepoint.com/sites/testsite/employeedetails/ThisItem.EmployeeName_ThisItem.EmployeeID_ThisItem.Created 

employee details in the link above is the root folder.   

How can I achieve this? Please help

@WarrenBelz @CNT @BCLS776 @Pstork1 

[KvB1]

Is the filename of the file created in the sharepoint library defined by variables you already have access to through the app? If so, you could just concatenate the URL and use Launch(

Otherwise you'd need to filter the library by folderpath, using the folderpath from the powerapp variables you use for your automate.

[SumanthDundi7]

@KvB1 I have access to the variables. As per my post, could you please help me the formula? As I said, I need to add ThisItem.EmployeeID_ThisItem.EmployeeName_ThisItem.Created to the URL at end. How can I achieve it?

[KvB1]

Launch(https://test.sharepoint.com/sites/sitename/libraryname/subfolder1/subfolder2/filenamewithextension)

You can also use a gallery filtering the library by folderpath

Filter(Libraryname,folderpath='subfolder1/subfolder2/')

Then add a control with Launch(ThisItem.linktoitem) within the gallery

[SumanthDundi7]

@KvB1, Looks like you didnt get my question clearly. Its simple. 

Basically, if it is static url I can simply use Launch("https://www.google.com"). But if it is suppose I will need to launch a url with details getting from App i.e., https://testsite.sharepoint.com/sites/testsite/employeedetails/ThisItem.EmployeeName_ThisItem.EmployeeID_ThisItem.Created  

Please note If I go paste this url in Launch function, "ThisItem" is considered as a part of URL. 

@WarrenBelz , Is my question clear? Help  

[KvB1 · Accepted solution]

Then you would need to take the dynamic aspect of your URL out of the parenthesis so they are fed by info from the app. 

Launch("https://test.sharepoint.com/sites/sitename/libraryname/"&ThisItem.EmployeeName&"_"&ThisItem.Created"/"&ThisItem.filename) 

Or something like that

[BCLS776 · Accepted solution]

I recommend creating a string using Concatenate() and then using Launch() to open it in a new browser window. Eg:

Launch(
    Concatenate(
        "https://test.sharepoint.com/sites/testsite/employeedetails/",
        ThisItem.EmployeeName,
        "_",
        ThisItem.EmployeeID,
        "_",
        ThisItem.Created
    )
)

However, implement some checks on those values you are passing to the string as URLs cannot handle some special characters. The Employee Name probably has a space character in some values and that won't work in a URL.

[SumanthDundi7]

@BCLS776 @KvB1 .. Thank you. Both worked. 

[Mordechay]

@BCLS776 

I would appreciate help with a similar problem.

I want to give the user a link to a PDF file that is in the Sharepoint folder, so far the solution you gave works well.
The thing is that sometimes I want to update the file that is in the folder and upload a more recent version of it, I tried to upload the new file with the same name as the old file and it doesn't accept it well.

Have you ever encountered such a case?