Show url link from file name

[baba_sawane]

Hi everyone !

I have multiples files that can be updated/added in a column. What I want for the user is to be able to see the link of the file. 

The files are generated from an automation and i'm using a column named [ReportName] for files names. My actual goal is to generate a link from a formula, that will change everytime my [ReportName] update.

This is the formula i'm using to create the link : 

When i open the link in the spreadsheet, i'm getting a "Bad request" error

Anyone have tips to fix this please ?

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CONCATENATE(‘https://www.appsheet.com/template/gettablefileurl?AppName=’,

ENCODEURL(CONTEXT(AppName)), ‘&tableName=’, ENCODEURL(CONTEXT(Table)), ‘&fileName=’,

ENCODEURL([ReportName]))

[Suvrutt_Gurjar]

You could elaborate your requirement. Does the URL never show? Or whenever you change the value of  [ReportName] the URL breaks? If so,  any specific reason we may know you wish to change the report name column ( and in turn the file name)  to change once file URL is created?

[baba_sawane]

Hi. 

To be more precise, some reports needs to be created within my app, and the reports sometimes needs to be updated. 

And because downloading files can take a lot of data, I want them to be able to see the created reports in the cloud OR be able to download.

With my actual formula, they can see the link, but the result is a bad request error. 

I don't know wether this is the right formula to use to reach my goal.

Thank you for your answer !

[baba_sawane · Accepted solution]

Sometimes i feel dumb 😓 

CONCATENATE(‘https://www.appsheet.com/template/gettablefileurl?appName=’, ENCODEURL(CONTEXT(AppName)), ‘&tableName=’, ENCODEURL(CONTEXT(Table)), ‘&fileName=’, ENCODEURL([Report_File]))