Question

Related action

Forum|Forum|2 years ago
February 2, 2024
9 replies
1 view

I need to create "start" button as related action .when i click on start it should check the status for that particular request if status is "Approved" then the api call should happen .and request will end .But we click on start then process model should start (will not any interface after clicking start)

Is it possible to create this related action without creating an interface. If yes,then how the start button will know which ID form it needs to call.

U

ujjwalr0002

Hi, we can simply pass the identifier from the related action itself to the process variable.

S

sabat0002Author

can you please share the documentation link

stefanhelzle0001

Brainy

https://docs.appian.com/suite/help/23.4/record-actions.html

konduruc733182

Hello [mention:85ca6a2722324ac4b08c9eda0b5e1851:e9ed411860ed4f2ba0265705b8793d05]

You can create a related action for this and you don't need to have any User interface. But instead of checking for the Status after triggering the process you can avoid showing the Related action until the status is Approved.

Since its a related action you can always use the context from the record and pass it to the Parameter of your process model

S

sabat0002Author

Hi [mention:2df2892a4df7486f9775b4d2f0e405fd:e9ed411860ed4f2ba0265705b8793d05] ,

Yes, the related action will be visible once status is approved

I'm getting below error when i click on related action button

can you please help me with this

harshitb6843

I think there is a difference in the names of the variable created in the process and mentioned in the processParameters. The R for a variable name ending with request is small at one place and capitalized in another. Can you check if that is causing this error?

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