Open specific record producer when click on UI action button

[Akella Lalitha]

I want to create a UI action button in agent workspace and native UI, when I click on the UI action button it will open the record producer portal view and when we submit the request, the new case should be added as a child to the parent case.

Please help me with your suggestions.

Thank you in advance!

[Aniket Bhanse]

@Akella Lalitha 

Follow this step by step in order to achieve your requirement:

1) Create an UI Action in global view. Make sure you are using this code "action.setRedirectURL(your record producer portal URL);" Pass the current case's sys_id in the URL as a parameter so that you can use it while considering Parent-Child combination.

2) Create an UI Action in the workspace view. Make sure you are using this code "action.setRedirectURL(your record producer portal URL);" Pass the current case's sys_id in the URL as a parameter so that you can use it while considering Parent-Child combination.

3) Create an onSubmit Client script and take out the parent sys_id from the URL from getURL method. Now when the case is submitted from the Portal, the onSubmit script will execute and with the help of current sysId (child) and parent sysId, you can call a script include in the onSubmit script to map the parent-child combination.

Let me know if this approach works for you.

[Akella Lalitha]

Could you please help me with the exact code and I need to create this UI action button in R&D scope as I am able to select the table in the R&D scope not in the global

[Akella Lalitha]

Could you please help me how to redirect to the record producer portal view when I click on the UI action button.