Opening Child Visualforce page based on parent field value

Question

I have a parent object Contract__c and a child record Risk__c. The parent object has a picklist: Classification: high, medium, low

Risk__c has two record types high and low. The fields for these record types are different, and I have created a visualforce page for each record type, because I need to modify the save&new functionality.

When the user enters high on Contract__c, the user needs to click on the related list button Create high risk, which opens the high risk visualforce page. I have another button called Create medium risk, which the user should click if medium was selected in the classification picklist.

The problem is, I want to only display the high button if high was selected in the picklist, and I want to be able to display the medium button, only if medium was selected. Is that possible?

Otherwise I would be happy to just have one button, which has some logic to decide which page to open. But I wouldn't know how to do that.

Tia.

Answer 1

• You can create one custom field on your Custom object "Contract__c" which holds the name of visualforce page lets call it as "Page__c".
• The value of Page__c will change according to value of Classification: high, medium, low
  e.g.

  if
      Classification == high
  then 
      page__c = 'HighVFPage';
  
  

• This page__c would be formula field which contains formula like

  CASE(Classification, "high", "HighVFPage", "medium", "MediumVFPage","LowVFPage")

• Create List button on child object("Risk__c") which contains below code.
  

  var pageValue = "{!Contract__c.Page__c}"; 
  var url = "/apex/"+pageValue; 
  parent.window.location = url; 
  

I hope this will work for you.