文本上的新 Visualforce 页面 onClick

Question

I am new to Apex. I am trying to create an object with 2 fields: Subject(Text) and Description(Rich Text Area). This should be displayed in a table format. The description should display only the first 50 characters, and onclick on the description, it should open a new Visualforce page to display the entire content.

<apex:page controller="Notes_Controller" >
<apex:form >
  <apex:pageBlock >
    <apex:pageBlockTable value="{!recordsList}" var="n">
        <apex:column value="{!n.Subject__c }" headerValue="Subject"/>

        <apex:column styleClass="slds-truncate" headerValue="Description"> 
            <apex:facet name="header">Description</apex:facet>

            <apex:outputLink value="{!n.Description__c}" target="_blank">{!LEFT(n.Description__c,50)}
            </apex:outputLink>
           </apex:column>
    </apex:pageBlockTable>
 </apex:pageBlock>
</apex:form>

So far I have tried the above, but I am unable to get the entire description to open up in a new tab. Any ideas?

Can this be done using a formula field? I tried using outputlink where value = description and label is the subject, such that click on the subject name will take you to the description. But I get "Url does not exist".

Any help is appreciated.

Answer 1

In the apex:outputLink you can try the value attribute specify URLFOR with the action and Id.

Something like

<apex:outputLink value="{!URLFOR($Action.n.Description__c)}" target="_blank">{!LEFT(n.Description__c,50)}
            </apex:outputLink>

Answer 2

The apex:outputlink value should be a url. URLFOR(Description__c) will not work since Description is a text field.

Try <apex:outputLink value="/{.n.id}" target="_blank">{,LEFT(n:Description__c,50)}</apex:outputLink>

See apex:outputLink