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How to show section on VF page on button click
I have a requirement to show a section only on button click. here is the code I have tried, but no luck. can someone tell me what is the error in the below code
VF code :
<apex:page controller="testRenderCtrl">
<apex:form>
<apex:commandButton value="Submit" action="{!submit}" reRender="panelId"/>
<apex:outputPanel rendered = "{!flag}" id = "panelId">
<apex:pageblock >
<apex:pageblockSection title = "Contact Information">
<apex:inputField value="{!con.firstname}"/>
<apex:inputField value="{!con.lastname}"/>
</apex:pageblockSection>
</apex:pageblock>
</apex:outputPanel>
</apex:form>
</apex:page>
Controller code
public class testRenderCtrl {
public Contact con {get;set;}
public boolean flag = false;
public boolean getFlag(){
return flag;
}
public void setFlag(boolean flg){
flag = flg;
}
public void submit(){
System.debug('****');
flag = true;
}
}
Thanks,
YP
Change your submit() method to the above, and it should work. action="" parameters require a method that takes no parameters and returns a pagereference. You don't even need the rerender on the commandButton.
Try this using javascript on the button click. Sample code is given below :
<apex:commandbutton onclick = "showsection()">
function showsection()
{
document.getElementById('{!$Component.pageid:formid:pageblocksectionid}').style.display = 'block';
}
Hope this helps.