Re: [flexcoders] Set Width of An Item Renderer Component Based on Visible Property of the Control - RESOLVED

[Angelo Anolin]

Was able to figure it out.

I set the width to a public function:

public function linkbutton_Width(isVisible:Boolean) :int
{
  if(isVisible)
  {
    return 100;
  }
  else
  {
    return 0;
  }
}

and in my MXML declaration:

width="{outerDocument. linkbutton_Width(this.visible)}"

Thanks.

angeLo



________________________________
From: Angelo Anolin <angelo_ano...@yahoo.com>
To: flexcoders@yahoogroups.com
Sent: Thu, 20 May, 2010 16:45:31
Subject: [flexcoders] Set Width of An Item Renderer Component Based on Visible 
Property of the Control

  
Hi FlexCoders,

I have declared an in-line item renderer in my datagrid:

<mx:DataGridColumn headerText=" Test" textAlign="center" sortable="false">
  <mx:itemRenderer>
    <mx:Component>
      <mx:HBox horizontalAlign= "center" >
        <mx:LinkButton id="lnk1" label="Link 1" visible={outerDocum 
ent.fncVisible( )}" width="{???} "/>
        <mx:LinkButton id="lnk1" label="Link 1" />
        <mx:LinkButton id="lnk1" label="Link 1" />
      </mx:HBox>
    </mx:Component>
  </mx:itemRenderer>
</mx:DataGridColumn>


as you may see, I am determining the visibility of the link button via a 
function.  What I want to achieve is that if the linkbutton contrl is 
invisible, the width should be 0, but if it is visible, then width should be 
100.

How do I pass in a function for the width of the control whether it's 
visibility is set to true or false?

Thanks.