Hi Ken,
There are several methods to accomplish this task. you can use onclick or
use javascript with onclick.
*Example: 1*
<form>
<input name="Block Supervisors" value=" "
onclick=window.open("testFileOne.php?id=<? echo $staff_id;
?>","Homepage","resizable=no,status=yes,scrollbars=yes,height=970,width=945,menubar=yes,addressbar=no")
type="button" />
</form>
*Example :2*
<script type="text/javascript" language="javascript" >
function popup() {
window.open("legend.php","Homepage","resizable=no,status=no,scrollbars=no,height=320,width=230,left=490,top=320,menubar=no,addressbar=no");
}
</script>
*1.* <a onclick="popup()" title="You will see a popup window">Click Me</a>
*2.*
<form>
<input name="Block Supervisors" value=" " onclick="popup()" type="button"
/>
</form>
You can always customize according to you need.
Hope i have helped you to solve the problem
Have a lovely day
Cheers
Mohamed Sathik.
On Thu, Dec 17, 2009 at 9:07 AM, kdubya <kenwin...@winanstech.com> wrote:
> I have an application where I want a "printer friendly" rendering. My
> print() action has it's own layout. The action is called from a form
> (method=POST). I need all the input from the form to generate the
> stuff to be printed. How do I get the resulting stuff to be rendered
> in a popup window instead of the existing window? Is it something I
> put in the layout? (I'm not really good with javascript stuff) Is
> there a way to specify this using redirect()? I have looked around
> quite a bit and I'm drawing a blank. (Note: I do know how to open a
> popup if I have a link like <a href="...>. I just can't figure out how
> to do it from an action that processes the data from a form)
>
> Thanks,
> Ken
>
> Check out the new CakePHP Questions site http://cakeqs.org and help others
> with their CakePHP related questions.
>
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Check out the new CakePHP Questions site http://cakeqs.org and help others with
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