Question

I have used this function to submit to any url or ‘self’,
with or without a querystring, many times without any problems.

function submitu(url, q) {
 var frm = document.<?php echo $formname ?>;
 if (url == '') {url = '<?php echo $thispage?>'; }
 frm.action = url+q; frm.submit(); }

If I try to move the PHP vars outside the function, as below, it stops working (frm undefined error)

var thispage = '<?php echo $thispage?>';
var frm = document.<?php echo $formname?>;
function submitu(url, q) {
 if (url == '') {url = thispage;}
 frm.action = url+q; frm.submit();}

I also tried var frm = document.forms[”];

I don’t have any other conflicting javascript,

(1). Can anyone tell me why this is not working?
(2). And why the first method also fails if the function is placed
inside and at the top of the jquery $(function() {…..} ready function?

Many thanks

Answer 1

It fails because when outside the function, var frm = document.formname;
will be run immediately when the page loads, i.e. before the form element has actually been constructed, so you get ‘undefined’. When inside the function, it is only run when the function is run, by which time the DOM is complete and it can find it.