why doesnt this work 2

richardko · Jun 20, 2006

<html>
<head>
<SCRIPT language="javascript" type="text/javascript">
function check()
{
var ret="";
for(var x=1;x<=2;x++)
{
var element="chektu"+x;
//alert("elas "+element);
if(document.getElementById(element).checked)
{
alert("insiede");
ret="no";
}
}
if(ret=="")
{
return true;
}else{
return false;
}
}
</script>
</head>
<body>
<form name="Test" action="<? $PHP_SELF ?>">
<input type="checkbox" id="chektu1">
<input type="submit" name="original" value="Send Original Items As indicated Above" onClick="return check();">
</form>
</body>
</html>

I dont want the form to be submitted if the button is checked but it does get submitted. Any idea why?
thanks

cLFlaVA · Jun 20, 2006

remove the call to your function in the submit button. instead, call the function from the form's onsubmit event:

Code:

<form ... onsubmit="return check();">

*cLFlaVA
----------------------------
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[URL unfurl="true"]http://www.coryarthus.com/[/url]

theniteowl · Jun 20, 2006

And one other small error. Your going to kick yourself for it.

You setup a for loop that goes from 1 to 2. This means the loop will execute once for 1 and a second time for 2.
The second time through will always evaluate false because there is no second checkbox so it will return true because of how your function is setup.

Drop the loop number to 1 or remove the loop entirely if there is only one box to test and it works.

Google, you're my hero!

richardko · Jun 20, 2006

thanks for the quick reply.

I was trying to avoid using onSubmit so I did not use the first option. Thank you cLFlaVA for the help though.
Thank you "theniteowl" for the reply. The loop correction worked good.

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why doesnt this work 2

richardko

Programmer

cLFlaVA

Programmer

theniteowl

Programmer

richardko

Programmer

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