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disabled and enadbled

Posted: Tue Apr 24, 2012 10:28 am
by Lphp
1)the show and hidden <div> work, but the disabled and enabled is not work
<script LANGUAGE="JavaScript">
function Show(){

if((document.fac.offer_type[2].checked==true)||(document.fac.offer_type[3].checked==true) ){
document.fac.getElementById("first").disabled='true'
document.fac.getElementById("sec").disabled='true'
document.fac.getElementById("four").disabled='true'
document.getElementById("five").disabled=true
document.getElementById("six").disabled=true
document.getElementById("eig").disabled=true
document.getElementById("nine").disabled=true

}
if(document.fac.offer_type[1].checked==true) {
document.getElementById("first").style.display='block';
document.getElementById("sec").style.display='block';
document.getElementById("four").disabled=true
document.getElementById("nine").disabled=true
document.getElementById("five").style.display='block';
document.getElementById("six").disabled=true
document.getElementById("eig").style.display='block';

}

if(document.fac.offer_type[0].checked==true) {
document.getElementById("first").style.display='block';
document.getElementById("sec").style.display='block';
document.getElementById("four").style.display='block';
document.getElementById("five").style.display='block';
document.getElementById("six").style.display='block';
document.getElementById("eig").style.display='block';
document.getElementById("nine").style.display='block';



}
}
</script>
2) is that possible to change <div id="first" #if(($flag1 =="unsuccessful")||($flag1 =="deferred"))style="display:none"#{else}style="display:block"#end> to disabled and enabled? :roll: 8)

Re: disabled and enadbled

Posted: Mon Apr 30, 2012 3:42 am
by Gopesh
Hi,Can u please explain in detail about the problem?