I have a problem with my checkbox. If I select a value and validate, I can't do that a second time, it's not working. Looking at other questions, I saw that I may have to use the prop() function with the "checked" property, which I did, but it's still not working.
Here is my website : http://nts.nuvision.co.za/index.php/training/
Here is my javascript :
$('#one').hide();
$('#two').hide();
$('#three').hide();
$('#show').click(function() {
$('#pgc-224-0-1').css("margin-top", "0px");
switch(document.getElementById('trainingcourses').value) {
case 'one':
$('#one').show();
$('#two').hide();
$('#three').hide();
$("#trainingcourses").prop('checked', false);
break;
case 'two':
$('#two').show();
$('#one').hide();
$('#three').hide();
$("#trainingcourses").prop('checked', false);
break;
case 'three':
$('#three').show();
$('#two').hide();
$('#one').hide();
$("#trainingcourses").prop('checked', false);
break;
}
});
And here my HTML :
<form id="trainingform">
<select id="trainingcourses" style="font-size: 12pt;">
<option style="font-size: 12pt;" value="one">Sugar and Suite CRM End User Training</option>
<option style="font-size: 12pt;" value="two">Sugar & Suite CRM Advanced End User Training</option>
<option style="font-size: 12pt;" value="three">Sugar and Suite CRM Administrator : Getting Started</option>
</select><input id="show" style="margin-top: -25px; margin-left: 20px;" name="show" type="submit" value="Show more information" onclick="location.href='http://nts.nuvision.co.za/index.php/training/#' + document.getElementById('trainingcourses').value; return false;"/></form>
<div id="one"> (...) </div>
<div id="two"> (...) </div>
<div id="three"> (...) </div>
You are setting checked
property to select box which is not valid. remove below lines from your switch case statement and it should work.
$("#trainingcourses").prop('checked', false);
Also not sure if you need a submit button here or not. If you need simple button which should not submit the page then change show button type="submit"
to type="button"
.
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