Reputation:
These are Dynamic dependent select controls. The sets of values of (1-3, 4-6, 7-9) determine the use hide/show divs function. The problem is the function i have only hide/show depending on the div id. How can i make the function hide/show div depended on the values(1-3, 4-6, 7-9) found in the selectbox?
Jquery
$('#select').change(function() {
$('#sub1, #sub2, #sub3').hide();
$('#sub' + $(this).find('option:selected').attr('id')).show();
});
Html Setup
<html>
<select size="6" id="category">
<option value="">categories 1-3 </option>
<option value="">----</option>
<option value="">----</option>
</select>
<div id="sub1" style="display:none">
<select name="subc1" size="6">
<option value="1">subcategories 4-6</option>
<option value="2">---</option>
<option value="3">---</option>
</select>
</div>
<div id="sub2" style="display:none">
<select name="subc2" size="6">
<option value="4">subcategories 7-9</option>
<option value="5">----</option>
<option value="6">----</option>
</select>
</div>
<div id="sub3" style="display:none">
<select name="subc3" size="6">
<option value="7">End</option>
<option value="8">----</option>
<option value="9">----</option>
</select>
</div>
</html>
Upvotes: 0
Views: 3726
Reputation: 87073
You nee to change minor change in your category select
<select size="6" id="category">
<option value="1">categories 1-3 </option>
<option value="2">----</option>
<option value="3">----</option>
</select>
$('select#category').on('change', function() {
$('div[id=^sub]:visible').hide();
$('div#sub' + this.value).show();
});
can also use .change()
instead of .on('change')
.
$('select#category').change(function() {
$('div[id=^sub]:visible').hide();
$('div#sub' + this.value).show();
});
NOTE:
$('div[id=^sub]:visible')
will point to all div
s that have id
start with sub
and visible
.
You are trying with $('#select')
which need to be $('#category')
or $('select#category')
.
According to your comment:
Complete solution will look like following:
function isOnlyDashed(text) {
return text.replace(/-/g, '').length === 0;
}
$('select#category').change(function() {
var text = $('option:selected', this).text();
if (!isOnlyDashed(text)) {
$('div[id=^sub]:visible').hide();
$('div#sub' + this.value).show();
}
});
$('select[name^=subc]').change(function() {
var text = $('option:selected', this).text();
if (!isOnlyDashed(text)) {
$(this).parent() // jump to parent div
.next('div[id^=sub]:hidden') // go to next hidden div
.show();
}
});
Upvotes: 0
Reputation: 19242
select the value from drop down change function and do the operation depends on the value of drop down, following is the sample code
$(function() // Shorthand for $(document).ready(function() {
$('select').change(function() {
if($(this).val() == 1)
{
$('#sub1').hide();
$('#sub2').show();
}
});
});
Upvotes: 1
Reputation: 34107
working demo http://jsfiddle.net/FwBb2/1/
I have made minor changes in your Jquery code as well as added value in your first select drop-down list which was missing rest hope this helps.
Please lemme know if I missed anything! B-)
code
$('select').change(function() {
$('#sub1, #sub2, #sub3').hide();
$('#sub' + $(this).val()).show();
});
HTML
<html>
<select id="category">
<option value="1">categories 1-3 </option>
<option value="2">----</option>
<option value="3">----</option>
</select>
<div id="sub1" style="display:none">
<select name="subc1" size="6">
<option value="1">subcategories 4-6</option>
<option value="2">---</option>
<option value="3">---</option>
</select>
</div>
<div id="sub2" style="display:none">
<select name="subc2" size="6">
<option value="4">subcategories 7-9</option>
<option value="5">----</option>
<option value="6">----</option>
</select>
</div>
<div id="sub3" style="display:none">
<select name="subc3" size="6">
<option value="7">End</option>
<option value="8">----</option>
<option value="9">----</option>
</select>
</div>
</html>
Upvotes: 0
Reputation: 34909
The problem with your code is that you have an incorrect selector.
$('#select')...
should be
$('select')...
Also, this part is wrong
$('#sub' + $(this).find('option:selected').attr('id')).show();
Replace it with this
$('#sub' + $(this).val()).show();
Finally, having different elements with the same name/id "sub1" is probably a bad idea, though technically not illegal. It is certainly confusing.
Upvotes: 0