'jQuery count checked checkbox in next element
My sample is counting every single checkbox on a page. How to count the checked checkboxes on next/sibling element with jQuery?
PS: I want this to work on multiple forms on a single page without calling the element(form) ID.
$(document).ready(function() {
function countChecked() {
var tCount = $("input:checked").length;
$(".totalchecked").text( tCount + ' selected');
}
countChecked();
var tCheck = $(".totalchecked");
$(tCheck).next().find(":checkbox").click(countChecked);
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div>form 1:
<div class="totalchecked">0 selected</div>
<div>
<input type="checkbox" class="class" name="1">
<input type="checkbox" class="class" name="2">
<input type="checkbox" class="class" name="3">
</div>
</div>
<div>form 2:
<div class="totalchecked">0 selected</div>
<div>
<input type="checkbox" class="class" name="4">
<input type="checkbox" class="class" name="5">
<input type="checkbox" class="class" name="6">
</div>
</div>
Solution 1:[1]
Check this working sample:
function checkTotalCheckedBoxes(){
$("div" ).each(function( ) {
if(this.children.length === 3){
var checked = 0;
$(this.children).each(function() {
this.checked ? checked++ : null;
$("input[name="+this.name+"]").parent().siblings().closest('.totalchecked')[0].innerText = (checked) + " selected";
});
}
});
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div>form 1:
<div class="totalchecked">0 selected</div>
<div>
<input type="checkbox" class="class" name="1" onclick="checkTotalCheckedBoxes()">
<input type="checkbox" class="class" name="2" onclick="checkTotalCheckedBoxes()">
<input type="checkbox" class="class" name="3" onclick="checkTotalCheckedBoxes()">
</div>
</div>
<div>form 2:
<div class="totalchecked">0 selected</div>
<div>
<input type="checkbox" class="class" name="4" onclick="checkTotalCheckedBoxes()">
<input type="checkbox" class="class" name="5" onclick="checkTotalCheckedBoxes()">
<input type="checkbox" class="class" name="6" onclick="checkTotalCheckedBoxes()">
</div>
</div>
Solution 2:[2]
$('input[type=checkbox]').change(function() {
var siblingsChecked = [];
$(this).siblings().each(function(el){
if($(this).is(':checked')) {
siblingsChecked.push(el);
}
});
console.log(siblingsChecked.length);
});
Hope this helps you !
Solution 3:[3]
Here is my own solution and I hope it helps someone.
$( document ).ready(function() {
var $SelectCount = $("input:checkbox")
$SelectCount.change(function(){
var countChecked = $(this).parent().find("input:checkbox").filter(':checked').length;
$(this).parent().siblings(".totalchecked").text(countChecked + " Selected");
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div>form 1:
<div class="totalchecked">0 selected</div>
<div>
<input type="checkbox" class="class" name="1">
<input type="checkbox" class="class" name="2">
<input type="checkbox" class="class" name="3">
</div>
</div>
<div>form 2:
<div class="totalchecked">0 selected</div>
<div>
<input type="checkbox" class="class" name="4">
<input type="checkbox" class="class" name="5">
<input type="checkbox" class="class" name="6">
</div>
</div>
Solution 4:[4]
$(document).ready(function() {
function countChecked() {
var tCount = $("input:checked").length;
$(".totalchecked").text( tCount + ' selected');
}
countChecked();
var tCheck = $(".totalchecked");
$(tCheck).next().find(":checkbox").click(countChecked);
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div>form 1:
<div class="totalchecked">0 selected</div>
<div>
<input type="checkbox" class="class" name="1">
<input type="checkbox" class="class" name="2">
<input type="checkbox" class="class" name="3">
</div>
</div>
<div>form 2:
<div class="totalchecked">0 selected</div>
<div>
<input type="checkbox" class="class" name="4">
<input type="checkbox" class="class" name="5">
<input type="checkbox" class="class" name="6">
</div>
</div>
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | Saurabh Yadav |
| Solution 3 | CocoSkin |
| Solution 4 | abid729 |

