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Home/ Questions/Q 9192757
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T21:00:11+00:00 2026-06-17T21:00:11+00:00

How can i check using Jquery that from every uniquely named checkboxes group atleast

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How can i check using Jquery that from every uniquely named checkboxes group atleast one option is checked ?
These checkboxes are generated dynamically and total numbers of uniquely named checkboxes will vary…

<input type="checkbox" class="mcqcb" name="ans_4[]" value="A">
<input type="checkbox" class="mcqcb" name="ans_4[]" value="B">
<input type="checkbox" class="mcqcb" name="ans_4[]" value="C">
<input type="checkbox" class="mcqcb" name="ans_4[]" value="D">

<input type="checkbox" class="mcqcb" name="ans_5[]" value="A">
<input type="checkbox" class="mcqcb" name="ans_5[]" value="B">
<input type="checkbox" class="mcqcb" name="ans_5[]" value="C">
<input type="checkbox" class="mcqcb" name="ans_5[]" value="D">

<input type="checkbox" class="mcqcb" name="ans_17[]" value="A">
<input type="checkbox" class="mcqcb" name="ans_17[]" value="B">
<input type="checkbox" class="mcqcb" name="ans_17[]" value="C">
<input type="checkbox" class="mcqcb" name="ans_17[]" value="D">

So far, I have tried this:

$('input:checkbox[class=mcqcb][name*="ans_*"]').not(':checked').length
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-17T21:00:13+00:00Added an answer on June 17, 2026 at 9:00 pm

    It seems like the only way to do this with jQuery is to use an array to store the grouping data for the checkboxes:

    JS:

    var detectionArry={};
    $('.mcqcb').each(function(){
        if (!detectionArry[$(this).attr("name")]){
        detectionArry[$(this).attr("name")]=$(this).prop("checked");
        }
    });
    for (var i in detectionArry){
        if (!detectionArry[i]){
         alert(i+" is not completed!");   
        }
    }
    

    Demo | Source

    Edit:
    To keep it within checkboxes only just replace the selector ($('.mcqcb')) with the selector that you have provided: $('input:checkbox[class=mcqcb]')

    Demo | Source

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