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Editorial Team
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Editorial Team
Asked: May 11, 20262026-05-11T18:13:14+00:00 2026-05-11T18:13:14+00:00

I have this HTML code for radios: <input type=’radio’ name=’a_27′ value=’Yes’ id=’a_27_0′ /> <input

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I have this HTML code for radios:

<input type='radio' name='a_27' value='Yes' id='a_27_0' />

<input type='radio' name='a_27' value='No' id='a_27_1' />

I’m trying to set the selected value of the radio using this code:

var field="a_" + this.id;
$('[name="' + field + '"]').val(this.value);
console.log("name is " + field + ", val is " + this.value);

However it doesn’t work, nothing happens when this runs. Here’s the output from Firebug’s console which occurs after the 3rd line:

name is a_27, val is Yes

Any ideas?

I would prefer a method which would also work on <select>s, so I wouldn’t need to write additional/seperate code for radios and selects.

Edit: A weird problem I’ve noticed that although my html code gives a different value (yes/no), in firebug it shows both radios as having the value ‘yes’. If I select no and click save, the javascript function still receives ‘yes’ instead of no. Am I doing something wrong?

Edit 2: The full function:

function processMultiOptAnswers()
{
    $.each(multiOpts,function()
        {
            var field="a_" + this.id;
            console.log("name is " + field + ", val is " + this.value);

            $('[name="' + field + '"]').val(this.value);
        }
    );
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-11T18:13:15+00:00Added an answer on May 11, 2026 at 6:13 pm

    your log should be if this.value is different.

    $('[name="' + field + '"]').val(this.value);
    console.log("name is " + field + ", val is " + $('[name="' + field + '"]').val());
    

    To make it selected

    $('[name="' + field + '"]').attr("checked", "checked");
    

    I haven’t tested this, but you might have to remove that attribute from the other ones.

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