CODEWITHSUNDEEP
javascriptc#htmlajaxasp.net-mvc
DinupaChamin
DinupaChamin

Reputation: 49

How do I get the 2nd defined radio button value?

This is my html code.

<form id="item">
<div class="form-group col-lg-3 col-md-3 col-sm-3 col-sx-3">
     <label for="itemtype">Hotel</label>
     <input type="radio" class="radio" id="itemtype" name="optradio" value="H" checked>
     <label for="itemtype">AirPort</label>
     <input type="radio" class="radio" id="itemtype" name="optradio" value="A">
</div>
</form>

And this is my jquery.

function save() {
    debugger

    $.ajax({
        type: "POST",
        data: { "item": { "itemid": $("#itemid").val(),  "itemtype": $('#itemtype').val() }},
        url: "@Url.Action("Save", "Item")",
        cache:false,
        success: function (data) {
            debugger
            if (data.isSuccess) {
                alert("Saved Succesfully!");
                window.location.href = "@Url.Action("Index", "Item")";
            }
        }, error: function (err) {
            alert("Saved Failed!")
            debugger
        }
});
}

All the time when I trying to save the item type call "A", it never happen. Always this query getting radio button value as "H". I need to pass the 2nd value "A".

Upvotes: 0

Views: 141

Answers (2)

Tohm
Tohm

Reputation: 305

$('input[name=optradio]:checked').val();

it is not recommended to have two identical ids on one page

other solution using class name:

<form id="item">
<div class="form-group col-lg-3 col-md-3 col-sm-3 col-sx-3">
     <label for="itemtype">Hotel</label>
     <input type="radio" class="radio itemtype" name="optradio" value="H" checked>
     <label for="itemtype">AirPort</label>
     <input type="radio" class="radio "itemtype" name="optradio" value="A">
</div>
</form>

$('.itemtype:checked').val();

In jquery the best way to send data form a formular is to use serialize() or serializeArray() methods.

Some exemple : serialize-form-data-to-json

Upvotes: 0

pbachman
pbachman

Reputation: 1059

You can get the correct value by using the name, please remove the duplicate id

$('input[name=optradio]:checked', '#item').val()

Upvotes: 2

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