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Ajax post serialize() does not include button name and value

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My $.ajax() is not serializing the button name and value.

I have a very simple form. It has a button and a textbox.

<form action="/MyController/MyAction" data-ajax-method="post" 
  data-ajax-mode="replace" data-ajax-target="#mypartial" id="myform" 
  method="post">
        <button type="submit" class="positive" name="button" value="click1">
            <img src="/Id/Images/Icons/16/enabled/tick.png" title="click1">
            Click
        </button>
        <input id="txtBlah" name="txtBlah" type="text" value="hello">
    </div>
</form>

When i call $(this).serialize(), the textbox is included in the string but not the button.

Debug.Log($(this).attr('id')); //== 'myform'
Debug.Log("data: " + $(this).serialize()); //== data: txtBlah=hello

I noted during research that with other questions the main reason was a missing name element on the button. I have a name element.

I have also tried making a very simple <input type="submit" name="mysubmit" /> which did not change anything.

solution i used

var buttonSubmit = (function (e)
{
    e.preventDefault();
    var form = $(this).closest('form');
    form.attr('data-button-name', $(this).attr('value'));
    form.closest('form').submit();
});
like image 496
Valamas Avatar asked Mar 26 '12 04:03

Valamas


2 Answers

jQuery's serialize() is pretty explicit about NOT encoding buttons or submit inputs, because they aren't considered to be "successful controls". This is because the serialize() method has no way of knowing what button (if any!) was clicked.

I managed to get around the problem by catching the button click, serializing the form, and then tacking on the encoded name and value of the clicked button to the result.

$("button.positive").click(function (evt) {
    evt.preventDefault();

    var button = $(evt.target);                 
    var result = button.parents('form').serialize() 
        + '&' 
        + encodeURI(button.attr('name'))
        + '='
        + encodeURI(button.attr('value'))
    ;

    console.log(result);
});
like image 87
slashingweapon Avatar answered Sep 20 '22 06:09

slashingweapon


Here's a catch-all solution that will look for an input in the button's containing form. If it exists, it'll set the value, otherwise it will create a hidden input and set its value. This can also be useful if you're not wanting to submit the form immediately.

$(document).on('click', '[name][value]:button', function(evt){
    var $button = $(evt.currentTarget),
        $input = $button.closest('form').find('input[name="'+$button.attr('name')+'"]');
    if(!$input.length){
        $input = $('<input>', {
            type:'hidden',
            name:$button.attr('name')
        });
        $input.insertAfter($button);
    }
    $input.val($button.val());
});
like image 24
stephenr85 Avatar answered Sep 19 '22 06:09

stephenr85