jQuery.POST - pass another parameter with Form.Serialize() - Asp.net MVC 3

Question

As I am working on Asp.Net MVC Application, in my application I am using jQuery.POST method to submit a form.

e.g.

jQuery.post('/Product/Save', jQuery(document.forms[0]).serialize(), 
       function (data) { alert('Product Added Successfully.); }
);

In above code snippet, I want to pass another parameter.. let's say.. ProductID.

So, the idea is, I want to pass both jQuery(document.forms[0]).serialize() and ProductID variable in jQuery.POST method, so I can get both Form and ProductID in my controller's action method.

Could anybody please let me know that I would I do this?

Thanks in Advance.

Community · Accepted Answer · 2017-05-23 10:33:11Z

13

You could use the following plugin to serialize the form into a JSON object and add another parameters:

$.fn.serializeObject = function()
{
    var o = {};
    var a = this.serializeArray();
    $.each(a, function() {
        if (o[this.name]) {
            if (!o[this.name].push) {
                o[this.name] = [o[this.name]];
            }
            o[this.name].push(this.value || '');
        } else {
            o[this.name] = this.value || '';
        }
    });
    return o;
};

like this:

var data = $('form').serializeObject();
data['ProductId'] = '123';
$.post('<%= Url.Action("Save", "Product") %>', data, function (data) { 
    alert('Product Added Successfully.'); 
});

edited May 23, 2017 at 10:33

CommunityBot

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answered Jan 24, 2011 at 17:38

Darin Dimitrov

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Comments

Jeff S · Accepted Answer · 2011-01-24 17:31:10Z

0

What about using the Url.Action method to build your calling URL? You can include the additional data you need that way.

<%: Url.Action('Save', 'Product', new { "ProductID", your product id }) %>

This would replace the hard-coded URL in your jQuery.post method call.

answered Jan 24, 2011 at 17:31

Jeff S

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Comments

WorldIsRound · Accepted Answer · 2011-01-24 17:43:52Z

0

How about:

 var url = '<%= Html.BuildUrlFromExpression<MyController>(c => c.Save(":productId")) %>';

You could later replace the url with the actual value in your post URL.

 url.replace(':productId', $productId);

answered Jan 24, 2011 at 17:43

WorldIsRound

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