Question

43

how to submit form as JSON object

rated 0 times [ 49] [ 6] / answers: 1 / hits: 31437 / 12 Years ago, tue, june 19, 2012, 12:00:00

what I am doing is creating a form using JSON this form can then be edited a and produce new JSON object. The problem I am having seems to be with getting the form id.
The code I am using to return a JSON object is:

form = document.forms[0];

$.fn.serializeObject = function()

{

    alert(start serializeObject);

    var o = {};

    var a = this.seralizeArray();

    $.each(a, function(){

        if (o[this.name] !== undefined) {

            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;

    alert(o);

};



$(function() {

    alert(here);

    form.submit(function(){

        result.append(JSON.stringify(form.serializeObject()));

        return false;

    });

});

This just refresh's the pageI am not sure why.
This program is not on a server and not be used on a server.
by this I mean It is only every going to be run locally on a local machine, with no apache2 setup.

Thanks.

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leilanichasel

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2 Years ago

answered 12 Years ago breap · Accepted Answer

You code can be written pretty easy. This is how I do it:

Ajax:

$('#formID').on('submit',function () {

    $.ajax({

        url: 'submit.php',

        cache: false,

        type: 'POST',

        data : $('#formID').serialize(),

        success: function(json) {

            alert('all done');

        }

    });

});

If you are not sending it with Ajax, why would you do this? If you are simply submitting the form, you can do it using PHP like this:

<?php

$json_object = json_decode($_POST);

?>