jQuery create and post array to a PHP script

Question

My experience with jQuery is very limited, and im on a very steep learning cure.

I have a dynamic form which is generated based on main and sub menu, placing marker points in each row which has an inputfield, I have managed to create a loop which looks for all the marker points and return the field_name and the field_value of any input field.

$('td[edit="1"]').each(function(i, el) {
    field_name = $(el).attr('fieldname');
    field_val = $('#'+field_name).val();

    alert(field_name + " = " + field_val);
});

What I am having trouble with now, is converting this in to an array or JSON and pushing it to a PHP file which can then pick up the results.

Here is an example of how I currently submit a forms data to a PHP file, however is not dynamic as I have to specify which fields and values I want to send.

$.ajax({
    type: "POST",
    url: "form.php",
    data: {
        title : title,
        age : age

    }).done(function(data) {
        results = $(data).find('data').html();

    }).fail(function() {
        alert("error");
    }
);

Gaganpreet Kaur · Accepted Answer · 2015-03-26 12:43:30Z

2

If i have clearly understand your problem then You will have to use serialize method of jquery to send fields to php file. You can do like

data:$('form').serialize(),

It will send all the fields to php file.

answered Mar 26, 2015 at 12:43

Gaganpreet Kaur

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YANTHO · Accepted Answer · 2015-03-26 12:42:59Z

0

As data you can user Jquery on the form to create an array of all the data: https://api.jquery.com/serializeArray/

    $.ajax({
        type: "POST",
        url: "form.php",
            data: $('#idOfForm').serializeArray()
    }).done(function(data) {
        results = $(data).find('data').html();
    }).fail(function() {
        alert("error");
    });

answered Mar 26, 2015 at 12:42

YANTHO

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1 Comment

YANTHO Over a year ago

With serializeArray the name of the field is used to be posted so <input name="fieldName" ... /> --> access in PHP $_POST['fieldName']

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