Ajax method in jquery using PHP as backend

Question

I have a script file where the ajax method has been implemented, which adds a record to a form ...and then i have a php file which serves as the backend. What I am trying to say is we dint declare the record variable in php, we only declared it in jquery part. But how did we actually accessed it using

$record = json_decode($_POST['record']);

What is json_decode and json_stringify in the script file.

main.js

$add_form.submit(function(e) {

        e.preventDefault();

        var fields = ['id', 'name', 'subject', 'theory', 'practical'];
        var record = {};

        for (var index in fields) {

            var field = fields[index];

            if (field == 'id' || field == 'theory' || field == 'practical')
                record[field] = parseInt( $('input#add_'+field).val() );

            else
                record[field] = $('input#add_'+field).val();

        }


        record.total = record.theory + record.practical;



        $.ajax({
            url: '/ab_batch/practice/db/action.php',
            type: 'POST',
            data: {
                action: 'ajaxAddRecord',
                record: JSON.stringify(record)
            },

            success: function(result) {

                if ( 'true' == result.trim() ) {

                    $add_modal.find('.ajax_add_result').text('Student Record Added...').css({
                        color: 'green',
                        display: 'block'
                    }).fadeOut(2500);



                }
                else {
                    $add_modal.find('.ajax_add_result').text('Error Adding Student Record!').css({
                        color: 'red',
                        display: 'block'
                    }).fadeOut(2500);
                }

            },

            error: function() {}

        });


    });

action.php

switch ($action) {
case 'ajaxAddRecord':

        $record = json_decode($_POST['record']);
        print ( $student->addRecord($record) ) ? 'true' : 'false' ;

        break;
}

Everything seems okay ! did you get any errors ?!! what's really the problem ? — Ismail RBOUH
– Ismail RBOUH, Commented Jul 4, 2016 at 15:21
Yes....I want somebody to explain me some part of the code...we dint declare the record variable in php, we only declared it in jquery part. But how did we actually accessed it using $record = json_decode($_POST['record']); — Abhijit Borkakoty
– Abhijit Borkakoty, Commented Jul 4, 2016 at 15:24

Louys Patrice Bessette · Accepted Answer · 2016-07-04 15:42:18Z

2

The action variable is declared (created) in main.js. You're right.
It is an object which is then used and filled with values.

The $.ajax({ code block sends it stringified to action.php.
"stringified" means converted to a string.
It has to be done to send it to the server-side PHP because an object (or an array) can't be sent directly without converting it to a string.

Then this string, received has $_POST['record'], has to be "decoded" to access the values.
This is what json_decode does... It creates an array with it.

Google theses keywords for more:
jQuery object
JSON.stringify()
json_decode()
PHP array
Ajax example tutorial

edited Jul 4, 2016 at 15:42

answered Jul 4, 2016 at 15:35

Louys Patrice Bessette

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2 Comments

Abhijit Borkakoty Over a year ago

Okay...so the data field in the ajax method is the part that we are sending to action.php or the part that we recieved from action.php ??

Louys Patrice Bessette Over a year ago

Yes, exactly data is the part sent. And the success is the "callback" actions to do with the result from action.php

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