send and decode JSON array

Question

I'm trying to encode and send JSON array to php page and add it to mysql:

var data = $('form').serialize();
$.ajax({
    type: 'POST',
    url: 'naujas.php',
    dataType: 'json',
    data: ({
        json: JSON.stringify(data)
    }),
    success: function () {
        $('#naujas').load('naujas.php');
    }
});

But I think its not working I'm getting response from php like that: pav=1&ppav=2&kiekis=3&kaina=4d&ppav=5&kiekis=6&kaina=7&ppav=8&kiekis=9&kaina=0

php file

<?php
    $json = json_decode($_POST['json']);
    echo $json;
?>

What I'm doing wrong?

.serialize() gives you a query string, .serializeArray() might be closer to what you want — Musa
– Musa, Commented Jul 11, 2012 at 19:04

Engineer · Accepted Answer · 2012-07-12 05:59:52Z

1

Try like this:

var data = $('form').serializeArray().reduce( function(obj,cur){
    obj[cur.name] = cur.value;
    return obj;
},{});

Explanation:

.serializeArray() returns an array, which has following structure:

[ {name:"inputname1",value:"inputvalue1"},
  {name:"inputname2",value:"inputvalue2"},
  //---------------------------------------
  {name:"inputnamen",value:"inputvaluen"} ]

.reduce() function converts that array to object:

{ "inputname1":"inputvalue1",
  "inputname2":"inputvalue2",
  //---------------------------------------
  "inputnamen":"inputvaluen" }

edited Jul 12, 2012 at 5:59

answered Jul 11, 2012 at 19:03

Engineer

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4 Comments

Osvalda Kazlaučiūnaitė Over a year ago

Your edited code gives me only last form fields values not all array.

Engineer Over a year ago

@OsvaldaKazlaučiūnaitė So you have multiple forms? Do I understand right?

Osvalda Kazlaučiūnaitė Over a year ago

well, form is one, but I can add additional rows set to it

Engineer Over a year ago

@OsvaldaKazlaučiūnaitė Make sure, that when you add/remove/modify new input elements, those elements are inside of the form tag.

user1454661 · Accepted Answer · 2012-07-11 19:07:31Z

0

dataType: 'json' already tell jQuery to post the data in a json format.

All you need to do is to post your data, something like this:

data: (data),

The problem comes from converting your object to a string representation (stringify).

answered Jul 11, 2012 at 19:07

user1454661

3 Comments

Engineer Over a year ago

dataType specifies the response-body type,not the request-body type !

MrB Over a year ago

If you only just do data: data you might run into a similar issue like I did here: stackoverflow.com/questions/11385668/…

user1454661 Over a year ago

@Engineer: Upvoted, you are right and I was off-track anyway with my answer, I misunderstood OP's question.

MrB · Accepted Answer · 2012-07-11 19:20:33Z

0

You might need to use php stripslashes

http://php.net/manual/en/function.stripslashes.php

I think it's something like this

json_decode(stripslashes($_POST['json']));

answered Jul 11, 2012 at 19:20

MrB

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