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, Jul 11 '12 at 19:04

Engineer · Accepted Answer · 2012-07-12T05:59:52.160

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 '12 at 05:59

answered Jul 11 '12 at 19:03

Engineer

47,849
12
88
91

Your edited code gives me only last form fields values not all array. – Osvalda Kazlaučiūnaitė Jul 12 '12 at 21:26
@OsvaldaKazlaučiūnaitė So you have multiple forms? Do I understand right? – Engineer Jul 13 '12 at 17:56
well, form is one, but I can add additional rows set to it – Osvalda Kazlaučiūnaitė Jul 13 '12 at 19:46
@OsvaldaKazlaučiūnaitė Make sure, that when you add/remove/modify new input elements, those elements *are inside of the `form` tag*. – Engineer Jul 13 '12 at 19:51

score 0 · Answer 2 · answered Jul 11 '12 at 19:07

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 '12 at 19:07

2

`dataType` specifies the *response-body* type,not the *request-body* type ! – Engineer Jul 11 '12 at 19:21
If you only just do data: data you might run into a similar issue like I did here: http://stackoverflow.com/questions/11385668/js-ajax-calling-php-and-getting-ajax-call-data – MrB Jul 11 '12 at 19:22
@Engineer: Upvoted, you are right and I was off-track anyway with my answer, I misunderstood OP's question. – Jul 11 '12 at 21:54

score 0 · Answer 3 · answered Jul 11 '12 at 19:20

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 '12 at 19:20

MrB

1,544
3
18
30

send and decode JSON array

3 Answers3