How to pass array from ajax to php?

Question

I'm trying to pass array from ajax to php (controller).

What is wrong with second code as var_dump($data) of first code returns appropriate content and second returns NULL?

FIRST. GOOD.

function myFunction() {
    var elementy = document.getElementsByClassName('inputISBN');
    var data = elementy[0].value;

    $.ajax({
           url: "{{ path('test') }}",
           type: "POST",
           data: { "data": data }

    });   

}

SECOND. BAD

function myFunction() {
    var elementy = document.getElementsByClassName('inputISBN');
    var data = [];    
    data[elementy[0].name] = elementy[0].value;

    $.ajax({
           url: "{{ path('test') }}",
           type: "POST",
           data: { "data": data }
    });   
}

THIRD. UGLY

var elementy = document.getElementsByClassName('inputISBN');
undefined
var data = []; 
undefined
data[elementy[0].name] = elementy[0].value;
"667"

Third one is line by line from the socond code written in browser console. And it's return what it should.

edit

and data is pulled out from here:

<input type="number" class="inputISBN" size="2" name="exampleName" 
                value="666" onchange="myFunction()">

Twisty · Accepted Answer · 2015-04-29 16:14:35Z

2

When passing an array to PHP, you want to include the array indicator: []. I thi8nk you need an Object: {}.

function myFunction() {
    var elementy = $('.inputISBN');
    var data = {};
    $.each(elementy, function(){
      data[$(this).attr('name')] = $(this).val();
    })

    $.ajax({
           url: "{{ path('test') }}",
           type: "POST",
           data: { "data": data }
    });   
}

At this point, you may also want to serialize the data (as was mentioned in the other answer by @Adelphia):

'data': JSON.stringify(data)

jsFiddle: https://jsfiddle.net/Twisty/cw77ann7/

You can call it in PHP: print_r($_POST['data']);

edited Apr 29, 2015 at 16:14

answered Apr 28, 2015 at 23:40

Twisty

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10 Comments

scrowler Over a year ago

What difference would that make for PHP?

Twisty Over a year ago

Depends... it's not clear from OPs code what he's trying to do. But that is how you pass data to be part of an array. And that's what he asked.

Sruj Over a year ago

@Twisty elementy[0].name are text type data allways, and elementy[0].value are number. Your hint with array indicator does not work. Still NULL.

Twisty Over a year ago

Added an update/edit. Let me know if this is more of what you're looking to do.

scrowler Over a year ago

@Twisty if data is an array in Javascript, it will be an array in PHP when it's sent via AJAX - you don't need the add [] to make that happen

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I wrestled a bear once. · Accepted Answer · 2015-04-29 00:25:54Z

1

You want to pass your data variable to PHP, which is an array, right? Why not data = JSON.stringify(data); and then on PHP's side, $data = json_decode($_POST['data'], true);

function myFunction() {
    var elementy = document.getElementsByClassName('inputISBN');
    i = elementy.length;
    data = [];
    while(i--) data[elementy[i].name] = elementy[i].value;
    data = JSON.stringify(data);

    $.ajax({
           url: "{{ path('test') }}",
           type: "POST",
           data: { "data": data }

    });   

}

edited Apr 29, 2015 at 0:25

answered Apr 28, 2015 at 23:51

I wrestled a bear once.

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8 Comments

Sruj Over a year ago

should I just paste your linedata = JSON.stringify(data); after line data[elementy[0].name] = elementy[0].value; (from my SECOND code) ? If yes then var_dump return array (size=1) 0 => string '[]' (length=2) instead of something like['name']=>'667'

I wrestled a bear once. Over a year ago

@Sruj make a jsfiddle and let me see what you're actually doing, I'll fix it up for you.

Sruj Over a year ago

@AdeIphia sorry for waisting your time, i don't belive it might help you jsfiddle.net/f3oLsu9m/1 .

I wrestled a bear once. Over a year ago

@Sruj - There is no array in your fiddle. I still don't understand what you're doing. I'd like to help, I just don't get it.

Sruj Over a year ago

@AdeIphia its test version to simplify my real version. I can do it again.

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