How to post an array using AJAX to PHP?

Question

I want to post an array to PHP using AJAX and on success returns the value back to JavaScript. Here's my code.

JavaScript:

$(document).ready(function(){
$.ajax({
    type: "POST",
    url: "phparray.php",
    data: {
        array1: phparray
    },

    success: function(data){
        alert("success");
        alert(data);
    }
});
});

HTML:

<html>
<head>
    <script type="text/javascript" src="jquery-2.0.2.js"></script>
    <script>
     var phparray = jQuery.makeArray();
     for(var i=0; i<10 ; i++){
        phparray.push(i); 
     }

    </script>
    <script type="text/javascript" src="phparraypost.js"></script>
</head>
<body>
</body>
</html>

PHP:

<?php
$n=$_POST['array1'];

echo $n;


?>

The data I get back says

<br /> <b>Notice</b>: Array to string conversion in <b>C:\xampp\htdocs\php\phparray.php</b> on line <b>4</b><br /> Array

and I don't have an idea what might be wrong with it.

The HTML, PHP and JavaScript code are in different files.

So what isn't working exactly. Other than using jQuery to create an empty array, there doesn't seem to be anything wrong? — adeneo
– adeneo, Commented Jul 27, 2013 at 13:30
The 'path' between a client (javascript) and the server (php) is HTTP. Http doens't care about arrays and only accepts strings. In your javascript you should format the array to a string (comma seperated, JSON) and parse it serverside to get the array back. — Joren
– Joren, Commented Jul 27, 2013 at 13:30
@adeneo that's why I didn't state is as an answer but as a comment. — Joren
– Joren, Commented Jul 27, 2013 at 13:32
the data i get back says " Notice: Array to string conversion in C:\xampp\htdocs\php\phparray.php on line 4 Array" and i don't have any idea about it. — L4reds
– L4reds, Commented Jul 27, 2013 at 13:37

user188654 · Accepted Answer · 2013-07-27 13:41:33Z

5

The problem is in your PHP - you are using echo on an array. Instead use var_dump.

<?php
    $n=$_POST['array1'];
    var_dump($n);
?>

answered Jul 27, 2013 at 13:41

user188654

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3 Comments

L4reds Over a year ago

how can i use $n in the form of an array in the php file? any suggestion, please!

user188654 Over a year ago

Well $n should already be an array so you could easily traverse it with foreach etc. Try var_dump($n) and see what kind of data type is reported back (it should be Array).

L4reds Over a year ago

var_dump($n) returns "array(10) { [0]=> string(1) "0" [1]=> string(1) "1" [2]=> string(1) "2" [3]=> string(1) "3" [4]=> string(1) "4" [5]=> string(1) "5" [6]=> string(1) "6" [7]=> string(1) "7" [8]=> string(1) "8" [9]=> string(1) "9" } " I think I might just get what i wanted.... thank you.

Konsole · Accepted Answer · 2013-07-27 14:09:58Z

I came up with the solution using $.param. This function is used internally to convert form element values into a serialized string representation. I am not sure if it fulfills your requirement.

   <script>
      $(document).ready(function(){
        var phparray = new Object();
         for(var i=0; i<10 ; i++){
            phparray['val' + i] = i; //store value in object
         }

        $.ajax({
            type: "POST",
            url: "phparray.php",
            data: {
                array1: $.param(phparray) //serialize data
            },

            success: function(data){
                alert("success");
                alert(data);
            }
        });
      });
    </script>

PHP

 $pieces = explode('&', $_POST['array1']); //explode passed serialized object
 $phparray = array();
 foreach($pieces as $piece){
   list($key, $value) = explode('=', $piece);
   $phparray[$key] = $value; //make php array
 }

 var_dump($phparray);

Garytje · Accepted Answer · 2013-07-27 13:38:45Z

0

http://api.jquery.com/serializeArray/ does this for you, since your already using jquery..

answered Jul 27, 2013 at 13:38

Garytje

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Comments

Naji Shmly · Accepted Answer · 2013-07-27 14:00:16Z

0

use this:

$("#formId").submit(function(event) {
        event.preventDefault();
        var url = $(this).attr('action');
        $.ajax({
            url : url,
            data : $(this).serialize(),
            cache : false,
            contentType : false,
            processData : false,
            type : 'POST',
            success : function(data) {
                $('#result').html('<div class="notes">Done</div>');
            }
        });
        return false;
    });

answered Jul 27, 2013 at 14:00

Naji Shmly

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