How to send an array to php with jquery ajax?

Question

I tried to create an array with serializeArray and post it to php. but my code doesn't work. I read this questions (question) but I didn't understand my mistake yet.

this is my ajax code

    var str = $("form").serializeArray();
    $.ajax({
        type: "POST",
        url: "myfile.php",
        data: str,
        success: function (value) {
            $("#mydata").html(value);
        }
    });

HTML Code

<form>
    <select name="num0">
        <option value="">num0</option>
        <option value="12">12</option>
        <option value="13">13</option>
    </select>
    <select name="num1">
        <option value="">num2</option>
        <option value="123">123</option>
        <option value="133">133</option>
    </select>
    <select name="num2">
        <option value="">num3</option>      
        <option value="12345">12345</option>
    </select>
</form>

PHP Code

$postarr = array();
$num=$_POST['num0'];
$postarr[]=$num;
$num=$_POST['num1'];
$postarr[]=$num;
$num=$_POST['num2'];
$postarr[]=$num;

it giving me the following error message:

Notice: Undefined index: num0 (and same message for other variables).

By the way, English is not my native language; please excuse typing errors.

I don't know if this is related or not, but your <form> tag needs a method. Since it isn't proper HTML, the serialize may not work. — Sablefoste
– Sablefoste, Commented Mar 25, 2014 at 16:33
use $("form").serialize(), not $("form").serializeArray(); see stackoverflow.com/a/10430571/689579 — Sean
– Sean, Commented Mar 25, 2014 at 16:36
@SableFoste: method and action are optional. By default, the form will submit to the current page via GET. — gen_Eric
– gen_Eric, Commented Mar 25, 2014 at 16:37
Works fine for me: jsfiddle.net/XN6aJ (Check the headers of the request) — gen_Eric
– gen_Eric, Commented Mar 25, 2014 at 16:41

Dinistro · Accepted Answer · 2014-03-25 16:38:52Z

0

With the name attribute of an input field you are able to create an array in your $_POST:

<form method="post">
    <select name="values[num0]">
        <option value="">num0</option>
        <option value="12">12</option>
        <option value="13">13</option>
    </select>
    <select name="values[num1]">
        <option value="">num2</option>
        <option value="123">123</option>
        <option value="133">133</option>
    </select>
    <select name="values[num2]">
        <option value="">num3</option>      
        <option value="12345">12345</option>
    </select>
</form>

In your php you can use it like this:

$postarr = $_POST['values'];
echo $postarr['num0'];

answered Mar 25, 2014 at 16:38

Dinistro

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2 Comments

Arvin Firouzi Over a year ago

Thanks @Dinistro ,but it giving me this error: values are undefined

Dinistro Over a year ago

@ArvinFirouzi did you also copy my html? By me, it works fine.

andrew · Accepted Answer · 2014-03-25 16:40:42Z

0

Try this:

   $.ajax({
     type: "POST",
     url: "myfile.php",
     data: $('form').serialize(),
     success: function (value) {
         $("#mydata").html(value);
     }
 });

and instead of

$num=$_POST['num0']

use

$num=filter_input(INPUT_POST,'num0');

edited Mar 25, 2014 at 16:40

answered Mar 25, 2014 at 16:35

andrew

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1 Comment

Arvin Firouzi Over a year ago

Thanks @andrew, but when I tried to echo $num, it doesn't echo $num value.

Shijin TR · Accepted Answer · 2014-03-25 17:10:54Z

0

Add an id to your form,

  <form id="my_form">
<select name="num0">
    <option value="">num0</option>
    <option value="12">12</option>
    <option value="13">13</option>
</select>
<select name="num1">
    <option value="">num2</option>
    <option value="123">123</option>
    <option value="133">133</option>
</select>
<select name="num2">
    <option value="">num3</option>      
    <option value="12345">12345</option>
</select>

Just change

  var str = $("#my_form").serialize();

In your script

answered Mar 25, 2014 at 17:10

Shijin TR

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