sending array to php function

Question

am trying send an array to php file using $_POST ,still always getting an undefined variable any idea !!

var all = [];
var textInput = $('input[type=text]');
var checkBox = $('input[type=checkbox]');
var radio = $('input[type=radio]');
textInput.each(function () {
    if ($(this).val() != "") {
        name = $(this).attr('name');
        value = $(this).attr('value');
        all[name] = value;
    }
});
checkBox.each(function () {
    if ($(this).attr("checked")) {
        name = $(this).attr('name');
        value = $(this).attr('value');
        all[name] = value;
    }
});
radio.each(function () {
    if ($(this).attr("checked")) {
        name = $(this).attr('name');
        value = $(this).attr('value');
        all[name] = value;
    }
});
var getSelect = $('#experince');
if (getSelect[0].selectedIndex != 0) {
    name = getSelect[0].name;
    value = getSelect[0].selectedIndex;
    all[name] = value;
}
$.post('test.php', all, function (data) {
    alert(data);
    $.each(data, function (key, value) {
        alert(value);
    });
}, 'json');

PHP FILE

    if(!isset($_POST['workF'])){
    $_POST['workF'] ="undefine";
}

if(!isset($_POST['workP'])){
    $_POST['workP']="undefine";

}

if(!isset($_POST['gender'])){
    $_POST['gender']="undefine";
}

if(!isset($_POST['experince'])){
    $_POST['experince']="undefine";
}

if (!isset($_POST['jobs'])){
    $_POST['jobs']="undefine";
}

if(!isset($_POST['location'])){
    $_POST['location']="undefine";

}

$work_type_f = $_POST['workF'];
$work_type_p = $_POST['workP'];
$gender = $_POST['gender'];
$experince = $_POST['experince'];
$jobs = $_POST['jobs'];
$location =$_POST['location'];

enter image description here

Use $('#FORMID').serialize() in data option. :) api.jquery.com/serialize — Prasenjit Kumar Nag
– Prasenjit Kumar Nag, Commented Jun 26, 2012 at 11:50
@Joy , i already tried using serialize() ,same all variable in php file is equal undefined, by the way am getting the values of input when user click on next page NOT when user click on submit button , and am not getting any error its just like , javascript not sending the values or am trying read value in php file in wrong way . — uno-2017
– uno-2017, Commented Jun 26, 2012 at 12:06
If you use .serialize() then you have to access the date with $_POST['FIELD_NAME']. Are all those keys your are using with $_POST[] field names in the form? — Prasenjit Kumar Nag
– Prasenjit Kumar Nag, Commented Jun 26, 2012 at 12:10
@Joy , yes , actually am already used this method in same page. — uno-2017
– uno-2017, Commented Jun 26, 2012 at 12:14

Community · Accepted Answer · 2017-05-23 10:09:18Z

1

all should be an object, not an array:

var all = {};

jQuery will do the necessary escaping for you when passing it to $.post, but it won't read the object properties of an array.

However as Joy stated, you're just trying to mimic a form here. Make sure the inputs are all contained inside a single form and use .serialize() to get a string which can be send along with a HTTP request.

edited May 23, 2017 at 10:09

CommunityBot

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answered Jun 26, 2012 at 11:55

Mattias Buelens

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1 Comment

uno-2017 Over a year ago

ok i tried as you sead i have changed (all) variable to an object then send it as $.post('test.php',all,function(data))... still cant see the variable in php page !!

NedStarkOfWinterfell · Accepted Answer · 2012-06-26 11:53:22Z

1

You better run JSON.stringify() on the dictionary, then use JSON_decode() in PHP to retrieve the array.

answered Jun 26, 2012 at 11:53

NedStarkOfWinterfell

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Comments

Thiem Nguyen · Accepted Answer · 2012-06-26 14:30:45Z

0

Sorry if I didn't understand well, but why you don't use the html attribute like this?

<input type="text" name="all[nameOfText]" />
<input type="text" name="all[nameOfText2]" />

<input type="radio" name="all[nameOfRadio]" />
...

edited Jun 26, 2012 at 14:30

Thiem Nguyen

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answered Jun 26, 2012 at 11:56

Joao Bueno

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