Passing data to Php with Jquery Ajax

Question

I want to pass two input data to PHP with JQuery ajax I read about json, serialize and similar but I couldn't quite get a grip around it. I got:

<p><input type="text" id="domain"></p>
<p><input type="text" id="domain1"></p>
<button id="order">Search</button>

<script type="text/javascript">
    function callAjax() {
        $.ajax({
            type:  "GET",
            cache: false,
            url:   'getip.php',
            data:  ???
    });
</script>

And also what do I need on PHP file because I only got this.

$name=($_GET['']);
$surname = 
echo($name) ;
endif;

thats all of html code. Body with two input, i'm learning this stuff so i don't have much to work with. — user3187715
– user3187715, Commented May 12, 2015 at 14:27
1.) you need to link to jQuery to load the library. 2.) you need to place your javascript/jQuery inside script tags or a .js file and link to it. 3.) You need to place your function in $( document ).ready(function() { // code here }); OR $(function() {//code here }); — NewToJS
– NewToJS, Commented May 12, 2015 at 14:33
you should really copy/past the full content of your html page and of your php page, this will help much, cuz like this, we don't even know if you included jquery lib or even if you call callAjax func — Bobot
– Bobot, Commented May 12, 2015 at 14:37
@Bob0t this is why I didn't edit the post and add the script tags because now you might have altered the source code to something the that user3187715 doesn't/might not have. user3187715 might not have known you need to place javascript into script tags and might not notice your edit. — NewToJS
– NewToJS, Commented May 12, 2015 at 14:40

Tushar · Accepted Answer · 2015-05-12 14:27:57Z

2

function callAjax() {

    // Get values to send to server
    var domain = $('#domain').val();
    var domain1 = $('#domain1').val();

    $.ajax({
        type: "GET",
        cache: false,
        url: 'getip.php',
        data: {
            domain: domain,
            domain1: domain1
        }

And in php you'll get it as

$domain = $_GET['domain'];
$domain1 = $_GET['domain1'];

Docs: http://api.jquery.com/jQuery.ajax/

answered May 12, 2015 at 14:27

Tushar

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Ahmed Ziani · Accepted Answer · 2015-05-12 14:38:59Z

2

$.ajax({ 
    type: "GET", 
    cache: false, 
    url: 'getip.php', 
    data = { 
        'domain': $('#domain').val(),
        'domain1': $('#domain1').val()
        };
)};

And in you php file php you can get data with :

$domain = $_GET['domain'];
$domain1 = $_GET['domain1'];

edited May 12, 2015 at 14:38

answered May 12, 2015 at 14:29

Ahmed Ziani

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Death-is-the-real-truth · Accepted Answer · 2015-05-12 14:30:42Z

1

Try this:-

<p><input type="text" id="domain" name = "domain"></p>
<p><input type="text" id="domain1" name ="domain1"></p>
<button id="order">Search</button>
function callAjax() {

$.ajax({
    type: "GET",
    cache: false,
    url: 'getip.php',
    data:{domain: $('#domain').val(),domain1: $('#domain1').val()}
});

In your php file get it like this:-

$domain = $_GET['domain'];
$domain1 = $_GET['domain1'];

Note:- Please try to read jquery a bit. It will help you.

answered May 12, 2015 at 14:30

Death-is-the-real-truth

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3 Comments

user3187715 Over a year ago

When i try to read documents on jquery its like reading Arabic or chinese, i always need to see live example, just my style of learning i will try this tough :). See how it works

Death-is-the-real-truth Over a year ago

then mark it as your answer. It helps other to get there solutions.

user3187715 Over a year ago

Oh i thought i did :).

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