CODEWITHSUNDEEP
phpjqueryajax
user3622910
user3622910

Reputation: 47

php function call with ajax

I am trying to call a php function when an HTML button is clicked.i have done some searching and found that its impossible to do this directly and i should use ajax. so this is my attempt so far which is not working.this is my test.php and the function is also in this page.

   <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.js"></script>
    <script type="text/javascript">
        $(document).ready(function() {
            $('.NextPage').on('click', function() {
                $.ajax({
                    url: 'test.php',
                    data: {x: 1},
                    type: 'POST',
                    dataType: 'JSON',
                    success: function(response) {
                        alert(response);
                    }
                });

            });
        });
    </script>
     </head>
     <body>

    <button type="button" class="NextPage">go to nextpage</button> 

    <?php
    if (isset($_POST['x'])) {
        if ($_POST['x'] == 1) {
            $data = function1();
            echo json_encode($data);
            exit;
        }
    }

    function function1() {
        return 'Hi user! im function #1';
    }

    function function2() {
        return 'Hi user! im function #2';
    }
    ?>

Upvotes: 3

Views: 263

Answers (4)

user1978142
user1978142

Reputation: 7948

First off, you need to set your button to type="button", then, make an AJAX request, then the missing part on your code is the backend part which processes the request. Then the backend responds to that call. After that you can just do what you please on that response. Consider this example: 

<?php

// this part handles the AJAX request
if(isset($_POST['x'])) {
    if($_POST['x'] == 1) {
        $data = function1();
        // a call has made, then give a response
        echo json_encode($data);
        exit;
    }
}

function function1() {
    // do some processing
    return 'Hi user! im function #1';
}

function function2() {
    return 'Hi user! im function #2';
}

?>

<!-- make its type "button" -->
<button type="button" class="NextPage">go to nextpage</button>
<div id="notification"></div>

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.js"></script>
<script type="text/javascript">
$(document).ready(function(){
    $('.NextPage').click(function(){
        $.ajax({
            url: 'test.php',
            data: {x: 1},
            type: 'POST',
            dataType: 'JSON',
            success: function(response) {
                $('#notification').html('Your ajax call has been successful<br/> and this is a notification').css({background: 'yellow'});
            }
        });

    });
});
</script>

Upvotes: 1

Bikas
Bikas

Reputation: 2759

I've practically used your code and got it working. Your PHP is just fine, as for your AJAX call, it must have success or done to benefit the returned data.

The code for reference is here

HTML

<!DOCTYPE html>
<html>
    <head>
        <title></title>
        <meta name="viewport" content="width=device-width, initial-scale=1">
        <meta name="mobile-web-app-capable" content="yes">
        <link rel="shortcut icon" sizes="196x196" href="">

        <link rel="stylesheet" type="text/css" href="" />
    </head>
    <body>
        <div id="abc"></div>
        <button type="submit" class="NextPage">go to nextpage</button>
        <script type="text/javascript" src="//cdnjs.cloudflare.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
        <script type="text/javascript" src="app.js"></script>
    </body>
</html>

JavaScript

$(document).ready(function() {
    $('.NextPage').click(function() {
        $.ajax({
            type:'post',
            url:'service.php',
            data:{x:1}
        }).done(function(data) {
            $("#abc").text(data);
        });
    });
});

PHP

<?php
    $x = $_POST['x'];

    if ($x == 1) {
        function1();
    }

    function function1() {
        echo "This is function 1";
    }

    function function2() {
        echo "This is function 2";
    }

Upvotes: 1

Girish Thimmegowda
Girish Thimmegowda

Reputation: 2169

get the value of the x from ajax call.

$x = $_POST['x'];

then use it.

EDIT

First you have to check weather your variable is set or not.. 

if(isset($_POST[x]))
{
$x = $_POST['x'];
}

try this

Upvotes: 2

adnan kamili
adnan kamili

Reputation: 9445

What makes you think it is impossible to call a php function directly. This is exactly what is done in CodeIgniter PHP MVC framework. You can call a php function with arguments inside a php class directly in CodeIgniter and that is actually what is done always.

Upvotes: -2

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