Calling php script from JavaScript

Question

I have spent all morning trying to figure out how I can call a PHP script from my JavaScript function and then return a value to output. I am not sure whether I can do this with just JavaScript or whether I have to include some Ajax aswell?

Here is the main page

<script>
    $("document").ready(function (){       
        $("select").change(function(){
            var currentValue = parseInt($(this).find("option:selected").val());
            //Call Script script.php here
            $("#output").html(result);
        })
    }); 
</script>   
</head>
<body>
    <select name="select">
        <option value='1' >1</option>
        <option value='2'>2</option>
        <option value='3'>3</option>
    </select>
    <div id="output"><!---OUTPUT HERE----></div>

And here is the script I want to call (I have simplified it all down to make it easier to understand)

<?php
    $result = $currentValue + 1;
?>

Basically I want it to read in the currentValue variable and output the result variable back to the webpage, to be outputted again.

What's the use of increasing a value like that? You need to store the value using for example like MySQL! — Alvin Wong
– Alvin Wong, Commented Jul 21, 2012 at 11:49
What do you mean by increasing it like that, the php script i have written is very long, too long to post here that involves mysql and all that, i just made it simple to show the fact, that i need to get the currentValue into the php script and get the result value out and back to the main page — Al Hennessey
– Al Hennessey, Commented Jul 21, 2012 at 11:51
look at this example I wrote .. you can download the files devpazzi.com/how-to-get-data-from-php-script-with-ajax-jquery — Skecci Dev
– Skecci Dev, Commented May 8, 2015 at 4:54

take · Accepted Answer · 2012-07-21 12:30:49Z

3

// Javascript:
$.ajax({
            type: "GET",
            url: "example.com/script.php?currentValue="+currentValue ,
            dataType: "json",
            statusCode: {
                200: function (result)
                {
                    $("#output").html(result.value);
                }
            }
        });


// PHP 
<?php 
$result = $_GET["currentValue"] + 1;
echo json_encode(array("value" => $result));

 ?>

edited Jul 21, 2012 at 12:30

answered Jul 21, 2012 at 11:53

take

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7 Comments

Al Hennessey Over a year ago

Hi, i am trying your code now, but i get an error on the last line echo json_encode(array("value" => $result);

Al Hennessey Over a year ago

ok, i just tried it, but whenever i select another option from the dropdown, nothing appears? Thanks for your help so far

take Over a year ago

Do you replaced "//Call Script script.php here" with the $.ajax ?

Al Hennessey Over a year ago

yes i replaced it exactly like that, but still it doesnt display anything

take Over a year ago

Edited my answer, updated line $("#output").html(result.value)

|

mrsrinivas · Accepted Answer · 2012-07-21 11:52:22Z

1

$.post('<url>', {currentValue : currentValue}, function(){}, 'json');

< url> example ajax/phpScript

Here ajax is controller and phpScript is the method in that controller

public function phpScript()
{
 $phpScript = $_POST['phpScript'];

//add your code 

}

answered Jul 21, 2012 at 11:52

mrsrinivas

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