How can I call PHP functions by JavaScript?

Question

I am trying to call a PHP function from an external PHP file into a JavaScript script. My code is different and large, so I am writing a sample code here.

This is my PHP code:

<?php
function add($a,$b){
  $c=$a+$b;
  return $c;
}
function mult($a,$b){
  $c=$a*$b;
  return $c;
}

function divide($a,$b){
  $c=$a/$b;
  return $c;
}
?>

This is my JavaScript code:

<script>
  var phpadd= add(1,2); //call the php add function
  var phpmult= mult(1,2); //call the php mult function
  var phpdivide= divide(1,2); //call the php divide function
</script>

So this is what I want to do.

My original PHP file doesn't include these mathematical functions but the idea is same.

If some how it doesn't have a proper solution, then may you please suggest an alternative, but it should call values from external PHP.

php is server side js is client side. You'll need to use ajax or page refresh with gets/posts or try creating a js equivalent function. — Class, Apr 02 '13 at 06:43

score 98 · Accepted Answer · edited Feb 19 '15 at 22:10

Yes, you can do ajax request to server with your data in request parameters, like this (very simple):

Note that the following code uses jQuery

jQuery.ajax({
    type: "POST",
    url: 'your_functions_address.php',
    dataType: 'json',
    data: {functionname: 'add', arguments: [1, 2]},

    success: function (obj, textstatus) {
                  if( !('error' in obj) ) {
                      yourVariable = obj.result;
                  }
                  else {
                      console.log(obj.error);
                  }
            }
});

and your_functions_address.php like this:

    <?php
    header('Content-Type: application/json');

    $aResult = array();

    if( !isset($_POST['functionname']) ) { $aResult['error'] = 'No function name!'; }

    if( !isset($_POST['arguments']) ) { $aResult['error'] = 'No function arguments!'; }

    if( !isset($aResult['error']) ) {

        switch($_POST['functionname']) {
            case 'add':
               if( !is_array($_POST['arguments']) || (count($_POST['arguments']) < 2) ) {
                   $aResult['error'] = 'Error in arguments!';
               }
               else {
                   $aResult['result'] = add(floatval($_POST['arguments'][0]), floatval($_POST['arguments'][1]));
               }
               break;

            default:
               $aResult['error'] = 'Not found function '.$_POST['functionname'].'!';
               break;
        }

    }

    echo json_encode($aResult);

?>

Can you explain this code to me? And how to use it under different functions and areguments? — KingsInnerSoul, Jul 23 '14 at 15:07
Yup, nice answer... I'm just adding to the note about jQuery, that this tends to be slower. Another alternative can be using browser native XMLHttpRequest like for example here https://blog.garstasio.com/you-dont-need-jquery/ajax/#posting — Kuba, Feb 20 '19 at 18:20
header('Content-Type: application/json'); I don't think you need this line. — Nguai al, Jan 17 '20 at 09:15
i was looking for calling php function when an input change, so this give me a clue i ended up using fetch, Thank you — Nald Dev, Jan 17 '21 at 04:01
@Victor I tried your code but it never goes to my PHP functions — Heba, Dec 01 '21 at 12:29

Sandeep Kapil · Answer 2 · 2013-04-02T06:53:55.473

46

Try This

<script>
  var phpadd= <?php echo add(1,2);?> //call the php add function
  var phpmult= <?php echo mult(1,2);?> //call the php mult function
  var phpdivide= <?php echo divide(1,2);?> //call the php divide function
</script>

edited Apr 02 '13 at 06:53

answered Apr 02 '13 at 06:46

Sandeep Kapil

984
8
14

6

Although this solution works, it is not a clean and a good way to do it. It has tight dependency between client and server code. Much better design would be using ajax to send request to a php page that has the php function call. – Shadi Feb 01 '17 at 13:12
1

Is it possible to call a function this way?? – A.A Noman Feb 26 '17 at 05:01
2

@Shadi - I'm curious, what's wrong with having a *"tight dependency between client and server code"*? Is there a potential problem, or is this only a matter of "tidy coding"? – ashleedawg Jan 20 '20 at 00:25
2

@ashleedawg good question, as if using Ajax somehow doesn't require a "tight dependency" between client side and server side.. – Robert Sinclair May 05 '20 at 08:24
The above solution works. But here is a problem with code organization. This method forces add(), mult(), divide() functions to be declared in the same file as the script. With AJAX, you are forced to write on a different file. If the functions are short, this does not matter. But if function is lengthy, it is much better to write on a different file. – Nguai al Mar 05 '21 at 05:28
Every ajax call uses like 10 times more resources than a simple `let jsVal = =$phpVar?>` Ajax is useful if the information is on a different page. If the information is already present on the page, is better/faster/cleaner, and easier to reuse it. – Daniele Rugginenti Jun 23 '22 at 20:30
2

@statosdotcom your code doesn't work. You can't use a JS var in PHP in that way. – Daniele Rugginenti Jun 23 '22 at 20:36

score 13 · Answer 3 · edited Jul 06 '15 at 21:24

13

use document.write for example,

<script>
  document.write(' <?php add(1,2); ?> ');
  document.write(' <?php milt(1,2); ?> ');
  document.write(' <?php divide(1,2); ?> ');
</script>

edited Jul 06 '15 at 21:24

Arjun Tuli

1,257
1
11
13

answered Nov 25 '13 at 05:53

Anitha Mani

863
1
9
17

4

Make sure when using this that you join up the ' – Mikeys4u Feb 24 '14 at 12:26

score 8 · Answer 4 · answered Apr 02 '13 at 10:49

8

You need to create an API : Your js functions execute AJAX requests on your web service

  var mult = function(arg1, arg2)
    $.ajax({
      url: "webservice.php?action=mult&arg1="+arg1+"&arg2="+arg2
    }).done(function(data) {
      console.log(data);
    });
  }

on the php side, you'll have to check the action parameter in order to execute the propre function (basically a switch statement on the $_GET["action"] variable)

answered Apr 02 '13 at 10:49

Romain Durand

783
1
6
22

why are you using function expression? this is more fitting for function declaration. – Nguai al Jan 17 '20 at 09:17

score 7 · Answer 5 · answered Aug 05 '13 at 02:11

index.php

<body>
...
<input id="Div7" name="Txt_Nombre" maxlenght="100px" placeholder="Nombre" />
<input id="Div8" name="Txt_Correo" maxlenght="100px" placeholder="Correo" />
<textarea id="Div9" name="Txt_Pregunta" placeholder="Pregunta" /></textarea>

<script type="text/javascript" language="javascript">

$(document).ready(function() {
    $(".Txt_Enviar").click(function() { EnviarCorreo(); });
});

function EnviarCorreo()
{
    jQuery.ajax({
        type: "POST",
        url: 'servicios.php',
        data: {functionname: 'enviaCorreo', arguments: [$(".Txt_Nombre").val(), $(".Txt_Correo").val(), $(".Txt_Pregunta").val()]}, 
         success:function(data) {
        alert(data); 
         }
    });
}
</script>

servicios.php

<?php   
    include ("correo.php");

    $nombre = $_POST["Txt_Nombre"];
    $correo = $_POST["Txt_Corro"];
    $pregunta = $_POST["Txt_Pregunta"];

    switch($_POST["functionname"]){ 

        case 'enviaCorreo': 
            EnviaCorreoDesdeWeb($nombre, $correo, $pregunta);
            break;      
    }   
?>

correo.php

<?php
    function EnviaCorreoDesdeWeb($nombre, $correo, $pregunta)
    { 
       ...
    }
?>

score 7 · Answer 6 · 2017-01-06T12:49:55.873

This work perfectly for me:

To call a PHP function (with parameters too) you can, like a lot of people said, send a parameter opening the PHP file and from there check the value of the parameter to call the function. But you can also do that lot of people say it's impossible: directly call the proper PHP function, without adding code to the PHP file.

I found a way:

This for JavaScript:

function callPHP(expression, objs, afterHandler) {
        expression = expression.trim();
        var si = expression.indexOf("(");
        if (si == -1)
            expression += "()";
        else if (Object.keys(objs).length > 0) {
            var sfrom = expression.substring(si + 1);
            var se = sfrom.indexOf(")");
            var result = sfrom.substring(0, se).trim();
            if (result.length > 0) {
                var params = result.split(",");
                var theend = expression.substring(expression.length - sfrom.length + se);
                expression = expression.substring(0, si + 1);
                for (var i = 0; i < params.length; i++) {
                    var param = params[i].trim();
                    if (param in objs) {
                        var value = objs[param];
                        if (typeof value == "string")
                            value = "'" + value + "'";
                        if (typeof value != "undefined")
                            expression += value + ",";
                    }
                }
                expression = expression.substring(0, expression.length - 1) + theend;
            }
        }
        var doc = document.location;
        var phpFile = "URL of your PHP file";
        var php =
            "$docl = str_replace('/', '\\\\', '" + doc + "'); $absUrl = str_replace($docl, $_SERVER['DOCUMENT_ROOT'], str_replace('/', '\\\\', '" + phpFile + "'));" +
            "$fileName = basename($absUrl);$folder = substr($absUrl, 0, strlen($absUrl) - strlen($fileName));" +
            "set_include_path($folder);include $fileName;" + expression + ";";
        var url = doc + "/phpCompiler.php" + "?code=" + encodeURIComponent(php);
        $.ajax({
            type: 'GET',
            url: url,
            complete: function(resp){
                var response = resp.responseText;
                afterHandler(response);
            }
        });
    }

This for a PHP file which isn't your PHP file, but another, which path is written in url variable of JS function callPHP , and it's required to evaluate PHP code. This file is called 'phpCompiler.php' and it's in the root directory of your website:

<?php
$code = urldecode($_REQUEST['code']);
$lines = explode(";", $code);
foreach($lines as $line)
    eval(trim($line, " ") . ";");
?>

So, your PHP code remain equals except return values, which will be echoed:

<?php
function add($a,$b){
  $c=$a+$b;
  echo $c;
}
function mult($a,$b){
  $c=$a*$b;
  echo $c;
}

function divide($a,$b){
  $c=$a/$b;
  echo $c;
}
?>

I suggest you to remember that jQuery is required:
Download it from Google CDN:

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>

or from Microsoft CDN: "I prefer Google! :)"

<script src="https://ajax.aspnetcdn.com/ajax/jQuery/jquery-3.1.1.min.js"></script>

Better is to download the file from one of two CDNs and put it as local file, so the startup loading of your website's faster!

The choice is to you!

Now you finished! I just tell you how to use callPHP function. This is the JavaScript to call PHP:

//Names of parameters are custom, they haven't to be equals of these of the PHP file.
//These fake names are required to assign value to the parameters in PHP
//using an hash table.
callPHP("add(num1, num2)", {
            'num1' : 1,
            'num2' : 2
        },
            function(output) {
                alert(output); //This to display the output of the PHP file.
        });

Let's try my solution, for me it works! After you comment for any probem! — , Jan 06 '17 at 12:55

score 6 · Answer 7 · answered Jan 10 '21 at 07:47

Void Function

<?php
function printMessage() {
    echo "Hello World!";
}
?>

<script>
    document.write("<?php printMessage() ?>");
</script>

Value Returning Function

<?php
function getMessage() {
    return "Hello World!";
}
?>

<script>
    var text = "<?php echo getMessage() ?>";
</script>

Lemon Drop · Answer 8 · 2014-05-04T20:12:40.950

5

If you actually want to send data to a php script for example you can do this:

The php:

<?php
$a = $_REQUEST['a'];
$b = $_REQUEST['b']; //totally sanitized

echo $a + $b;
?>

Js (using jquery):

$.post("/path/to/above.php", {a: something, b: something}, function(data){                                          
  $('#somediv').html(data);
});

edited May 04 '14 at 20:12

answered Apr 02 '13 at 06:49

Lemon Drop

2,113
2
19
34

score 3 · Answer 9 · answered Oct 18 '13 at 13:03

I wrote some script for me its working .. I hope it may useful to you

<?php
if(@$_POST['add'])
{
function add()
{
   $a="You clicked on add fun";
   echo $a;
}
add();
}
else if (@$_POST['sub']) 
{
function sub()
{
   $a="You clicked on sub funn";
echo $a;  
}
sub();  
}
?>
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="POST">

<input type="submit" name="add" Value="Call Add fun">
<input type="submit" name="sub" Value="Call Sub funn">
<?php echo @$a; ?>

</form>

score 1 · Answer 10 · answered Apr 02 '13 at 08:14

1

Try looking at CASSIS. The idea is to mix PHP with JS so both can work on client and server side.

answered Apr 02 '13 at 08:14

radmen

1,584
9
13

score 1 · Answer 11 · answered Dec 24 '19 at 10:39

I created this library JS PHP Import which you can download from github, and use whenever and wherever you want.

The library allows importing php functions and class methods into javascript browser environment thus they can be accessed as javascript functions and methods by using their actual names. The code uses javascript promises so you can chain functions returns.

I hope it may useful to you.

Example:

<script>
$scandir(PATH_TO_FOLDER).then(function(result) {
  resultObj.html(result.join('<br>'));
});

$system('ls -l').then(function(result) {
  resultObj.append(result);
});

$str_replace(' ').then(function(result) {
  resultObj.append(result);
});

// Chaining functions 
$testfn(34, 56).exec(function(result) { // first call
   return $testfn(34, result); // second call with the result of the first call as a parameter
}).exec(function(result) {
   resultObj.append('result: ' + result + '<br><br>');
});
</script>

score 1 · Answer 12 · answered Jul 31 '22 at 16:32

1

I made a version only using js, without using any dependencies. I think this is the shorest solution but probably not the best one since it doens't check for any errors.

javascript

var a = 1;
var b = 2;
function add(){
  var xmlHttp = new XMLHttpRequest();
  xmlHttp.open( "GET", "YOUR_SERVER/function.php?a="+a+"&b="+b, false );
  xmlHttp.send( null );
  return xmlHttp.responseText; 
}
var c = add(a, b)

function.php file

<?php echo $_GET["a"] + $_GET["b"]?>

c = 3

answered Jul 31 '22 at 16:32

SuninDarknes

11
1

If you have a new question, please ask it by clicking the [Ask Question](https://stackoverflow.com/questions/ask) button. Include a link to this question if it helps provide context. - [From Review](/review/late-answers/32376730) – Matt Pengelly Aug 04 '22 at 16:57

score 0 · Answer 13 · answered Apr 02 '19 at 08:14

I created this library, may be of help to you. MyPHP client and server side library

Example:

<!DOCTYPE html>
<html>
<head>
    <meta charset="utf-8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <title>Page Title</title>
    <meta name="viewport" content="width=device-width, initial-scale=1">
</head>
<body>

    <!-- include MyPHP.js -->
    <script src="MyPHP.js"></script>

    <!-- use MyPHP class -->
    <script>
        const php = new MyPHP;
        php.auth = 'hashed-key';

        // call a php class
        const phpClass = php.fromClass('Authentication' or 'Moorexa\\Authentication', <pass aguments for constructor here>);

        // call a method in that class
        phpClass.method('login', <arguments>);

        // you can keep chaining here...

        // finally let's call this class
        php.call(phpClass).then((response)=>{
            // returns a promise.
        });

        // calling a function is quite simple also
        php.call('say_hello', <arguments>).then((response)=>{
            // returns a promise
        });

        // if your response has a script tag and you need to update your dom call just call
        php.html(response);

    </script>
</body>
</html>

How can I call PHP functions by JavaScript?

13 Answers13

This work perfectly for me:

Linked

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