How do you input javascript variable into php script?

Question

I'm trying to get a function called that calls a php function with an input.

javascript function (picNum is an integer):

function hello(picNum) {
   var pictureNumber = picNum;

   var phpFunc = "<?php 
      include 'otherfile.php';
      otherFileFunc(" + pictureNumber + ") //This is where the problem is, the input(pictureNumber) wont go through
   ?>";
   echo phpFunc;
}

otherfile.php

<?php
   function otherFileFunc($i) {
      $final = $i + 1;
      echo $final;
   }
?>

this code pretty much says if you do onclick="hello(1)" then the output or phpFunc should be 2 because you add one in the otherfile.php, but no matter the input the output is always 1 so I'm guessing the input at where I marked just isn't going through.

DONT TELL ME IT DOESNT WORK BECAUSE IT DOES. if i put an integer instead of " + pictureNumber + " it works perfectly!

any help is appreciated :)

You are confused about the difference between server-side and client-side. Put simply there is no way this will work. Read up on Ajax since that's what you need. — developerwjk
– developerwjk, Commented Apr 3, 2014 at 22:51
Once PHP is ran on a page it won't run again. The reason it works perfectly with an integer is because you run the PHP function with a number and the function runs. The pictureNumber variable won't change if you call it from JS since PHP can only run once. — MrHunter
– MrHunter, Commented Apr 3, 2014 at 23:35

Tyler Scott · Accepted Answer · 2014-04-03 22:52:04Z

1

Unfortunately you won't be able to call php from javascript.

Php is run from the server and javascript is run on a client (usually, the exception being node.js. However even in the instance of node.js, php is not used as javascript has replaced its functionality)

If you need to have javascript "call" a server function you will need to look into ajax requests so that the server can then run a function and return it to the client.

answered Apr 3, 2014 at 22:52

Tyler Scott

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Comments

Herland Cid · Accepted Answer · 2014-04-04 13:43:58Z

You have to use Ajax bro:

Javascript:

 function hello(picNum) {
   var pictureNumber = picNum;
   $.ajax({
     url: "otherfile.php",
     data: {"picNum":pictureNumber},
     type:'post',
     dataType:'json',
     success: function(output_string){
       PictureNumber = output_string['picturenumber'];
       alert(PictureNumber);
     }
   });
 }

PHP otherfile.php:

$picNum = $_POST['picNum'];
function otherFileFunc($pic){
  $final = $pic + 1;
  return $final;
}
$outputnumber = function($picNum);
$array = ('picturenumber' => $outputnumber);
echo json_encode($array);

Note: Untested

EDIT, tested:

javascript:

function hello(picNum) {
   var pictureNumber = picNum;
   $.ajax({
     url: "otherfile.php",
     data: {"picNum":pictureNumber},
     type:'post',
     dataType:'json',
     success: function(output_string){
       pictureNumber = output_string['picturenumber'];
       alert(pictureNumber);
     }
   });
 }
hello(1); //sample

PHP otherfile.php:

$picNum = $_POST['picNum'];
$picNum = 1;
function otherFileFunc($pic){
  $final = $pic + 1;
  return $final;
}
$outputnumber = otherFileFunc($picNum);
$array = array('picturenumber' => $outputnumber);

echo json_encode($array);

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