Calling a PHP function from JavaScript

DigiMartKe · November 4, 2015

i have this on my drop box;

(Part)

Then i have(java script);

function LSCJ(){

<?php LSC();?>

}

</script>

Then i have (PHP function)

<?php

function LSC() {

echo "Hello world Changed!";

}

?>

when i change my choice in dropbox the code LSC should run

How can i achieve this?

what am i doing wrong?

Please not that i have run a very complex php code . This is just a sample.

Any help given will be highly appreciated

.

Regards

Azhar

Ch0cu3r · November 4, 2015

You cannot call PHP functions from javascript. PHP and javascript are two completely different languages, they both run at different times. PHP runs on the server when the .php file is requested. Javascripts runs in browser when it receives the response from the request.

The only way you can do this is via ajax, whereby you send a http request to your php script. In the PHP script you have PHP code which then will call the LCS PHP function. Example JavaScript code, using JQuery ajax method

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script>
function LSC() {
    // using jquery ajax method for calling the php script
    $.ajax({
        // set this to the url of your php script for calling the LSC function
        url: 'lsc_script.php',

        // if the result of the ajax request is ok then this function is called
        success: function(response) {
            // the variable 'response' will contain the output from your php script
            // as an example we'll use a javascript alert to show the output of the response
            alert(response);
        }
    });
}
</script>

Example code for your PHP script

<?php

function lsc() {
   echo "Hello world has changed";
}

// call lsc function
lsc();

DigiMartKe · November 4, 2015

i have tried running this code.

I saved the first part in ajax.html

and

the second to lsc_script.php

But when i run ajax.html, Nothing happens.Blank Page loads.

What am i doing wrong?

Edited November 4, 2015 by DigiMartKe

Ch0cu3r · November 4, 2015

But when i run ajax.html, Nothing happens.Blank Page loads.

In ajax.html you do have your <select></select> drop down menu right?

Also open your browser developers console (should open when pressing F12 and clicking the console tab). It will mention any javascript errors.

DigiMartKe · November 4, 2015

Understood , but now i have another two issues;

1.The echo is not displayed on the form rather as a popup. I have pasted the screen shot. See Error1.pdf

2. My header page had three pictures , one showing after the other , Now all showing at once , some java script is conflicting. Can it be so?

Thank you

Ch0cu3r · November 4, 2015

1.The echo is not displayed on the form rather as a popup. I have pasted the screen shot. See Error1.pdf

Its displaying as popup because of the use of alert(response);

Where in the form do you want the response to show? You need to show your form code.

2. My header page had three pictures , one showing after the other , Now all showing at once , some java script is conflicting. Can it be so?

No idea. My code has nothing to do with your header.

DigiMartKe · November 4, 2015

This is my full code

<?php require_once('Connections/Connshifa.php'); ?>
<?php
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") 
{
  if (PHP_VERSION < 6) {
    $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;
  }
 
  $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);
 
  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;    
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? doubleval($theValue) : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}
}
 
mysql_select_db($database_Connshifa, $Connshifa);
$query_RsSymptom = "SELECT * FROM symptoms ORDER BY symptoms.name";
$RsSymptom = mysql_query($query_RsSymptom, $Connshifa) or die(mysql_error());
$row_RsSymptom = mysql_fetch_assoc($RsSymptom);
$totalRows_RsSymptom = mysql_num_rows($RsSymptom);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Cure by Symptom</title>
<?php
// Start the session
session_start();
?>
 
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<link href="css/style.css" rel="stylesheet" type="text/css" />
<link rel="stylesheet" type="text/css" href="css/coin-slider.css" />
<style type="text/css">
.main .content .content_resize .mainbar .article label {
font-weight: bold;
}
</style>
<script type="text/javascript" src="js/cufon-yui.js"></script>
<script type="text/javascript" src="js/cufon-aller.js"></script>
<script type="text/javascript" src="js/jquery-1.4.2.min.js"></script>
<script type="text/javascript" src="js/script.js"></script>
<script type="text/javascript" src="js/coin-slider.min.js"></script>
</head>
<body>
<div class="main">
  <div class="header">
    <div class="header_resize">
      <div class="slider">
        <div id="coin-slider"></div>
        <div class="clr"></div>
      </div>
      <div class="clr"></div>
    </div>
  </div>
  <div class="content">
    <div class="content_resize">
      <div class="mainbar">
        <div class="article">
          <h2>Find Your Cure</h2>
          <p class="infopost"> </p>
          <div class="clr"></div>
          <div class="clr"></div>
        </div>
        <div class="article">
          <h2>Every Disease has a Cure</h2>
<?php
 
 
echo "Welcome " . $_SESSION["username"] . " !" . ".<br>";
 
?> 
 
   
          <form id="form1" name="form1" method="post" action="">
 
   
          <label for="Uname">User Name/Email </label>
            <input name="Uname" type="text" id="Uname" value="<?php echo $_SESSION["username"] ?>" readonly="readonly" />
            <form method="post">
          </form>
 
            <label for="Symptom"><br />
              Select your Symptom
              <br />
              <br />
            Symptom</label>
                        
              <select name="Symptom" id="Symptomid" onchange="LSC()">
              
              
                        
                <?php
do {  
?>
                <option value="<?php echo $row_RsSymptom['name']?>"><?php echo $row_RsSymptom['name']?></option>
                <?php
} while ($row_RsSymptom = mysql_fetch_assoc($RsSymptom));
  $rows = mysql_num_rows($RsSymptom);
  if($rows > 0) {
      mysql_data_seek($RsSymptom, 0);
 $row_RsSymptom = mysql_fetch_assoc($RsSymptom);
  }
?>
              
              <br />
              <br />
              <br />
              <br />
<br />
      <p id="demo"></p>
<script>
                    function Symselect() {
                        var x = document.getElementById("Symptomid").value;
                        document.getElementById("demo").innerHTML = "You selected: " + x;
                    }
       </script>
 
    
     <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script>
function LSC() {
    // using jquery ajax method for calling the php script
    $.ajax({
        // set this to the url of your php script for calling the LSC function
        url: 'LoadSymptomDetails.php',
 
        // if the result of the ajax request is ok then this function is called
        success: function(response) {
            // the variable 'response' will contain the output from your php script
            // as an example we'll use a javascript alert to show the output of the response
            alert(response);
        }
    });
}
</script>
            
           
 
               <?php
 
//require("classes\LoadSymptomDetails.php");
?>
              <br />
            System</label>
          </form>
          <select name="System" id="System">
          </select>
          <p class="infopost"> </p>
          <p class="infopost"> </p>
        </div>
      </div>
    </div>
  </div>
  <div class="footer">
    <div class="footer_resize">
      <div style="clear:both;"></div>
    </div>
  </div>
</div>
<div align=center>This template  downloaded form <a href='http://all-free-download.com/free-website-templates/'>free website templates</a></div></body>
</html>
<?php
mysql_free_result($RsSymptom);
?>

and what i to call when a symptom is chosen is(as i want the system other columns to be displayed). I need to do this as when the columns are displayed in a table and a system chosen , the cures also shown in a table. So this code is to be used in two places; Although the code i have below is based on a post , i need it to run based on the combobox change.

<?php
$data = array();
if(isset($_POST['search'])){
$symptoms_name = $_POST['Symptom'];
$sql1 = mysql_query("SELECT name,description , comments
FROM symptoms WHERE symptoms.name ='$symptoms_name'") or die(mysql_error());
 
while($row1 = mysql_fetch_assoc($sql1)){
$data[] = $row1;
}
}
?>
<?php
if($data){
?>
<table border="1">
<tr>
<td>name</td>
<td>description</td>
<td>comments</td>
</tr>
 
<?php
if($data){
foreach($data as $sy){
?>
<tr>
<td><?php echo $sy['name']; ?> </td>
<td><?php echo $sy['description']; ?> </td>
<td><?php echo $sy['comments']; ?> </td>
 
</tr>
<?php
}
}
?>
 
 
</table>

Edited November 4, 2015 by Ch0cu3r

Ch0cu3r · November 4, 2015

Oh so you also need to have the selected symptom submitted too?

First change onchange="LSC()" to be onchange="LSC(this)".

Now change your LSC function to be

     <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script>
function LSC(option) {
    // using jquery ajax method for calling the php script
    $.ajax({
        // set this to the url of your php script for calling the LSC function
        url: 'LoadSymptomDetails.php',
        // send POST request
        method: 'POST',
        // set the Symptom name value to be submitted
        data: {Symptom: option.value},
 
        // if the result of the ajax request is ok then this function is called
        success: function(response) {
            // the variable 'response' will contain the output from your php script
            // as an example we'll use a javascript alert to show the output of the response
            alert(response);
        }
    });
}
</script>

And I assume you want the response of the ajax request to be added to the <p id="demo"></p> element? In that case replace alert(response); with

$('p#demo').html(response);

Edited November 4, 2015 by Ch0cu3r

DigiMartKe · November 6, 2015

I have tried to follow your instructions , but have not been able to achieve. Nothing happens on change of dropbox value.

I have made the following changes;(maybe based on this you can tell me what i have missed.)

dropbox

<select name="Symptom" id="Symptomid" onchange="LSC(this)>

HTML

have added the line

Java script

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script>
function LSC(option) {
    // using jquery ajax method for calling the php script
    $.ajax({
        // set this to the url of your php script for calling the LSC function
        url: 'LoadSymptomDetailsfridaytest.php',
        // send POST request
        method: 'POST',
        // set the Symptom name value to be submitted
        data: {Symptom: option.value},
        // if the result of the ajax request is ok then this function is called
        success: function(response) {
            // the variable 'response' will contain the output from your php script
            // as an example we'll use a javascript alert to show the output of the response
            $('p#demo').html(response);
        }
    });
}
</script>

and the php LoadSymptomDetailsfridaytest.php (a bit simplified)

<?php require_once('Connections/Connshifa.php'); ?>
 
 
<?php
 
$data = array();
{
mysql_select_db("shifa",$Connshifa );
$sql1 = mysql_query("SELECT name,description , comments
FROM symptoms  ") or die(mysql_error());
 
while($row1 = mysql_fetch_assoc($sql1)){
$data[] = $row1;
}
}
?>
<?php
if($data){
?>
<table border="1">
<tr>
<td>name</td>
<td>description</td>
<td>comments</td>
</tr>
 
<?php
if($data){
foreach($data as $sy){
?>
<tr>
<td><?php echo $sy['name']; ?> </td>
<td><?php echo $sy['description']; ?> </td>
<td><?php echo $sy['comments']; ?> </td>
 
</tr>
<?php
}
}
?>
<?php
 
}
?>
 
 
</table>

Kindly assist.

Edited November 6, 2015 by Ch0cu3r

DigiMartKe · November 6, 2015

Thanks , sorted

should have been

DigiMartKe · November 6, 2015

see also http://www.w3schools.com/php/php_ajax_database.asp

DigiMartKe · November 6, 2015

one question ,

the code

if the site is down ? How can this be handled?

Ch0cu3r · November 6, 2015

You can just download jquery and link to it locally rather the linking to a remote source if you want to.

Also note when posting code to wrap it within

tags or click <> button in the editor.

DigiMartKe · November 6, 2015

I still have a problem further down,

I have now changed the code for LoadSymptomDetails.php

and it works.

<?php require_once('Connections/Connshifa.php'); ?>
 
<?php
 
//$q = 1
 
?>
 
<?php
 
$data = array();
 
 
{
mysql_select_db("shifa",$Connshifa );
 
//This works 
$sql1 = mysql_query("SELECT name,description , comments
FROM symptoms" ) or die(mysql_error());
 
 
 
while($row1 = mysql_fetch_assoc($sql1)){
$data[] = $row1;
}

when i change the php code to

<?php require_once('Connections/Connshifa.php'); ?>
<?php
$data = array();
{
mysql_select_db("shifa",$Connshifa );
$symptoms_name = $_POST['Symptom'];
$sql1 = mysql_query("SELECT name,description , comments
FROM symptoms WHERE symptoms.name ='$symptoms_name'") or die(mysql_error());
 
while($row1 = mysql_fetch_assoc($sql1)){
$data[] = $row1;
}
}
?>
<?php
if($data){
?>
<table border="1">
<tr>
<td>name</td>
<td>description</td>
<td>comments</td>
</tr>
 
<?php
if($data){
foreach($data as $sy){
?>
<tr>
<td><?php echo $sy['name']; ?> </td>
<td><?php echo $sy['description']; ?> </td>
<td><?php echo $sy['comments']; ?> </td>
 
</tr>
<?php
}
}
?>
<?php
}
 
?>
 
 
</table>

When i make these changes , it stops working.

I get an error Notice: Undefined index: Symptom in C:\xampp\htdocs\sh\server\LoadSymptomDetails.php on line 6

So how can i pass the selected value in the dropbox to the where part of the select statement.

Kindly assist

Edited November 6, 2015 by Ch0cu3r

Ch0cu3r · November 6, 2015

So how can i pass the selected value in the dropbox to the where part of the select statement.

The code I gave earlier should be submitting the POST Symptom value via ajax when you make a change in the dropdown menu. I have tested it and it words ok for me.

To debug. in my code, add

        beforeSend: function(xhr, obj) {
            alert(obj.data);
        },

After

 data: {Symptom: option.value},

When you make a selection in the drop down it should display an alert with text like "Symption=<the value you have selected>"

This the last time I'm going to say this

Also note when posting code to wrap it within
tags or click <> button in the editor.

Next time will issue warning.

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Calling a PHP function from JavaScript

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