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Warning: mysql_fetch_assoc() expects parameter 1 to be resource

ryuvely

By ryuvely
in PHP Coding Help

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ryuvely Newbie

ryuvely

Posted
Posted (edited)
we did an exercise in IT class it was about how to inject php code in css code we took a template as an example to work on. i injected php code in the page "services.php"as instructed but it didn't work and i keep getting this error : 

 

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\Entreprise\Entreprise\employee.class.php on line 21

 

unfortunately, the professor didn't correct this example so i hope someone will help here.

 

these are the codes i'm working on:

 

employee.class.php file:

 




<?php

   class Employee {

         public $codemp;

         public $nomemp;

         public $datnais;

         public $codfct;



function getEmployees() {

    $sql = "SELECT codemp, nomemp, datnais, codfct ";

    $sql .= "FROM employee ";

    $req = mysql_query($sql);

    return $req;

 }



function showHtmlEmployees($resource){

    print "<table border='1' width='60%'>";

    print "<tr>";

         print "<td>Employee Name</td>";

         print "<td>Birthday Date</td>";

    print "</tr>";



while ($data= mysql_fetch_assoc($resource)) {

   print "<li>";

       print "<h2>".$data['nomemp']."</h2>";

       print "<p>".$data['datnais']."</p>";

  print "</li>";

}



print "</table>"; 

}



}


 

Connection file:

 




<?php

//parameters

$server = "localhost" ;

$user = "root";

$password= "" ;

$database= "entreprise" ;



//connection

$db= mysql_connect($server,$user,$password);

$result= mysql_select_db($database,$db);

?>


 

 

 

services.php file:

 




<!DOCTYPE html>

<!-- Website template by freewebsitetemplates.com -->

<?php

include "connection.php";

include "employee.class.php";

$employee= New employee ();



?>

<html>

   <head>

     <title>Services - Chronokeep Website Template</title>

         <meta charset="utf-8">

         <link href="css/style.css" rel="stylesheet" type="text/css">

   <!--[if IE 7]>

         <link href="css/ie7.css" rel="stylesheet" type="text/css">

   <![endif]-->

  </head>



<body>

   <div id="background">

<div id="page">

  <div id="header"> <a href="index.html" id="logo"><img src="images/logo.png" width="295" height="55" alt="Chronokeep Website Template"></a>

   <ul class="navigation">

     <li>

       <a href="index.html">Home</a>

    </li>

    <li>

       <a href="about.html">About</a>

    </li>

     <li>

     <a href="blog.html">Blog</a>

    </li>

    <li>

      <a class="active" href="services.html">Services</a>

   </li>

   </ul>

 </div>

<!-- start of body-->

<div id="body">

  <div id="content">

     <h2>Services</h2>

       <ul id="article">

 <?php

$result=$employee->getEmployees();

$employee->showHtmlEmployees($result);

?>

       </ul>

   </div>

 </div>

</body>

</html>


 

 

the template is not mine ! we downloaded it to just work on it in class to better understand how php code is injected in a css file. 

I hope someone can help me with this. 

Edited by Ch0cu3r
Added code tags
Ch0cu3r Newbie

Ch0cu3r

Posted
Posted (edited)

Usually this type of error means your mysql query failed due to an error. You can use mysql_error to find out why its failing. Change lines 44 and 45 in services.php to the following

$result=$employee->getEmployees();

// if the query executed
if($result)
{
    $employee->showHtmlEmployees($result);
}
else
{
   // lets find out why the query failed
   trigger_error('Cannot get Employees from database: ' . mysql_error());
}

Please note when posting code wrap it within 



 tags (or click the <> button in the editor)



			
				


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