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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
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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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