displayind data to textboxes from phpmyadmin

[michael booz]

i have this code below and i want the details of the employees to be displayed when i click the 'search button' ..... the errors are :: - "Notice: Undefined variable: result in C:\xampp\htdocs \michael\sign.p hp on line 85

Warning: mysql_fetch_arr ay() expects parameter 1 to be resource, null given in C:\xampp\htdocs \michael\sign.p hp on line 85"


the code is as follows :

<?PHP
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("cdaleave", $con);
 
    if (isset($_GET['search'])){
    $pno=isset($_GET['P_NO']) ? $_GET['P_NO'] : '';
    $fname=isset($_GET['F_name']) ? $_GET['F_name'] : '';
    $sname=isset($_GET['S_name']) ? $_GET['S_name'] : '';
    $department=isset($_GET['Department']) ? $_GET['Department'] : '';
    $password=isset($_GET['Password']) ? $_GET['Password'] : '';
    $ldb=isset($_GET['Balance']) ? $_GET['Balance'] : '';
    $dbf=isset($_GET['Forward']) ? $_GET['Forward'] : '';
    $ela=isset($_GET['Allowance']) ? $_GET['Allowance'] : '';
    $edr=isset($_GET['Resumption']) ? $_GET['Resumption'] : '';
 
$result = mysql_query("SELECT * FROM employee WHERE P_NO='$pno'");
}
while($row = mysql_fetch_array($result))........line 85
{
      echo "<input type='text' name='txtFname' value='".$row['F_name']."'>";
   //... this is a section of indicating the other details to be displayed in the textboxes....// 
 
}
mysql_close($con);
?> 
</form>
</body>

any assistance would be of great help to me

thanks in advance

[Dormilich]

that’s correct, if the search URL parameter is not given, $result will not be created, but you always try to loop over it.