Joomla Problem With PHP Code

[Sanfoor]

Hi everyone!

 

I made a php code to show a staff list when choosing a job title from a drop-down menu, everything work's fine when i use the staff.php file alone

BUT when i put it in JOOMLA (with JUMI plugins) i only can see the drop-down list and when i choose a job title from the drop-down menu i don't get anything from the staff list info.

 

it must something with JOOMLA, i mean i must have to change something in the code to work with JOOMLA, because it's working well when its alone

 

BEST REGARDS TO YOU ALL

 

===================

staff.php

===================

<?php

 

defined( '_JEXEC' ) or die( 'Restricted access' );

 

// Connect To MySQL

//mysql_connect("localhost", "UserName", "PassWord") or die(mysql_error());

//mysql_select_db("DatabaseName") or die(mysql_error());

//$YrkeSlkt="SELECT Yrke FROM TableName order by Yrke ASC";

 

$db =& JFactory::getDBO();

 

 

// Connect database

 

// If submitted, check the value of "select". If its not blank value, get the value and put it into $select.

if(isset($select)&&$select!="")

{

$select=$_GET['select'];

}

?>

 

 

<form id="YrkeForm" name="YrkeForm" method="get" action="<? echo $PHP_SELF; ?> ">

  <div align="center">

    <select name="select" onChange="submit()">

  <option value="">--- Välj Yrke ---</option>

  <?

// Get records from database (table "TableName").

$list=mysql_query("SELECT DISTINCT(Yrke) as Yrke FROM TableName ORDER BY Yrke ASC;");

 

// Show records by while loop.

while($row_list=mysql_fetch_assoc($list)){

?>

  <option value="<? echo $row_list['Yrke']; ?>" <? if($row_list['Yrke']==$select){ echo "selected"; } ?>><? echo $row_list['Yrke']; ?></option>

  <?

// End while loop.

}

?>

    </select>

   

  </div>

</form>

<hr>

<p>

<?

// If you have selected from list box.

if(isset($select)&&$select!=""){

 

// Get records from database (table "TableName").

$result1=mysql_query("select * from TableName where Yrke='$select'");

$row=mysql_fetch_assoc($result1);

?>

 

<?

// End if statement.

}

 

// Close database connection.

$query = sprintf("SELECT * FROM TableName WHERE Yrke='%s' order by FNamn ASC", mysql_real_escape_string($row['Yrke'])) or die ("Unable to Make the Query:" . mysql_error() );

 

 

// Perform Query

$result2 = mysql_query($query);

echo "<table border='1' align='center'>";

echo "<tr> <th>Fornamn</th> <th>Efternamn</th> <th>PersonNummer</th> <th>Stad</th> <th>Address</th> <th>PostNr</th> <th>PostAd</th> <th>Mobil</th> <th>Yrke</th> <th>Motte</th> </tr>";

// keeps getting the next row until there are no more to get

while($row = mysql_fetch_array( $result2 )) {

// Print out the contents of each row into a table

echo "<tr><td>";

echo $row['FNamn'];

echo "</td><td>";

echo $row['ENamn'];

echo "</td><td>";

echo $row['PersonNmr'];

echo "</td><td>";

echo $row['Stad'];

echo "</td><td>";

echo $row['Address'];

echo "</td><td>";

echo $row['PostNr'];

echo "</td><td>";

echo $row['PostAd'];

echo "</td><td>";

echo $row['Mobil'];

echo "</td><td>";

echo $row['Yrke'];

echo "</td><td>";

echo $row['Motte'];

echo "</td></tr>";

}

echo "</table>";

mysql_close();

?>

</p>