[Bobd314]

Aight I am trying to get this function to work everything in the function looks right but when I do for example TutorialDisplay( $_GET['id'],PHP ) it displays all the tutorials in the database when I only want it to display one type.. can any one help?

PHP Code:

function TutorialDisplay( $id , $type ) {

$db = mysql_connect("localhost", "USERNAME", "PASSWORD");

mysql_select_db("DATABASE",$db);

// display individual record

if ($id) {

  $result = mysql_query("SELECT * FROM Tutorial WHERE id=$id AND Category=$type",$db);

   $myrow = mysql_fetch_array($result);

   printf("<b>%s\n<br></b>", $myrow["Title"]);

   printf("<p align=\"center\">%s</p>", $myrow["Tutorial"]);

} else {

    // show employee list

   $result = mysql_query("SELECT * FROM Tutorial",$db);

    if ($myrow = mysql_fetch_array($result)) {

      // display list if there are records to display

      do {

        printf("<a href=\"%s?id=%s\">%s</a><br>\n", $PHP_SELF, $myrow["id"], $myrow["Title"]);

      } while ($myrow = mysql_fetch_array($result));

    } else {

      // no records to display

      echo "Sorry, no records were found!";

    }

}

} 

[tltran]

First what I suggest is to make a debug

are you sure that your that you have an Id ?

you can try a debug mode like:

if($id){
echo "id= $id";
} else {
echo "No id";
}

then try step by step to debug your script I have read your SQL query and everything seem to be good.

To make sure that your function work well is to do some debug mode.

[Bobd314]

The whole thing works it's just that it displays all tutorials and the $type part dont work right.