SoS

Hi, All: I'm a beginner of PHP. While trying the code from a tutorial, I encountered the following problem. The variable $id can not be transfered to my server. You can find the code at the end of this email. When I visit http://mydomain.com/test.php?id=1 it always shows the list of the database instead of a perticular record. I appreciate your kind helps! Zhan Xu EECS Department Case Western Reserve University <html> <body> <?php $db = mysql_connect("localhost", "root"); mysql_select_db("mydb",$db); // display individual record if ($id) { $result = mysql_query("SELECT * FROM employees WHERE id=$id",$db); $myrow = mysql_fetch_array($result); printf("First name: %s\n<br>", $myrow["first"]); printf("Last name: %s\n<br>", $myrow["last"]); printf("Address: %s\n<br>", $myrow["address"]); printf("Position: %s\n<br>", $myrow["position"]); } else { // show employee list $result = mysql_query("SELECT * FROM employees",$db); if ($myrow = mysql_fetch_array($result)) { // display list if there are records to display do { printf("<a href=\"%s?id=%s\">%s %s</a><br>\n", $PHP_SELF, $myrow["id"], $myrow["first"], $myrow["last"]); } while ($myrow = mysql_fetch_array($result)); } else { // no records to display echo "Sorry, no records were found!"; } } ?> </body> </html>