Re: SoS

Hi. If your id is not a numeric field you should change your query this way: $result = mysql_query("SELECT * FROM employees WHERE id='$id'",$db); <zxx10@cwru.edu> wrote in message news:84db9b84fea8.84fea884db9b@cwru.edu... > 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>