On Wed, May 11, 2005 10:19 pm, Mark Sargent said: > I'll try this differently(I hope). With the below code, variables are > named/set twice ($db, $num, $result, $myrow) etc. In ASP, from what I > remember, that would be a no no. Currently, this code only populates the I don't think ASP would issue an error if you re-used a variable... In fact, I can almost guarantee I re-used variables all the time in ASP. There might have been some kind of Programming Guidelines where you worked that said not to do that, though... > 1st select box. Coupla Qs. Do I have to connect to the db again, in the > second block of code..? Should the variables in the 2nd block of code, > be named differently to the 1st block.? In ASP the query variables would > be named uniquely, identifying to each query like so, prod_query_num, > prod_query_result and so forth. I must admit, my recollection of ASP is > very vague. Anyway, I hope this explains what I'm looking to understand. > Cheers. > > Mark Sargent. > > <html> > <body> > <hr> > <h1 align=center>JUMBO STATUS</h1><p> > <center>Used Hardware Specialist</center> > <hr> > <h2 align=center>Admin</h2> > <table align="center" border="2"> > <tr> > <td><h3 align=center>PRODUCT TYPE</h3></td> > </tr> > <tr><td><center> > <select name="poduct_type"> > <?php > $db = mysql_connect("localhost", "root", "grunger"); > mysql_select_db("status",$db); > $result = mysql_query("SELECT ProductTypes.product_type_detail, > ProductTypes.product_type_id FROM ProductTypes",$db); > $num = mysql_num_rows($result); I would move all of this stuff OUTSIDE the <select> because after you add your error messages, you'll never see them if they are inside the <select>. > for ($i=0; $i<$num; $i++){ > $myrow=mysql_fetch_array($result); > $product_type=mysql_result($result,$i,"product_type_detail"); > $product_type_id=mysql_result($result,$i,"product_type_id"); > $maker=mysql_result($result,$i,"maker_detail"); > $maker_id=mysql_result($result,$i,"maker_id"); > echo "<option value=\"$product_type_id\">$product_type</option><br>"; You are not using maker, nor maker_id, so why put them in the query, and why get them from your result set? > } > ?> > </select> > </center> > </td></tr> > </table> > <table align="center" border="2"> > <tr> > <td><h3 align=center>Maker</h3></td> > </tr> > <tr><td><center> > <select name="slect_maker"> > <?php > $db = mysql_connect("localhost", "root", "grunger"); Don't connect to the database twice. That's the most expensive (time-wise) thing in your whole script, probably. > mysql_select_db("status",$db); And you don't need this either. > $result = mysql_query("SELECT Makers.maker_id Makers.maker_detail FROM > Makers",$db); > $num = mysql_num_rows($result); Again, move these outside the <select> > for ($i=0; $i<$num; $i++){ > $myrow=mysql_fetch_array($result); > $maker=mysql_result($result,$i,"maker_detail"); > $maker_id=mysql_result($result,$i,"maker_id"); > echo "<option value=\"$maker_id\">$maker</option><br>"; > } > ?> > </select> > </center> > </td></tr> > </table> > <p> > <p> > <p> > <hr><center>email: ???????????<p> > Telephone:????????<p> > Fax: ????????? > </center> > <hr> > </body> > </html> -- Like Music? http://l-i-e.com/artists.htm -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php