In article <200211011751.03080.James.Hatridge@epost.de>, James.Hatridge@epost.de says... > HI all, > > In the code below I'm trying to get the last column to show 1, 2, 3, or 4= > =20 > according to which quarter of the year it is. But all it shows in that co= > lumn=20 > is " Resource ID # X". The X starts with #3 and goes to 18. There are (at= > the=20 > moment) 15 items in the table. Any ideas what's wrong? > > Thanks > > JIM > > ############################# > <?php > echo "<table border=3D1> \n"; > $i=3D1; > while ($myrow =3D mysql_fetch_array($result)) { > if($i % 2) { //this means if there is a remainder > echo "<TR bgcolor=3D\"yellow\">\n"; > } else { //if there isn't a remainder we will do the else > echo "<TR bgcolor=3D\"white\">\n"; > } > $qdate=3D$myrow["date"]; > $sql =3D "select quarter($qdate)" or die("not work #3"); > $yyy =3D mysql_query ($sql) or die("not work #4"); > > printf("<td><a href=3D\"%s?id=3D%s&delete=3Dyes\">Delete</a></td>", $PHP_= > SELF,=20 > $myrow["id"]); > printf("<td><a href=3D\"%s?id=3D%s&submit=3Dyes\">Update</td><td>%s</td><= > td> =20 > %s</td><td> %s</td></a></tr>",=20 > =09"update-inv.php", $myrow["id"], $myrow["name"], $myrow["details"], $yy= > y); You are echoing/printing $yyy here, which is the mysql result pointer; you need to use mysql_fetch_array or similar to get the value of the quarter. Hint - you might want to use an alias in the query where you select the quarter. > $i=3D$i+1; > } > echo "</table>\n"; > } > ?> > ############################# > -- David Robley Temporary Kiwi! Quod subigo farinam -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
