try this:
-js
<table border=1>
<?
$i = 0;
while ( $myrow = mysql_fetch_array($result) )
{
$c = ( ++$i % 2 ? 'yellow' : 'white' );
echo "<tr bgcolor=\"$c\">";
printf('<td><a href="%s?id=%s&delete=yes">Delete</a></td>',
$PHP_SELF, $myrow['id'] );
$q = ceil( 4/(int)date('n', strtotime($mydata['date']) ));
printf('<td><a
href="%s?id=%s&submit=yes">%s</a></td><td>%s</td><td>%s</td></tr>',
'update-inv.php', $myrow['id'], $myrow['name'], $myrow['details'], $q);
}
?>
</table>
Jim Hatridge wrote:
> HI all,
>
> In the code below I'm trying to get the last column to show 1, 2, 3, or 4
> according to which quarter of the year it is. But all it shows in that column
> is " Resource ID # X". The X starts with #3 and goes to 18. There are (at the
> moment) 15 items in the table. Any ideas what's wrong?
>
> Thanks
>
> JIM
>
> #############################
> <?php
> echo "<table border=1> \n";
> $i=1;
> while ($myrow = mysql_fetch_array($result)) {
> if($i % 2) { //this means if there is a remainder
> echo "<TR bgcolor=\"yellow\">\n";
> } else { //if there isn't a remainder we will do the else
> echo "<TR bgcolor=\"white\">\n";
> }
> $qdate=$myrow["date"];
> $sql = "select quarter($qdate)" or die("not work #3");
> $yyy = mysql_query ($sql) or die("not work #4");
>
> printf("<td><a href=\"%s?id=%s&delete=yes\">Delete</a></td>", $PHP_SELF,
> $myrow["id"]);
> printf("<td><a href=\"%s?id=%s&submit=yes\">Update</td><td>%s</td><td>
> %s</td><td> %s</td></a></tr>",
> "update-inv.php", $myrow["id"], $myrow["name"], $myrow["details"], $yyy);
> $i=$i+1;
> }
> echo "</table>\n";
> }
> ?>
> #############################
>
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