RE: [PHP-DB] Quarter question..
$sql = select quarter($qdate) or die(not work #3); change to $sql = select quarter($qdate) as my_quarter; //Added an alias to quarter($qdate). Easier to access that way... And there's no need for an 'or die()' here. You're just assigning a variable and most likely this will allways work! $yyy = mysql_query ($sql) or die(not work #4); change to $yyy = mysql_query ($sql) or die(not work #4: . mysql_error()); //Standard MySQL trouble-shooting... printf(tda href=\%s?id=%sdelete=yes\Delete/a/td, $PHP_SELF, $myrow[id]); printf(tda href=\%s?id=%ssubmit=yes\Update/tdtd%s/tdtd %s/tdtd %s/td/a/tr, update-inv.php, $myrow[id], $myrow[name], $myrow[details], $yyy); and here you actually echo out the resource id instead of it's value... change $yyy to $yyy[my_quarter] HTH Joakim -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Quarter question..
printf(tda href=\%s?id=%sdelete=yes\Delete/a/td, $PHP_SELF, $myrow[id]); printf(tda href=\%s?id=%ssubmit=yes\Update/tdtd%s/tdtd %s/tdtd %s/td/a/tr, update-inv.php, $myrow[id], $myrow[name], $myrow[details], $yyy); My bad. Before this printf statement you need $my_var = mysql_fetch_array($yyy); and then in the printf statement change $yyy to $my_var[my_quarter] That should work... Regards Joakim -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Quarter question..
On Monday 04 November 2002 19:58, [EMAIL PROTECTED] wrote: $sql = select quarter($qdate) or die(not work #3); change to $sql = select quarter($qdate) as my_quarter; //Added an alias to quarter($qdate). Easier to access that way... And there's no need for an 'or die()' here. You're just assigning a variable and most likely this will allways work! On the contrary. If you've made a mistake in your query you wouldn't know what's happening. So having the below is a very good idea. $yyy = mysql_query ($sql) or die(not work #4: . mysql_error()); //Standard MySQL trouble-shooting... -- Jason Wong - Gremlins Associates - www.gremlins.com.hk Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* Ask not what's inside your head, but what your head's inside of. -- J.J. Gibson */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Quarter question..
On Monday 04 November 2002 19:58, [EMAIL PROTECTED] wrote: $sql = select quarter($qdate) or die(not work #3); change to $sql = select quarter($qdate) as my_quarter; //Added an alias to quarter($qdate). Easier to access that way... And there's no need for an 'or die()' here. You're just assigning a variable and most likely this will allways work! On the contrary. If you've made a mistake in your query you wouldn't know what's happening. So having the below is a very good idea. $yyy = mysql_query ($sql) or die(not work #4: . mysql_error()); //Standard MySQL trouble-shooting... Hi Jason, I might have been unclear in my reply. The original code had an 'or die()' on $sql = select... aswell and that is, in my opinion, rather unnecessary. Using 'or die(mysql_error())' on mysql_query should be mandatory. At least during development. Regards Joakim -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Quarter question..
On Monday 04 November 2002 22:56, [EMAIL PROTECTED] wrote: I might have been unclear in my reply. Sorry, I was following this thread on and off and probably misinterpreted your post. The original code had an 'or die()' on $sql = select... aswell and that is, in my opinion, rather unnecessary. Of course ;-) Using 'or die(mysql_error())' on mysql_query should be mandatory. At least during development. That should be drummed into people's heads from the word go! I've made a request in the users' comment of the mysql section of the manual hoping that the example code is changed so that instead of just die(), it dies with a mysql_error(). -- Jason Wong - Gremlins Associates - www.gremlins.com.hk Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* Maybe you can't buy happiness, but these days you can certainly charge it. */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Quarter question..
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(tda href=\%s?id=%sdelete=yes\Delete/a/td, $PHP_SELF,
$myrow[id]);
printf(tda href=\%s?id=%ssubmit=yes\Update/tdtd%s/tdtd
%s/tdtd %s/td/a/tr,
update-inv.php, $myrow[id], $myrow[name], $myrow[details], $yyy);
$i=$i+1;
}
echo /table\n;
}
?
#
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