can you show the code around it? are you using single quotes in defining the
sql statement? or double quotes?
Bastien
From: "ReClMaples" <reclmaples@xxxxxxxxxxxxx>
To: "Jason Walker" <desktophero@xxxxxxx>
CC: "PHP" <php-db@xxxxxxxxxxxxx>
Subject: RE: Can someone help me out?
Date: Tue, 7 Jun 2005 19:00:59 -0500
Jason,
After looking at this a little more, the variable aren't being changed
into their value.
The echo looks like this:
select a.name,a.position,a.team,sum(b.yards) as Results from player
a,passing b where a.player_num = b.player_num and a.position = $pos and
b.year >= $year and a.team_div = $div group by a.name order by Results desc
limit '$display'
As you can see the variables aren't being changed, any ideas?
Thanks
-Rich
-----Original Message-----
From: Jason Walker [mailto:desktophero@xxxxxxx]
Sent: Tuesday, June 07, 2005 6:54 PM
To: 'ReClMaples'
Subject: RE: Can someone help me out?
In your PHP page, can you echo the actual query variable to the browser and
send that to the mail group?
I don't necessary see anything 'wrong' with your query but see the three
variables, as they are interpreted by PHP, may help.
Thanks,
-----Original Message-----
From: ReClMaples [mailto:reclmaples@xxxxxxxxxxxxx]
Sent: Tuesday, June 07, 2005 4:05 PM
To: PHP
Subject: Can someone help me out?
I am having a problem getting a sql statement to run.
I have this in a table called stat_categories, in a field called sql2
select a.name,a.position,a.team,sum(b.yards) as Results from player
a,passing b where a.player_num = b.player_num and a.position = '$pos'
and b.year >= '$year'
group by a.name
order by Results desc
limit '$display'
when I try to display this I get this error:
You have an error in your SQL syntax. Check the manual that corresponds to
your MySQL server version for the right syntax to use near ''$display'' at
line 8
I tried removing the ''s but this didn't help.
If I put in the sql instead of the variable, it works fine. What am I
doing
wrong?
Any help would be greatly appreciated.
Thanks
-Rich
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