Please post the table definition to the mailing list so we can help you
more quickly. Also, is there a table called RuneRunner_1 or are you
meaning to use RuneRunner_1 as an alias?
Odds are, your problem is not PHP related, it's an error in your SQL syntax.
Jon
water_foul wrote:
i fixed those things and it didn't fix it :( :( :( :( :( :( :( :( :( is
there a icq chatroom for php?)
"Shahmat Bin Dahlan" <shahmatd@xxxxxxxxxxxx> wrote in message
news:16780f91a.f91a16780@xxxxxxxxxxxxxxxxxxxxxxxxx
Your SQL statement:
'SELECT Rune, username FROM RuneRunner
RuneRunner_1 WHERE (User_ID = 3)'
What is "RuneRunner" and "RuneRunner_1"? Are these two different tables.
If it is, you might want to separate them with a comma.
Why not try getting rid of the brackets surrounding "User_ID=3"? And
also wrap single quotes around the digit "3".
----- Original Message -----
From: "water_foul" <phpdbnews@xxxxxxxxxxxxxxxxxxx>
Date: Wednesday, June 23, 2004 8:58 am
Subject: Re: problem....
I checked em all they were right
"Daniel Clark" <dclark@xxxxxxxxxx> wrote in message
news:18323.65.203.232.102.1087951969.squirrel@xxxxxxxxxxxxxxxxxxxxxxx
Sounds like it doesn't like your SQL statement. Perhaps a field
or table
name is incorrect?
Warning: mysql_fetch_array(): supplied argument is not a valid
MySQL> > result
resource in .......
"Daniel Clark" <dclark@xxxxxxxxxx> wrote in message
news:16692.65.203.232.102.1087950853.squirrel@xxxxxxxxxxxxxxxxxxxxxxx>
What error are you getting?
why doesn't this work:
$pic=mysql_query('SELECT Rune, username FROM
RuneRunner> >> > RuneRunner_1 WHERE (User_ID = 3)',$connection);
$pic=mysql_fetch_array($pic);
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