I think here's your problem:
$query="INSERT INTO game_of_the_day VALUES
('',$curr_yday,'$gameone_genre',$gameone_number,'$gametwo_genre',$gametwo_nu
mber)";
should look like :
$query="
INSERT INTO
game_of_the_day
(ID, curr_yday, gameone_genre, gameone_number, gametwo_genre,
gametwo_number)
VALUES
('',$curr_yday,'$gameone_genre',$gameone_number,'$gametwo_genre',$gametwo_nu
mber)
";
try it.
Sincerely,
Maxim Maletsky
Founder, Chief Developer
PHPBeginner.com (Where PHP Begins)
[EMAIL PROTECTED]
www.phpbeginner.com
-----Original Message-----
From: Brian Rue [mailto:[EMAIL PROTECTED]]
Sent: Thursday, April 26, 2001 12:24 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP] MySQL problem...
Here's all the code that uses MySQL...
$db = mysql_connect("localhost","user","pass");
mysql_select_db("db",$db);
$gmdquery="SELECT * FROM game_of_the_day";
$the_info = mysql_query($gmdquery,$db);
while ($myrow = mysql_fetch_row($the_info)) {
(get info from the result)
}
... (decide whether or not to conduct the following operation)
if (true) {
$query="SELECT id FROM games WHERE rating >= 7";
$result=mysql_query($query,$db);
$numgames=mysql_num_rows($result);
$z=0;
while ($row=mysql_fetch_row($result)){
$gotd_cand[$z]=$row[0];
$z++;
}
(at this point, i randomly select 2 games from the db)
$query="SELECT genre,number FROM games WHERE id=$game1_to_get";
$gameinfo=mysql_query($query,$db);
while($row=mysql_fetch_row($gameinfo)){
(use the result)
}
(do the same thing as before, but for the second game)
}
(update the db)
$query="DELETE FROM game_of_the_day";
$result=mysql_query($query,$db);
$query="INSERT INTO game_of_the_day VALUES
('',$curr_yday,'$gameone_genre',$gameone_number,'$gametwo_genre',$gametwo_nu
mber)";
$result=mysql_query($query,$db);
}
Keep in mind that this only happens some of the time... sometimes it works,
and sometimes it just doesn't.
Today, I noticed that it stored the first game into the db twice (the code
doesn't allow for the same game to be selected twice...)
Thanks for your time
""Peter Houchin"" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> some code would be nice to have a look at :)
>
> Other than that, check table names, database names, also your result
lines, I've found i get that error by not calling a result or calling the
incorrect table/database
>
> Peter
>
> -----Original Message-----
> From: Brian Rue [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, April 26, 2001 10:28 AM
> To: [EMAIL PROTECTED]
> Subject: [PHP] MySQL problem...
>
>
> MySQL doesn't like me....
>
> Sometimes, my pages that connect to the database get the error "Warning:
> Supplied argument is not a valid MySQL result resource..." repeated over
and
> over again (something like 1000 times...)
>
> What's causing this error? Obviously, PHP isn't getting a result back from
> MySQL... and it keeps trying to get it.
>
> Any help?
>
>
> Thanks,
> Brian Rue
>
>
>
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>
>
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