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Dec 21st, 2003, 10:01 AM
#1
Thread Starter
Hyperactive Member
mySql Query **SOLVED**
Hey all, it been a while since my last php project, and I guess I frogot everything........
I am trying to get a result from db in mySql. But i always get this error Message:
Warning: Supplied argument is not a valid MySQL result resource in ...
Heres the relevant code:
PHP Code:
$queryTeams = "SELECT * FROM teams WHERE Group = 'A'";
$resultTeams = mysql_query($queryTeams);
while($Team = mysql_fetch_array($resultTeams)) // this is where the error is pointing
Thanks
Stephan
Last edited by Sgt-Peppa; Dec 21st, 2003 at 11:06 AM.
Keep Smiling - even if its hard 
Frankie Says Relax, wossname Says Yeah!
wossname:--Currently I'm wearing a gimp suit and a parachute.
C# - Base64 Blog
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Dec 21st, 2003, 11:07 AM
#2
Thread Starter
Hyperactive Member
myQuery ws wrong
SELECT * FROM `teams` WHERE `Group` = 'A'
that one works
Finally,
Thanks,
Stephan
Keep Smiling - even if its hard 
Frankie Says Relax, wossname Says Yeah!
wossname:--Currently I'm wearing a gimp suit and a parachute.
C# - Base64 Blog
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