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mysql_result should be used once for a variable. Then the assigned var becomes an ordinary variable and you can use it WITHOUT mysql_result.. So you used
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So replace line echo mysql_result($numberguest,0); to echo $numberguest; BTW you else syntax is wrong either: it should be: PHP Code:
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I think MAYBE it's because of $cookietimeout variable... it's not defined any places... what do you think?
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]well... maybe I'm tired :) maybe... but I have to get it working... :)
by now I have no errors anymore! YAY! but visual results are: Active Users 0 Active Members Active Guests 0 And forum tells me this: Active Users 46 Active Members 24 Active Guests 22 So... maybe you are tired too :) Because Active Users and Active Guests working only at 50% (show wrong results) and Active Members doesn't show anything at all... :( grrrrrrrrrrrr... why should everything be so difficult? :/ |
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]you're the man, Logician! :) worked perfectly...
as you see, it's not easy to get it working... u have to use different ways... well... that's PHP... anyway, THANX A LOT!!! I really appreciate your help :) |
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