male and female online

[brand]

dont want ur crappy coding lol i said the basic v2 script that u can download ere. U got ur own server but u cant even code seperate pages 4 for male and female online goodluck

the plain v2 is on here go look for it and i got the pages i need thank you very much i just having problems getting it to show things

i have now fixed this thanks to everyone who helped please close this now thanks

[fullcircle]

dont want ur crappy coding lol i said the basic v2 script that u can download ere. U got ur own server but u cant even code seperate pages 4 for male and female online goodluck

[brand]

if u wana setup a free hosting account and upload the basic v2 script i can help ya, cant b assed 2 do it myself lol

fullcircle do u think i am daft i have my own server and my coding is staying there thanks

[fullcircle]

if u wana setup a free hosting account and upload the basic v2 script i can help ya, cant b assed 2 do it myself lol

[brand]

thanks ori

[ori]

looks like you need to join 3 tables in the query ill get it soon for you lol

[phzone]

great coding mate......

[brand]

hmm.....

$noi=mysql_fetch_array(mysql_query("SELECT COUNT(online.userid) FROM online LEFT JOIN profiles ON online.userid = profile.id WHERE profile.sex='M'"));

thats does not work thanks for helping me gunslone

[GumSlone]

hmm.....

$noi=mysql_fetch_array(mysql_query("SELECT COUNT(online.userid) FROM online LEFT JOIN profiles ON online.userid = profile.id WHERE profile.sex='M'"));

[brand]

can anyone help me with this please

[brand]

oops sorry .... yeah needs to be JOIN .... whats the error you getting when it was the other way?

the error is where its trying to pick up just male and female members so eg:

male members:

tom
micky
harry

female online:
joe
lisa
wendy

i want it to show it on two pages

so this bit is trying to get just the female members

    $noi=mysql_fetch_array(mysql_query("SELECT COUNT(DISTINCT a.userid) FROM online JOIN  profiles ON a.userid = b.id WHERE sex='F'"));

and this is to try and get male members

    $noi=mysql_fetch_array(mysql_query("SELECT COUNT(DISTINCT a.userid) FROM online JOIN  profiles ON a.userid = b.id WHERE sex='M'"));

[something else]

oops sorry .... yeah needs to be JOIN .... whats the error you getting when it was the other way?

[brand]

is that instead of a JOIN mate

this is the error message i got

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/*************/public_html/online2/tester4.php on line 38

[something else]

Inner join

(but in caps coding talk wont let me write it in caps :/ )

[brand]

can some one please help me i am trying to sort out so my online list can pick up both male and female online but on two diffrent pages

 $male=mysql_fetch_array(mysql_query("SELECT COUNT(DISTINCT a.userid) FROM online a JOIN profiles b ON a.userid = b.id WHERE b.sex='M'"));
  $main.="Male Online:<a href=\"online.php?action=male&amp;sid=$sid\">".$male[0]."</a><br/>";

  $female=mysql_fetch_array(mysql_query("SELECT COUNT(DISTINCT a.userid) FROM online a JOIN profiles b ON a.userid = b.id WHERE b.sex='F'"));
  $main.="Female Online:<a href=\"online.php?action=female&amp;sid=$sid\">".$female[0]."</a><br/>";

i keep getting errors on

 $male=mysql_fetch_array(mysql_query("SELECT COUNT(DISTINCT a.userid) FROM online a JOIN profiles b ON a.userid = b.id WHERE b.sex='M'"));
 
  $female=mysql_fetch_array(mysql_query("SELECT COUNT(DISTINCT a.userid) FROM online a JOIN profiles b ON a.userid = b.id WHERE b.sex='F'"));

can someone please help me with this i am trying to do it for the wapdesire v_2
please

thanks to who ever helps me with it