현재접속자 관련 부분 질문좀 드릴게요.. 정보
현재접속자 관련 부분 질문좀 드릴게요..본문
//회원 이름 정보출력
$sql = " select a.mb_id, mb_sex, b.mb_nick, b.mb_name, b.mb_email, b.mb_homepage, b.mb_open, a.lo_ip, a.lo_location, a.lo_url
from $g4[login_table] a left join $g4[member_table] b on (a.mb_id = b.mb_id)
where a.mb_id <> '$config[cf_admin]'
GROUP BY a.mb_id
HAVING cnt >=1
order by a.lo_datetime desc";
$sql = " select a.mb_id, mb_sex, b.mb_nick, b.mb_name, b.mb_email, b.mb_homepage, b.mb_open, a.lo_ip, a.lo_location, a.lo_url
from $g4[login_table] a left join $g4[member_table] b on (a.mb_id = b.mb_id)
where a.mb_id <> '$config[cf_admin]'
GROUP BY a.mb_id
HAVING cnt >=1
order by a.lo_datetime desc";
이 소스가 현재접속자 제 접속자 스킨인데요...
오류가 나네요 ㅜ.ㅜ
select a.mb_id, mb_sex, b.mb_nick, b.mb_name, b.mb_email, b.mb_homepage, b.mb_open, a.lo_ip, a.lo_location, a.lo_url from g4_login a left join g4_member b on (a.mb_id = b.mb_id) where a.mb_id <> 'fhgjhh' GROUP BY a.mb_id HAVING cnt >=1 order by a.lo_datetime desc
1054 : Unknown 칼럼 'cnt' in 'having clause'
error file : /gnuboard4/1index.php
이런 오류가 나요 ..
어느 부분이 잘못된것인지 지적좀 해주세요 ..
댓글 전체
$sql = " select a.mb_id, count(*) cnt, mb_sex, b.mb_nick, b.mb_name, b.mb_email, b.mb_homepage, b.mb_open, a.lo_ip, a.lo_location, a.lo_url
from $g4[login_table] a left join $g4[member_table] b on (a.mb_id = b.mb_id)
where a.mb_id <> '$config[cf_admin]'
GROUP BY a.mb_id
HAVING cnt >=1
order by a.lo_datetime desc";
from $g4[login_table] a left join $g4[member_table] b on (a.mb_id = b.mb_id)
where a.mb_id <> '$config[cf_admin]'
GROUP BY a.mb_id
HAVING cnt >=1
order by a.lo_datetime desc";
앗 감사 합니다 로로님 행복한 주말 되세요^^
