我想知道'member_seq'的计数以匹配'R_INFO.nick_name'的值。
如何获取计数数据?
SQL
SELECT
nick_name,
REPORT.seq_no,
report_type,
report_item,
contents,
DATE_FORMAT(report_dt, '%Y-%m-%d %H:%i:%s') AS report_dt,
REPORT.status,
question,
E_INFO.status AS game_status,
entry_fee,
REPORT.game_seq,
Count(member_seq)
FROM
REPORT
LEFT JOIN R_INFO ON
REPORT.member_seq = R_INFO.seq_no
LEFT JOIN E_INFO ON
REPORT.game_seq = E_INFO.seq_no
GROUP BY R_INFO.nick_name;
答案 0 :(得分:0)
我不能保证您的JOIN逻辑是否正确,但是要纠正错误,如其文字所示,您需要删除所有未汇总或分组的列。所有这些:
REPORT.seq_no,
report_type,
report_item,
contents,
DATE_FORMAT(report_dt, '%Y-%m-%d %H:%i:%s') AS report_dt,
REPORT.status,
question,
E_INFO.status AS game_status,
entry_fee,
REPORT.game_seq
查询将变为:
SELECT nick_name, Count(member_seq)
FROM REPORT
LEFT JOIN R_INFO ON REPORT.member_seq = R_INFO.seq_no
LEFT JOIN E_INFO ON REPORT.game_seq = E_INFO.seq_no
GROUP BY R_INFO.nick_name;
答案 1 :(得分:0)
我因为英语不好而感到困惑。我很努力地找出并以这种方式解决。
SELECT
nick_name,
REPORT.seq_no,
report_type,
report_item,
contents,
DATE_FORMAT(report_dt, '%Y-%m-%d %H:%i:%s') AS report_dt,
REPORT.status,
question,
E_INFO.status AS game_status,
entry_fee,
REPORT.game_seq,
(SELECT COUNT(member_seq) REPORT WHERE member_seq = R_INFO.seq_no) AS count_member,
FROM
REPORT
LEFT JOIN R_INFO ON
REPORT.member_seq = R_INFO.seq_no
LEFT JOIN E_INFO ON
REPORT.game_seq = E_INFO.seq_no