摘要: SELECT b.level, a.cnt FROM (SELECT `level`,COUNT(*) AS cnt FROM wlsjlog GROUP BY level) a LEFT JOIN `levelName` b ON a.level = b.id; SELECT `behavirou 阅读全文
posted @ 2016-11-15 19:05 ComeComeGo 阅读(98) 评论(0) 推荐(0) 编辑