select DATE_FORMAT(f.upload_time,'%Y%u') weeks,count(*),sum(p.download_times),sum(p.collection_times),sum(p.click_times) from file_base f left join file_property p on f.property_id = p.id group by weeks; select DATE_FORMAT(f.upload_time,'%Y-%m-%d') d
表TESTER 字段:id -- INT date -- TIMESTAMP 1.如何按年.月.日分组查询? select DATE_FORMAT(date,'%Y-%m-%d') time, count(*) count from TESTER group by year(date), month(date), day(date); 其中year().month().day()分别是提取date中的年.月.日字段. 2.时间分组查询的效率? 在不建立索引时,我100W行数据进行测试,用
1.GROUP BY 与聚合函数 2.GROUP BY 与 HAVING 3.GROUP BY 扩展分组 3.1.GROUP BY ROLLUP 3.2.GROUP BY CUBE 3.3.GROUP BY GROUPING SETS 4.GROUP BY 扩展函数 4.1.GROUPING 函数 4.2.GROUPING_ID 函数 5.本文小结 1.GROUP BY 与聚合函数 GROUP BY 是一种能将查询结果划分为多个行组的查询语句的子句,其目的通常是为了在每个组上执行一个或多个聚合运