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mysql group Count

可直接用嵌套查询。 方法如下: 如数据: id name 1 a 1 b 2 c 2 d 3 e 3 f 3 g 3 h select count(t.counts) from(select id,count(*) counts from 表名 group by id) t这样得到的结果就是3。

select count(*) from p_score_sx_zhishanglihepiao21024 a where user_id in (SELECT user_id, COUNT(*) AS cnt FROM p_score_sx_zhishanglihepiao21024 WHERE create_dt LIKE '2010-05-06%' GROUP BY user_id HAVING cnt >= 100) 就是效率低了...

这个是把数据表按a2 分组,统计买家买东西的天数,distinct 去重复 select UID,A2,SUM(JF),COUNT(distinct DATE_FORMAT(A23,'%Y-%m-%d')) FROM taobao GROUP BY A2 ORDER BY a23 ASC

select IFNULL(count1,0) from dual; 用 ifnull(a1,a2)函数 a1为你count的值,a2为你需要输出的值,count函数会输出null? 不会吧

select uid,count(*), (select count(*) from b where b.newid=a.newid) , (select count(*) from c where c.neid=a.neid and c.type='Student')

打印结果 array (size=2) 'count(*)' => string '25' (length=2) 'dtime' => string '2015-06-20' (length=10) 竟然是string型,保持观望

group by 按照字段中取出值判断是不是相同的组就可以了 order by 排序按照字母、文字、数字排序 count 将结果集放到集合看集合有多大 但是效率在多数情况下不如直接在数据库里

select max(cc.数量) from (SELECT COUNT(*) as 数量 FROM fn_fixed GROUP BY CODE) cc 应该这么用

SELECT category_Sn,isnull(count,0) FROM category left join (SELECT count(*) as count,category_Sn FROM [article] group by category_Sn) AS [article] on category.[category_Sn]=article.[category_Sn] 一、COUNT()函数 COUNT()函数进行...

select count(if(isCheck=1,1,0)), count(if(isPay=1,1,0)),count(id) from table group by group;

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