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sql查询 次数

select Name as '姓名', count(base) as '咨询次数' from A inner join B on A.a = B.b where Date between '2015-06-15 00:00:00' and '2015-06-21 23:59:59' and qykf='签约客户' and Name in ('李雷','韩梅梅') group by Name order by Name;

我觉得好像不用楼上那么复杂吧? 1 select count(*) from (select distinct 列名 from 表名) 子查询中使用distinct查询出所有不同的值,然后用select count(*)查询子查询返回到行数。

select count(name) ,name from table group by name having count(name)>1

select B, count(*) -- 计算个数,速度快于count(字段). 因为 count(字段)会先判断这个字段是否为空,只计算非空的个数。所以这里直接用count(*)。 from A group by B

select [产品类型], count([产品类型]), [仓库编号] from [表名称] where [仓库编号] = '01' group by [产品类型], [仓库编号]

特别推荐: select xingming , count(*) from biao group by xingming having count(*) > 2

遍历所有表?你的需求是什么? 如果只是查一个表中所有人名出现的次数,可以用以下查询。 SELECT name ,COUNT(NAME) FROM a group by name order by name

需要用group by中的having子句。 1、如test表中有如下数据: 2、现要查出dept出现2次以上的dept。可用如下语句: select dept from test group by dept having count(*) >2;3、查询结果:

select 字段,count(*) from 表明 group by 字段

使用count函数即可。 例如: 表名test id name 1 a 1 b 1 f 2 c 2 d 3 e select id from test group by id having count(*)=(select top1 count(*) as counts from test group by id order by counts desc)

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