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SQL Count 多表总数求和问题

可以这样写: select (select count(*) from bumen)+(select count(*) from mrs) as sum_count 如果两个表结果相同的话,也可以用 select count(*) as sum_count from ( select * from A union all select * from B ) as tmp

在最后的地方 随便加个字符, ) a就可以了。缺少一个表名,相当于。 select sum(tmpcount) from ( select count(*) as tmpcount from tab1 union all select count(*) as tmpcount from tab2) a具体如下: 1、简介 结构化查询语言(Structured Qu...

select a. col1, b.col2 from (select count(id) as col1 from table1) as a, (select count(id) as col2 from table2) as b -------------这样写。

可以有两种解决方法。 方法1: SELECT paperName , COUNT (1) AS 总题数 , sum (CASE WHEN statu = 1 THEN 1 ELSE 0 END) AS 审核题数 FROM question GROUP BY paperNme 方法2: select s.总题数, s.审核题数, s.paperName from ( select COUNT(...

办法是有,不过要自己在数据库上写程序通过写游做标循环来完成。 相当于在20多个表之间做循环,没循环一次,就有一个count(*),这样放到一个变量中做累加,最终肯定能得到你要的结果。 一个SELECT语句肯定是无法实现的,因为一张表就涉及到一个S...

有2个方法: 1、select *,(select count(u_id) from h_travel_line where u_id = a.u_id) 已线路数 from h_travel a 2、SELECT c.ttt AS 已线路数, h_travel.* FROM h_travel INNER JOIN (SELECT COUNT(*) AS ttt, u_id FROM h_travel_line GROU...

首先这个应该是外连接查询。你知道你错在何处吗?你要查询所有的news,为什么还要限定n.newsid=@ID,你这样查询的只有一条记录了。 select top 10 n.newid,count(c.*) from new as n left jion conmment as c on n.newid=c.newid group by n.newid...

declare @t table (hm varchar(10),fy int) insert @t values ('23456',12) insert @t values ('56423',13) insert @t values ('56321',15) insert @t values ('89546',25) insert @t values ('78965',85) insert @t values ('56789',88) select...

SELECT DB1.A,SUM(DB2.A) as B FROM DB1 LEFT JOIN DB2 ON DB1.B=DB2.B GROUP BY DB1.A 这是没有去重的,看你db1表数据有重复的,假如想去重的话可以加上DISTINCT SELECT DISTINCT DB1.A,SUM(DB2.A) as B FROM DB1 LEFT JOIN DB2 ON DB1.B=DB2.B...

可以有两种解决方法, 所需工具:SQL 查询两个count的方法1: SELECT paperName , COUNT (1) AS 总题数 , sum (CASE WHEN statu = 1 THEN 1 ELSE 0 END) AS 审核题数FROM questionGROUP BY paperNme 查询两个count的方法2: select s.总题数, s.审...

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