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orAClE hAving Count

譬如如下数据 id value 1 2 1 3 2 3 3 5 3 6 可以写个语句统计value的分组 select id,sum(value) from table group by id having sum(value)>=5 这样的结果就是 1 5 3 11 其实这句的意思就是 select id,sum(value) from table where sum(value)>...

--1,查询hm有重复的记录select hm,count(*) from a group by hm having count(*)>1--2,查询hm和xm都有重复select hm,xm count(*) from a group by hm,xm having count(*)>1

select a,count(*) from table group by a having count(*)>1 嗯。是的。这样是查出来该字段重复2次以上的列。

61=C或D 62=C SELECT Sno , AVG(Grade) FROM SC GROUP BY Sno HAVING COUNT(Sno)>1;

select tabtype,count(*) from tab where onwer='abcde' group by tabtype having count(*) > 2

加个where 条件不行么,where status 0 如果不行的话,就写子查询, 先筛选where status 0 再group by

第一种方式: select distinct x, y from t; 第二种方式: select x,y from t group by x,y 推荐第二种方式

当然不是了 比如 select name,count(*) from tabname group by name having count(*) > 1 这个肯定无法改到where子句中的。

1) select * from tab t where exists (select 1 from tab where id = t.id and (name t.name or age t.age) 2) select * from tab where id in (select id from tab group by id having count(*) > 1)

用group by col_name 一下,然后having count(col_name) > 1。如果查出来数据,就是是有相同值

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