Question

我有两张桌子：

MED

create table t1(id int,cust int, [order] int)
go
create table t2(id int,cust int,parts1 int,parts2 int)
go

insert t1 values(1,1,1)
insert t1 values(2,1,1)
insert t1 values(3,1,1)
insert t1 values(4,2,1)
go
insert into t2 values(1,1,5,3)
go

select * from t1 
select * from t2 

select a.cust ,a.orders,isnull(b.parts1,0) +isnull(b.parts2,0)   from 
    (select cust ,SUM([order] ) as orders from t1 group by cust  ) a
    left join t2 b on a.cust =b.cust

即

CREATE TABLE med 
(id_med INT AUTO_INCREMENT PRIMARY KEY,
name_m VARCHAR(50),
surname_m VARCHAR(50),
spec VARCHAR(50),
UNIQUE (surname_m,name_m));

我需要为每个规范选择fullname和count（id_p）的前3个值。以下查询给出了第一个值，但我不知道下一步该做什么。

CREATE TABLE viz
(id_m INT NOT NULL,
id_p INT  NOT NULL,
id_c INT  NOT NULL,
dt DATETIME)

任何答案都将受到高度赞赏。谢谢。

结果应该是这样的：

  (SELECT distinct(concat(name_m,' ',surname_m)) AS fullname, COUNT(id_p) as no, (select distinct(spec))
    FROM med m JOIN viz v ON m.id_med = v.id_m
    WHERE year(dt)=2005
    GROUP BY spec
    ORDER BY spec;

...

我目前的结果显示：

  fullname       count(id_p)          spec
    name1              1000            a
    name2               900            a
    name3               890            a
    name4              2000            b
    name5               600            b
    name6               200            b
    name7               100            c
    name8                60            c
    name9                59            c

Answer 1

加入一个子查询，该子查询返回spec列的前三个值。

SELECT DISTINCT CONCAT(name_m, ' ', 'surname_m) AS fullname, COUNT(id_p) AS no, m1.spec
FROM med AS m1
JOIN viz AS v ON m1.id_med = v.id_m
JOIN (SELECT DISTINCT spec
      FROM med
      ORDER BY spec
      LIMIT 3) AS m2 ON m1.spec = m2.spec
WHERE YEAR(v1.dt) = 2005
GROUP BY fullname, m1.spec
ORDER BY fullname, m1.spec

选择前三个值按列排序

1 个答案: