按列顺序选择前三个值[重复]

Question

我有 2 张桌子：

医学

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));

即

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

我需要为每个规范选择全名和 count(id_p) 的前 3 个值。 以下查询给了我第一个值，但我只是看不出下一步该做什么。

  (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;

任何答案将不胜感激。谢谢。

结果应该是这样的：

  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

....

我目前的结果显示：

    fullname       count(id_p)        spec
    name1              1000            a
    name4              2000            b
    name7               100            c

Answer 1

加入一个返回spec 列的前三个值的子查询。

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