Question

我有2个表Mutations和Medications。

我需要在医学名称中同时拥有mut1和mut2的那些成员的数量

SELECT 
    med.Name AS medicine, COUNT(*) AS count
FROM 
    Mutations mut
INNER JOIN 
    Medications med ON med.MemberId = mut.MemberId
WHERE 
    mut.Mutation IN ('mut1','mut2')
GROUP BY 
    med.Name
HAVING 
    COUNT(DISTINCT mut.Mutation) = 2

如何获取同时具有突变的成员的计数？

Answer 1

一种方法是两种聚合级别：

select med.name, count(*)
from (select med.name, m.memberid
      from medications med join
           mutations m
           on m.memberid = med.memberid
      where m.mutation in ('mut1', 'mut2')
      group by med.name, m.memberid
      having count(*) = 2
     ) m
group by med.name;

Answer 2

目前尚不清楚您需要什么计数，您的查询实际上是在获取具有两种突变的药物...

如果您希望对给定查询进行计数，请使用子查询

SELECT 
    med.Name AS medicine, 
    COUNT(CASE Mutation WHEN 'Mut1' THEN 1 END) AS CountMut1
    COUNT(CASE Mutation WHEN 'Mut2' THEN 1 END) AS CountMut2
FROM 
    Mutations mut
INNER JOIN 
    Medications med ON med.MemberId = mut.MemberId
WHERE 
    mut.Mutation IN ('mut1','mut2')
GROUP BY 
    med.Name
HAVING 
    COUNT(DISTINCT mut.Mutation) = 2

如果您想将Mut1和mut2用例的数量分开

JSON.parse

SQL：获取具有两个突变的数据计数

2 个答案: