MySQL - 選擇基於一列中的重複會議需求的另一個

Question

對於兩列;人和衣服;

person clothing 
A   shirt 
B   shoes 
A   pants 
A   shoes 

如何選擇只有誰擁有這三種類型的服裝（上衣，褲子和鞋子）的人給予回覆：

person 
A 

編輯：我不知道是否有某種迭代可以做

即return person if for (i=0, i<number of rows, i++) all three types are found.

Answer 1

另一種解決方案（未例如花式作爲Gordon的一個）是利用某種MySQL的交點。

SELECT DISTINCT pc.person 
FROM personclothing pc 
INNER JOIN personclothing pc2 on pc.person = pc2.person AND pc2.clothing = 'shirt' 
INNER JOIN personclothing pc3 on pc.person = pc3.person AND pc3.clothing = 'pants' 
INNER JOIN personclothing pc4 on pc.person = pc4.person AND pc4.clothing = 'shoes' 

或使用基於評論IN

SELECT pc.person 
FROM personclothing 
WHERE person IN (SELECT person FROM personclothing WHERE clothing = 'shirt') AND 
     person IN (SELECT person FROM personclothing WHERE clothing = 'pants') AND 
     person IN (SELECT person FROM personclothing WHERE clothing = 'shoes') 

編輯。如果你需要有shirt和其中的一個人：pants或shoes

SELECT pc.person 
FROM personclothing 
WHERE person IN (SELECT person FROM personclothing WHERE clothing = 'shirt') AND 
    (  
     person IN (SELECT person FROM personclothing WHERE clothing = 'pants') OR 
     person IN (SELECT person FROM personclothing WHERE clothing = 'shoes') 
    ) 

Answer 2

一種方法是：

select person 
from personclothing 
where clothing in ('shirt', 'pants', 'shoes') 
group by person 
having count(distinct clothing) = 3; 

如果您沒有重複項，則使用count(*)而不是count(distinct)。

Answer 3

另一種方法是：

select 
person, 
count(distinct(clothing)) as clothing_count 
from 
personclothing 
where 
clothing in ('shirt', 'pants', 'shoes') 
group by 
person 
having 
clothing_count = 3;