SQL查询到Count()多个表

Question

我有一个表与其他表有几个一对多的关系.假设主表是一个人,其他表代表宠物,汽车和儿童.我想要一个查询,返回人的详细信息,宠物,汽车和孩子的数量,例如

Person.Name   Count(cars) Count(children) Count(pets)

John Smith    3           2               4
Bob Brown     1           3               0

做这个的最好方式是什么？

Answer 1

子查询因子(9i +):

WITH count_cars AS (
    SELECT t.person_id
           COUNT(*) num_cars
      FROM CARS c
  GROUP BY t.person_id),
     count_children AS (
    SELECT t.person_id
           COUNT(*) num_children
      FROM CHILDREN c
  GROUP BY t.person_id),
     count_pets AS (
    SELECT p.person_id
           COUNT(*) num_pets
      FROM PETS p
  GROUP BY p.person_id)
   SELECT t.name,
          NVL(cars.num_cars, 0) 'Count(cars)',
          NVL(children.num_children, 0) 'Count(children)',
          NVL(pets.num_pets, 0) 'Count(pets)'
     FROM PERSONS t
LEFT JOIN count_cars cars ON cars.person_id = t.person_id
LEFT JOIN count_children children ON children.person_id = t.person_id
LEFT JOIN count_pets pets ON pets.person_id = t.person_id

使用内联视图:

   SELECT t.name,
          NVL(cars.num_cars, 0) 'Count(cars)',
          NVL(children.num_children, 0) 'Count(children)',
          NVL(pets.num_pets, 0) 'Count(pets)'
     FROM PERSONS t
LEFT JOIN (SELECT t.person_id
                  COUNT(*) num_cars
             FROM CARS c
         GROUP BY t.person_id) cars ON cars.person_id = t.person_id
LEFT JOIN (SELECT t.person_id
                  COUNT(*) num_children
             FROM CHILDREN c
         GROUP BY t.person_id) children ON children.person_id = t.person_id
LEFT JOIN (SELECT p.person_id
                  COUNT(*) num_pets
             FROM PETS p
         GROUP BY p.person_id) pets ON pets.person_id = t.person_id

Answer 2

你可以使用COUNT(distinct x.id)synthax:

SELECT person.name, 
       COUNT(DISTINCT car.id) cars, 
       COUNT(DISTINCT child.id) children, 
       COUNT(DISTINCT pet.id) pets
  FROM person
  LEFT JOIN car ON (person.id = car.person_id)
  LEFT JOIN child ON (person.id = child.person_id)
  LEFT JOIN pet ON (person.id = pet.person_id)
 GROUP BY person.name