[英]SQL: Select records where ALL joined records satisfy some condition
How can I write an SQL query that returns a record only if ALL of the associated records in a joined table satisfy some condition. 只有当连接表中的所有关联记录满足某些条件时,如何编写仅返回记录的SQL查询。
For example, if A has many B, I want to SELECT * FROM A WHERE all related B's for a given A have B.some_val > value 例如,如果A有很多B,我想SELECT * FROM A WHERE给定A的所有相关B都有B.some_val>值
I know this is probably a pretty basic question, so thanks for any help. 我知道这可能是一个非常基本的问题,所以感谢您的帮助。 Also, if it makes a difference, I'm using postgres.
此外,如果它有所作为,我正在使用postgres。
Sam 山姆
Assuming no need for correlation, use: 假设不需要关联,请使用:
SELECT a.*
FROM A a
WHERE EXISTS(SELECT NULL
FROM B b
HAVING MIN(b.some_val) > a.val)
If you do need correlation: 如果你确实需要相关性:
SELECT a.*
FROM A a
WHERE EXISTS(SELECT NULL
FROM B b
WHERE b.id = a.id
HAVING MIN(b.some_val) > a.val)
The EXISTS
evaluates on a boolean, based on the first match - this makes it faster than say using IN, and -- unlike using a JOIN -- will not duplicate rows. EXISTS
根据第一个匹配对布尔值进行求值 - 这比使用IN更快,而且 - 与使用JOIN不同 - 不会重复行。 The SELECT portion doesn't matter - you can change it to EXISTS SELECT 1/0 ...
and the query will still work though there's an obvious division by zero error. SELECT部分无关紧要 - 您可以将其更改为
EXISTS SELECT 1/0 ...
并且查询仍然有效,尽管存在明显的零错误除法。
The subquery within the EXISTS
uses the aggregate function MIN to get the smallest B.some_val - if that value is larger than the a.val value, the a.val is smaller than all of the b values. EXISTS
的子查询使用聚合函数MIN来获得最小的B.some_val - 如果该值大于a.val值,则a.val小于所有b值。 The only need for a WHERE
clause is for correlation - aggregate functions can only be used in the HAVING
clause. WHERE
子句的唯一需要是用于关联 - 聚合函数只能在HAVING
子句中使用。
select * from A
where -- at least one record in B is greater than some_val
exists (select null from B
where B.some_val > :value
and A.join_column = B.join_column)
and -- no records in B are not greater than some_val
not exists (select null from B
where B.some_val <= :value
and A.join_column = B.join_column)
The following should work: 以下应该有效:
SELECT *
FROM a
JOIN b ON a.key = b.key AND a.value > b.value
Because this does an inner join and not a outer join , records from A will only be included if they have records in B that satisfy the condition. 因为它执行内连接而不是外连接 ,所以只有在B中满足条件的记录时,才会包含A中的记录。
I don't use PostGRE, so I can't guarantee that the syntax is exactly correct. 我不使用PostGRE,所以我无法保证语法完全正确。
You are wanting an INNER JOIN: 你想要一个INNER JOIN:
SELECT
A.*
FROM
A
INNER JOIN B
ON A.identifier = B.identifier
WHERE
B.some_val > value
You will want to ensure that there is a foreign key from A to B, or some other common identifier. 您需要确保存在从A到B的外键或其他常用标识符。
select * from a where a.key = b.a_key where b.value > condition
Use bookstores and books as an example. 以书店和书籍为例。
bookstoreID, bookID bookstoreID,bookID
bookID, price bookID,价格
I suppose you want to return all the bookstores in which all the books have a price greater than X. 我想你想要归还所有书籍价格都高于X的书店。
select *
from Bookstore bs1
where bs1.bookstoreID not exist
(
select bs.bookstoreID
from Bookstore bs, Book b
where bs.bookID= b.bookID
b.price < x; -- your value
)
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