Question

如果我有一个 1到N 的关系，那么products表是这样的：

  products
+----+------+
| id | name |
+----+------+
| 1  | foo  |
| 2  | bar  |
+----+------+

和options这样的表：

  options
+----+------------+--------+--------+
| id | product_id | option | active |
+----+------------+--------+--------+
| 10 | 1          | size   | true   |
| 11 | 1          | color  | false  |
| 12 | 2          | size   | true   |
| 13 | 2          | color  | true   |
+----+------------+--------+--------+

如何通过products获取具有所有选项的active = true？请注意，如果某个产品至少有一个active = false选项，那么我不希望该查询检索该产品。

在这种情况下，想要的结果将是产品2。

Answer 1

您可以使用子查询从选项表中获取所有活动的product_ids，使用Postgres的布尔聚合bool_and，如果所有值均为true

，则仅返回true

select *
from products
where id in (select product_id
             from options
             group by product_id
             having bool_and(active));

在线示例：http://rextester.com/ILPX66228

Answer 2

您只需使用not exists运算符

即可

select p.* from  products p
where not exists (select 1 from options where product_id = p.id and active = false)

这假定active具有boolean类型，否则您需要使用单引号（即'string'）进行比较

Answer 3

我猜你正在寻找这个

SELECT  P.id, P.name, O.option, O.active from products P
INNER JOIN options O
ON P.id = O.product_id
WHERE active <> 'false'

Answer 4

这对你有用......

Select distinct P.name from products P 
join options O on O.id=P.id
where O.active=true

Answer 5

您可以将GROUP BY与HAVING子句一起使用条件聚合：

SELECT p.id, p.name
FROM products AS p
LEFT JOIN options AS o ON ON p.id = o.product_id
GROUP BY p.id, p.name
HAVING COUNT(CASE WHEN active = true THEN 1 END) = COUNT(*)

Answer 6

类似这样的事情

 select * from products 
 where id not in (select product_id from options where active = false)

Answer 7

将GROUP BY与HAVING子句一起使用：

SELECT pr.id, pr.name
FROM products AS pr
LEFT JOIN options AS op ON ON pr.id = op.product_id
GROUP BY pr.id
HAVING COUNT(CASE WHEN active = false THEN 1 else 0 END) <= 0

选择所有关系数据

7 个答案: