Question

我陷入了SQL子查询选择。现在，我有一个表产品：

 id | name  |     description      
----+-------+----------------------
  6 | 123   | this is a           +
    |       | girl.
  7 | 124   | this is a           +
    |       | girl.
  8 | 125   | this is a           +
    |       | girl.
  9 | 126   | this is a           +
    |       | girl.
 10 | 127   | this is a           +
    |       | girl.
 11 | 127   | this is a           +
    |       | girl. Isn't this?
 12 | 345   | this is a cute slair
 13 | ggg   | this is a           +
    |       | girl
 14 | shout | this is a monster
 15 | haha  | they are cute
 16 | 123   | this is cute

我想要做的是找到（ 记录总数 和 前5条记录 ）'1'或'this'列中包含name或description。

我能弄清楚的是如此丑陋：

SELECT *, (select count(id)
           from (SELECT * from products
                 where description like any (array['%1%','%this%'])
                           or name like any (array['%1%','%this%'])
                ) as foo
          ) as total
from (SELECT * from products
      where description like any (array['%1%','%this%'])
                or name like any (array['%1%','%this%']))
     ) as fooo
limit 5;

Answer 1

假设您使用的是postgresql 9.0+，可以使用CTE来实现此目的。例如

WITH p AS (
        SELECT *
        FROM products
        WHERE description LIKE ANY (ARRAY['%1%','%this%']) OR name LIKE ANY (ARRAY['%1%','%this%'])
        )
SELECT  *,
        (select count(*) from p) as total
FROM p
ORDER BY id LIMIT 5;

Answer 2

您可以使用聚合函数count()作为窗口函数来计算同一查询级别的总计数：

SELECT id, name, description, count(*) OVER () AS total
FROM   products p
WHERE  description LIKE ANY ('{%1%,%this%}'::text[])
           OR name LIKE ANY ('{%1%,%this%}'::text[])
ORDER  BY id
LIMIT  5;

引用manual on window functions：

除了这些函数之外，任何内置或用户定义的聚合函数都可以用作窗口函数

这很有效，因为在窗口函数之后应用了LIMIT。

我还使用了数组文字的替代语法。一个和另一个一样好。对于较长的阵列，这个较短。有时候需要一个明确的类型转换。我在这里假设text。

在我的测试中，它比具有CTE的版本更简单，更快。

顺便说一下，这个带有正则表达式的WHERE子句更短 - 但更慢：

WHERE description ~ '(1|this)' OR name ~ '(1|this)'

丑陋，但很快

还有一个测试：我发现原始版本（类似于你已经的版本）甚至更快：

SELECT id, name, description , (SELECT count(*) FROM products p WHERE description LIKE ANY ('{%1%,%this%}'::text[]) OR name LIKE ANY ('{%1%,%this%}'::text[]) ) AS total FROM products p WHERE description LIKE ANY ('{%1%,%this%}'::text[]) OR name LIKE ANY ('{%1%,%this%}'::text[]) ORDER BY id LIMIT 5;

仅选择部分结果，但获取总行数

2 个答案:

丑陋，但很快