SQL 查询：Select 子选择的最大值

Question

我有一个已经有效的 sql 语句，但我认为一定有比我的更好的解决方案。

我正在尝试获取文章 最高价格 其中有 从未售出.

有了这个select我得到了所有还没有售出的物品（数量+价格）：

select anr, price
from article a 
where not exists(
    select 1 from OrderItems o
    where o.artnr = a.anr 
  )

货号+价格结果好像

| Anr | Price |
| 1   | 300.0 |
| 4   | 340.0 |
| 5   | 340.0 |
| 3   | 200.0 |

我临时获取价格最高的文章的解决方案是：

select anr, price 
from article 
where anr in(
  select anr
  from article a 
  where not exists(
    select 1 from OrderItems o
    where o.artnr = a.anr 
  )
)
and price = (
  select max(price)
  from article a 
  where not exists(
    select 1 from OrderItems o
    where o.artnr = a.anr 
  )
)

正确的解法是：

| Anr | Price |
| 4   | 340.0 |
| 5   | 340.0 |

有没有办法避免相同的 subselect 两次？

此处用于测试的是带有我的插入值的缩短的创建 Table 脚本：

CREATE TABLE Article
(
  Anr Int Primary Key,
  Price Numeric(9,2) Not Null
);

CREATE TABLE Orders
(
  OrderNr Int Primary Key
)

CREATE TABLE OrderItems
(
  OrderNr Int References Orders On Delete Cascade, 
  ItemNr  Int,
  Artnr   Int References Article Not Null,
  Amount  Int Not Null Check(Amount >= 0),
  Primary Key(OrderNr, ItemNr)
)

-- articles without an order
Insert into Article (Anr, Price) values(1,300.0);
Insert into Article (Anr, Price) values(4,340.0);
Insert into Article (Anr, Price) values(5,340.0);
Insert into Article (Anr, Price) values(3,200.0);

-- articles for order with orderNr '1'
Insert into Article (Anr, Price) values(2,340.0);
Insert into Article (Anr, Price) values(6,620.0);

-- insert test order that contains the two articles
Insert into Orders (OrderNr) values (1);
Insert into OrderItems(OrderNr, ItemNr, Artnr, Amount) values(1,1,2,4);
Insert into OrderItems(OrderNr, ItemNr, Artnr, Amount) values(1,2,6,2);

我也看了题目但我认为就我而言，必须有更短的方法来完成 select.

Answer 1

这是一个解决方案，它避免了您拥有的一个相关子查询，而是将其替换为 LEFT JOIN:

SELECT a.*
FROM article a LEFT JOIN OrderItems o ON a.anr = o.artnr
WHERE o.artnr IS NULL AND
    a.price = (SELECT MAX(a.price)
               FROM article a LEFT JOIN OrderItems o ON a.anr = o.artnr
               WHERE o.artnr IS NULL)

此解决方案应符合 ANSI-92，这意味着它应该与 MySQL、Oracle、SQL 服务器以及您可能遇到的任何其他类型的快餐兼容。

Answer 2

您可以将原始查询与 window 函数结合使用来获得您想要的结果：

select anr, price
from (
    select anr, price, max(price) over () as max_price
    from article a 
    where not exists (select 1 
                      from OrderItems o
                      where o.artnr = a.anr)
) t
where price = max_price;

这可能更快，因为只需要对 article table 进行一次扫描。

或者，您可以使用左连接解决方案来查找从未被订购的文章，但如果 Oracle 会为此使用不同的执行计划，我会感到惊讶：

select anr, price
from (
    select anr, price, max(price) over () as max_price
    from article a 
      left join OrderItems o ON o.artnr = a.anr
    where o.artnr is null
) t
where price = max_price;

SQLFiddle 示例：http://sqlfiddle.com/#!15/1eb69/1

Answer 3

SELECT a.*
FROM article a LEFT JOIN OrderItems o ON a.anr = o.artnr
WHERE o.artnr IS NULL 
AND a.price = (SELECT TOP 1 a.price
           FROM article a LEFT JOIN OrderItems o ON a.anr = o.artnr
           WHERE o.artnr IS NULL
           Order By a.price Desc
           )

试试这个....

SQL 查询：Select 子选择的最大值

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