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我的工作是运行 SQL Server 2008，并且我花费大量时间查询数据库以获取信息，作为我工作的一部分。如果我需要与我的数据集不在同一聚合级别的信息，我会在 select 语句中使用嵌入式查询。通常是相同数字的 2 或 3 个略有不同的版本，因此它们都查询相同的表。（见下面的例子）

问题是嵌入在 select 语句中的子查询的别名的范围是什么。我想到的两个选择是：

  1. 在脚本级别并且必须对所有子查询和表都是唯一的？
  2. 在子查询级别，并且可以在每个中共享相同的别名。

我知道对于 From 语句中别名的表，它们必须是唯一的。我认为查询在每一行生成上执行的事实可能是不同的情况。

示例（完全编造，让我知道任何明显的错误，我会纠正它们）：

脚本级别 - 所有子查询和表的唯一别名：

Select
  p.purchaseid, p.purchasedate, 
  s.storename, c.customerid, 
(select count(p2.purchaseid) 
  from purchases p2 inner join 
       store s2 on p2.storeid = s2.storeid
  where s2.storeid = s.storeid
    and p2.purchasedate = p.purchasedate) as 'Store Daily Total Purchases',
(select count(p3.purchaseid) 
  from purchases p3 inner join 
       store s3 on p3.storeid = s3.storeid
  where p3.customerid = p.customerid
    and p3.purchasedate = p.purchasedate) as 'Customer Daily Total Purchases'
from 
  purchases p inner join   
  customer c on p.customerid = c.customerid
  store s on p.storeid = s.storeid

查询级别 - 子查询的常用别名可以：

Select
  p.purchaseid, p.purchasedate, 
  s.storename, c.customerid, 
(select count(p2.purchaseid) 
  from purchases p2 inner join 
       store s2 on p2.storeid = s2.storeid
where s2.storeid = s.storeid
  and p2.purchasedate = p.purchasedate) as 'Store Daily Total Purchases',
(select count(p2.purchaseid) 
  from purchases p2 inner join 
       store s2 on p2.storeid = s2.storeid
where p2.customerid = p.customerid
  and p2.purchasedate = p.purchasedate) as 'Customer Daily Total Purchases'
from 
  purchases p inner join   
  customer c on p.customerid = c.customerid
  store s on p.storeid = s.storeid
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1 回答 1

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别名是在查询级别，所以你的第二个例子很好。

from 子句中对表的引用会产生这些相关的子查询，以防您想阅读有关该主题的更多信息。

一般来说，我建议您将查询移至 from 子句并将它们作为连接进行管理：

Select p.purchaseid, p.purchasedate, s.storename, c.customerid,
       pdate.cnt as [Store Daily Total Purchases],
       pcust.cnt as [Customer Daily Total Purchases]
from purchases p inner join   
     customer c
     on p.customerid = c.customerid join
     store s on p.storeid = s.storeid left outer join
     (select s2.storeid, p2.purchasedate, count(p2.purchaseid) 
      from purchases p2 inner join 
           store s2 on p2.storeid = s2.storeid
      group by s2.storeid, p2.purchasedate
     ) pdate
     on pdate.purchasedate = p.purchasedate and
        pdate.storeid = s.storeid left outer join
     (select s2.customerid, p2.purchasedate, count(p2.purchaseid) 
      from purchases p2 inner join 
           store s2 on p2.storeid = s2.storeid
      group by s2.storeid, p2.purchasedate
     ) pcust
     on pcust.purchasedate = p.purchasedate and
        pcust.customerid= s.customerid

但是，就您而言，我认为仅使用聚合有一个更简单的解决方案：

Select p.purchaseid, p.purchasedate, s.storename, c.customerid,
       count(*) over (partition by s.storeid, p.purchasedate) as  as [Store Daily Total Purchases],
       count(*) over (partition by c.customerid, p.purchasedate) as [Customer Daily Total Purchases]
from purchases p inner join   
     customer c
     on p.customerid = c.customerid join
     store s on p.storeid = s.storeid
于 2012-06-23T00:39:58.870 回答