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SQL：将子查询转换为连接：如何在嵌套连接where子句中引用外部表？

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我试图将我的子查询更改为一个连接，它只选择子查询中的一个记录 . 它似乎为每个找到的记录运行子查询，花费一分钟来执行：

select afield1, afield2, (
    select top 1 b.field1
    from   anothertable as b
    where  b.aForeignKey = a.id
    order by field1
) as bfield1
from sometable as a

如果我尝试仅选择相关记录，则它不知道如何在嵌套选择中绑定a.id.

select afield1, afield2, bfield1
from   sometable a left join (
    select top 1 id, bfield, aForeignKey 
    from   anothertable
    where  anothertable.aForeignKey = a.id
    order by bfield) b on
       b.aForeignKey = a.id
-- Results in the multi-part identifier "a.id" could not be bound

如果我在嵌套的where子句中硬编码值，则选择持续时间从60秒降至5以下 . 任何人都有关于如何连接两个表而不处理内部表中的每个记录的任何建议？

编辑：

我最后补充说

left outer join (
    select *, row_number() over (partition by / order by) as rank) b on
    b.aforeignkey = a.id and b.rank = 1

从22秒开始，从约50秒变为8秒 .

sql sql-server-2005 left-join

2 回答

  • 2

    试试这个：

    WITH qry AS
(
    SELECT afield1, 
           afield2, 
           b.field1 AS bfield1,
           ROW_NUMBER() OVER(PARTITION BY a.id ORDER BY field1) rn
      FROM sometable a LEFT JOIN anothertable b
        ON b.aForeignKey = a.id
)
SELECT *
  FROM qry
 WHERE rn = 1
    回复于
  • 0

    试试这个

    select afield1, 
afield2, 
bfield1 
from sometable a 
left join 
(select top 1 id, bfield, aForeignKey from  anothertable where  aForeignKey in(a.id)  order by bfield) b on b.aForeignKey = a.id
    回复于

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