SELECT与多个子查询到同一个表

时间：2010-10-11 14:55:04

标签： sql sql-server tsql subquery

我一遍又一遍地使用相同的SQL模式，我知道必须有更好的方法，但我无法将它拼凑在一起。这是模式的一个简单版本，我将撤回学生的信息和他们检出的最后一本书（如果存在的话）：

SELECT TStudents.*,
       BookName = (SELECT TOP 1 BookName 
                     FROM TBookCheckouts 
                    WHERE StudentID = TStudents.ID 
                 ORDER BY DateCheckedOut DESC),
       BookAuthor = (SELECT TOP 1 BookAuthor 
                       FROM TBookCheckouts 
                      WHERE StudentID = TStudents.ID 
                   ORDER BY DateCheckedOut DESC),
       BookCheckout = (SELECT TOP 1 DateCheckedOut 
                         FROM TBookCheckouts 
                         WHERE StudentID = TStudents.ID 
                     ORDER BY DateCheckedOut DESC)
   FROM TStudents

（为了这个例子，请忽略TBookCheckouts应该分成TCheckouts和TBooks的事实）

我想说明的是：我倾向于为同一个表中的列提供大量子查询。我也倾向于需要按日期对这些子查询表进行排序以获得最新记录，因此对于LEFT JOIN来说，这并不是那么简单（至少对我而言）。但是请注意，除了返回哪个字段之外，我基本上做了3次相同的子查询。 SQL Server可能足够聪明，可以优化它，但我不这么认为（我肯定需要在阅读执行计划方面做得更好......）。

虽然以这种方式构造它可能有好处（如果我有大量子查询和子表，有时最终会更具可读性），但这似乎并不是特别有效。

我已经考虑从派生表中执行LEFT JOIN，可能包含ROW_NUMBER（）和PARTITION BY，但我似乎无法将它们拼凑在一起。

8 个答案:

答案 0 :(得分：12)

如果您使用的是SQL Server 2005及更高版本，则可以使用类似的排名函数：

With LastCheckout As
    (
    Select StudentId, BookName, BookAuthor, DateCheckedOut 
        , Row_Number() Over ( Partition By StudentId Order By DateCheckedOut Desc) As CheckoutRank
    From TBookCheckouts
    )
Select ..., LastCheckout.BookName, LastCheckout.BookAuthor, LastCheckout.DateCheckedOut
From TStudents
    Left Join LastCheckout 
        On LastCheckout.StudentId = TStudents.StudentId
                And LastCheckout.CheckoutRank = 1

答案 1 :(得分：9)

2005年及以后，OUTER APPLY是你的朋友：

SELECT TStudents.*,
       t.BookName ,
       t.BookAuthor ,
       t.BookCheckout
   FROM TStudents
  OUTER APPLY(SELECT TOP 1 s.* 
                     FROM TBookCheckouts AS s
                    WHERE s.StudentID = TStudents.ID 
                 ORDER BY s.DateCheckedOut DESC) AS t

答案 2 :(得分：3)

使用：

   SELECT s.*,
          x.bookname,
          x.bookauthor,
          x.datecheckedout
     FROM TSTUDENTS s
LEFT JOIN (SELECT bc.studentid,
                  bc.bookname,
                  bc.bookauthor,
                  bc.datecheckedout,
                  ROW_NUMBER() OVER(PARTITION BY bc.studentid
                                        ORDER BY bc.datecheckedout DESC) AS rank
             FROM TSBOOKCHECKOUTS bc) x ON x.studentid = s.id
                                       AND x.rank = 1

如果学生没有结帐任何图书，bookname，bookauthor和datecheckedout将为空。

答案 3 :(得分：0)

尝试

    ;WITH LatestCheckouts
    AS
    (
        SELECT  DISTINCT
                A.StudentID
            ,   A.BookName   
            ,   A.BookAuthor
            ,   A.DateCheckedOut
        FROM    TBookCheckouts A
            INNER JOIN
        (   
            SELECT  StudentID
            ,   DateCheckedOut =  MAX(DateCheckedOut)
             FROM TBookCheckouts
            GROUP  BY
                StudentID
        ) B

        ON A.StudentID = B.StudentID
        AND A.DateCheckedOut =  B.DateCheckedOut
    )       
    SELECT students.*
        ,  BookName     = checkouts.BookName
        ,  BookAuthor   = checkouts.BookAuthor
        ,  BookCheckout = checkouts.DateCheckedOut

    FROM    TStudents students
        LEFT JOIN
         LatestCheckouts checkouts
    ON  students.ID = checkouts.StudentID

答案 4 :(得分：0)

如果您想使用公用表表达式，可以进行以下查询。在这种情况下，它不会给你带来任何好处，但对于未来：

;with LatestBookOut as 
(
    SELECT  C.StudentID, BookID, Title, Author, DateCheckedOut AS BookCheckout 
    FROM    CheckedOut AS C
    INNER JOIN ( SELECT StudentID, 
                        MAX(DateCheckedOut) AS DD 
                FROM Checkedout 
                GROUP BY StudentID) StuMAX                 
    ON StuMAX.StudentID = C.StudentID 
    AND StuMAX.DD = C.DateCheckedOut  
)

SELECT    B.BookCheckout,
        BookId, 
        Title,    
        Author, 
        S.*

FROM    LatestBookOut AS B
INNER JOIN Student  AS S ON S.ID = B.StudentID

答案 5 :(得分：0)

create table BookCheckout(StudentID int, CheckoutDate date, BookName varchar(10))

insert into BookCheckout values (1, '1.1.2010', 'a');
insert into BookCheckout values (1, '2.1.2010', 'b');
insert into BookCheckout values (1, '3.1.2010', 'c');
insert into BookCheckout values (2, '1.1.2010', 'd');
insert into BookCheckout values (2, '2.1.2010', 'e');

select *
from BookCheckout bc1
where CheckoutDate = (
    Select MAX(CheckoutDate) 
    from BookCheckout bc2
    where bc2.StudentID= bc1.StudentID)

StudentID    CheckoutDate    BookName
2    2010-01-02    e
1    2010-01-03    c

只需将连接添加到TStudent即可。 还有一个问题：如果学生有2个或更多的Bookcheckout，并且有相同的最长结账日期，则每个学生可以获得多个BookCheckout。

  select s.*, LastBookCheckout.*
  from TStudent s, 
    (select *
    from BookCheckout bc1
    where CheckoutDate = (
        Select MAX(CheckoutDate) 
        from BookCheckout bc2
        where bc2.StudentID= bc1.StudentID)) LastBookCheckout
  where s.ID = LastBookCheckout.StudentID

为避免重复：

select * 
from (
  select *, RANK() over (partition by StudentID order by CheckoutDate desc,BookName) rnk
    from BookCheckout bc1) x
where rnk=1

我使用“BookName”作为第二个排序标准。 =＆GT;请使用主键使其成为真正独特的标准。

答案 6 :(得分：0)

OMGPonies的答案很好。为了便于阅读，我会用Common Table Expressions编写它：

WITH CheckoutsPerStudentRankedByDate AS (
    SELECT bookname, bookauthor, datecheckedout, studentid,
        ROW_NUMBER(PARTITION BY studentid ORDER BY datecheckedout DESC) AS rank
    FROM TSBOOKCHECKOUTS
)
SELECT 
    s.*, c.bookname, c.bookauthor, c.datecheckedout
FROM TSTUDENTS AS s
LEFT JOIN CheckoutsPerStudentRankedByDate AS c
    ON s.studentid = c.studentid
    AND c.rank = 1

c.rank = 1可以被c.rank IN(1, 2)替换为最后2个签出，BETWEEN 1 AND 3替换为最后3个等等...

答案 7 :(得分：0)

希望这就是您想要的，这是我在这种情况下知道的简单方法

SELECT (SELECT TOP 1 BookName 
                 FROM TBookCheckouts 
                WHERE StudentID = TStudents.ID 
             ORDER BY DateCheckedOut DESC)[BOOK_NAME],
   (SELECT TOP 1 BookAuthor 
                   FROM TBookCheckouts 
                  WHERE StudentID = TStudents.ID 
               ORDER BY DateCheckedOut DESC)[BOOK_AUTHOR],
   (SELECT TOP 1 DateCheckedOut 
                     FROM TBookCheckouts 
                     WHERE StudentID = TStudents.ID 
                 ORDER BY DateCheckedOut DESC)[DATE_CHECKEDOUT]

当我遇到这样的问题时，这就是我解决的方法，我认为这将是解决您案子的方法。

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