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简化SQL语句的一般规则

简化SQL语句的一般规则

换种说法:使用JOIN,SUBSELECT和UNION进行(复杂)查询是否有可能(或不能)通过使用一些转换规则将其简化为更简单的等效sql语句,从而产生相同的结果?

这正是优化程序谋生的手段(不是说他们总是做得很好)。

由于sql是一种基于集合的语言,因此通常存在不止一种将一个查询转换为另一个查询方法

像这个查询

SELECT  *
FROM    mytable
WHERE   col1 > @value1 OR col2 < @value2

可以转换成这样:

SELECT  *
FROM    mytable
WHERE   col1 > @value1
UNION
SELECT  *
FROM    mytable
WHERE   col2 < @value2

或这个:

SELECT  mo.*
FROM    (
        SELECT  id
        FROM    mytable
        WHERE   col1 > @value1
        UNION
        SELECT  id
        FROM    mytable
        WHERE   col2 < @value2
        ) mi
JOIN    mytable mo
ON      mo.id = mi.id

,看起来比较丑陋,但可以产生更好的执行计划。

最常见的事情之一是替换此查询

SELECT  *
FROM    mytable
WHERE   col IN
        (
        SELECT  othercol
        FROM    othertable
        )

与此:

SELECT  *
FROM    mytable mo
WHERE   EXISTS
        (
        SELECT  NULL
        FROM    othertable o
        WHERE   o.othercol = mo.col
        )

在某些RDBMS(如Postgresql)中,DISTINCTGROUP BY使用不同的执行计划,因此有时最好将其中一个替换为另一个

SELECT  mo.grouper,
        (
        SELECT  SUM(col)
        FROM    mytable mi
        WHERE   mi.grouper = mo.grouper
        )
FROM    (
        SELECT  DISTINCT grouper
        FROM    mytable
        ) mo

SELECT  mo.grouper, SUM(col)
FROM    mytable
GROUP BY
        mo.grouper

在中PostgresqlDISTINCT排序和GROUP BY散列。

MysqL缺少FULL OUTER JOIN,因此可以将其改写为以下形式:

SELECT  t1.col1, t2.col2
FROM    table1 t1
LEFT OUTER JOIN
        table2 t2
ON      t1.id = t2.id

SELECT  t1.col1, t2.col2
FROM    table1 t1
LEFT JOIN
        table2 t2
ON      t1.id = t2.id
UNION ALL
SELECT  NULL, t2.col2
FROM    table1 t1
RIGHT JOIN
        table2 t2
ON      t1.id = t2.id
WHERE   t1.id IS NULL

,但请参阅我的博客中有关如何更有效地执行此操作的文章MysqL

此分层查询Oracle

SELECT  DISTINCT(animal_id) AS animal_id
FROM    animal
START WITH
        animal_id = :id
CONNECT BY
        PRIOR animal_id IN (father, mother)
ORDER BY
        animal_id

可以转换为:

SELECT  DISTINCT(animal_id) AS animal_id
FROM    (
        SELECT  0 AS gender, animal_id, father AS parent
        FROM    animal
        UNION ALL
        SELECT  1, animal_id, mother
        FROM    animal
        )
START WITH
        animal_id = :id
CONNECT BY
        parent = PRIOR animal_id
ORDER BY
        animal_id

,后者表现更好。

有关执行计划的详细信息,请参阅我的博客中的这篇文章

要查找与给定范围重叠的所有范围,可以使用以下查询

SELECT  *
FROM    ranges
WHERE   end_date >= @start
        AND start_date <= @end

,但是在sql Server这种更复杂的查询中,更快地产生相同的结果:

SELECT  *
FROM    ranges
WHERE   (start_date > @start AND start_date <= @end)
        OR (@start BETWEEN start_date AND end_date)

,不管您信不信,我的博客上也有一篇文章

sql Server 还缺少执行累积聚合的有效方法,因此此查询

SELECT  mi.id, SUM(mo.value) AS running_sum
FROM    mytable mi
JOIN    mytable mo
ON      mo.id <= mi.id
GROUP BY
        mi.id

可以使用,主帮助我更有效地重写,游标(你听我的权利:cursorsmore efficientlysql Server一个句子)。

请参阅我的博客中有关如何执行此操作的文章

在金融应用程序中通常会遇到一种查询查询某种货币的有效汇率,例如Oracle

SELECT  TO_CHAR(SUM(xac_amount * rte_rate), 'FM999G999G999G999G999G999D999999')
FROM    t_transaction x
JOIN    t_rate r
ON      (rte_currency, rte_date) IN
        (
        SELECT  xac_currency, MAX(rte_date)
        FROM    t_rate
        WHERE   rte_currency = xac_currency
                AND rte_date <= xac_date
        )

可以大量重写此查询以使用等于条件,HASH JOIN而该条件允许a而不是NESTED LOOPS

WITH v_rate AS
        (
        SELECT  cur_id AS eff_currency, dte_date AS eff_date, rte_rate AS eff_rate
        FROM    (
                SELECT  cur_id, dte_date,
                        (
                        SELECT  MAX(rte_date)
                        FROM    t_rate ri
                        WHERE   rte_currency = cur_id
                                AND rte_date <= dte_date
                        ) AS rte_effdate
                FROM    (
                        SELECT  (
                                SELECT  MAX(rte_date)
                                FROM    t_rate
                                ) - level + 1 AS dte_date
                        FROM    dual
                        CONNECT BY
                                level <=
                                (
                                SELECT  MAX(rte_date) - MIN(rte_date)
                                FROM    t_rate
                                )
                        ) v_date,
                        (
                        SELECT  1 AS cur_id
                        FROM    dual
                        UNION ALL
                        SELECT  2 AS cur_id
                        FROM    dual
                        ) v_currency
                ) v_eff
        LEFT JOIN
                t_rate
        ON      rte_currency = cur_id
                AND rte_date = rte_effdate
        )
SELECT  TO_CHAR(SUM(xac_amount * eff_rate), 'FM999G999G999G999G999G999D999999')
FROM    (
        SELECT  xac_currency, TRUNC(xac_date) AS xac_date, SUM(xac_amount) AS xac_amount, COUNT(*) AS cnt
        FROM    t_transaction x
        GROUP BY
                xac_currency, TRUNC(xac_date)
        )
JOIN    v_rate
ON      eff_currency = xac_currency
        AND eff_date = xac_date

尽管笨拙,但后者查询6速度要快几倍。

这里的主要思想是将替换<==,这需要构建一个内存日历表。要JOIN与。

SQLServer 2022/1/1 18:39:16 有547人围观

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