关于SQLServer2005的学习笔记——SQL查询解析步骤

最近一来因工作上的事情比较闹心，没心事再研究 SQLServer2005 的体系结构；一来关于体系结构确实过于复杂，远远不如应用来的直接明了，所以暂时搁笔。

出于工作上的需要，对应用开发需要更多的了解，所以把心事暂时放到这方面。

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先从最简单的 SQL 入手来分析一下 SQL 的执行步骤，为什么了解执行步骤，其实与 JOIN 后的 ON 条件和 WHERE 条件容易混淆有关系。

是先执行 ON 还是先执行 WHERE ，很大程度上会决定 SQL 的结果集正确与否。

CREATE TABLE Customers

(

? CustomerID? CHAR(5)???? NOT NULL PRIMARY KEY,

? City??????? VARCHAR(10) NOT NULL

);

CREATE TABLE Orders

(

? OrderID???? INT???? NOT NULL PRIMARY KEY,

? CustomerID? CHAR(5) NULL REFERENCES Customers(CustomerID)

);

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INSERT INTO Customers VALUES('FISSA','Madrid');

INSERT INTO Customers VALUES('FRNDO','Madrid');

INSERT INTO Customers VALUES('KRLOS','Madrid');

INSERT INTO Customers VALUES('MRPHS','Zion');

INSERT INTO Orders VALUES(1,'FRNDO');

INSERT INTO Orders VALUES(2,'FRNDO');

INSERT INTO Orders VALUES(3,'KRLOS');

INSERT INTO Orders VALUES(4,'KRLOS');

INSERT INTO Orders VALUES(5,'KRLOS');

INSERT INTO Orders VALUES(6,'MRPHS');

INSERT INTO Orders VALUES(7,NULL);
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试看看以上两个语句有什么不同，你就会发现很有趣的现象。

SELECT C.CustomerID,COUNT(O.OrderID) AS NumOrders

? FROM Customers C

? LEFT OUTER JOIN Orders O

??? ON C.CustomerID=O.CustomerID

? WHERE C.City='Madrid'

? GROUP BY C.CustomerID

? HAVING COUNT(O.OrderID)<3

? ORDER BY NumOrders;

?

SELECT C.CustomerID,COUNT(O.OrderID) AS NumOrders

? FROM Customers C

? LEFT OUTER JOIN Orders O

??? ON C.CustomerID=O.CustomerID

??? AND C.City='Madrid'

? GROUP BY C.CustomerID

? HAVING COUNT(O.OrderID)<3

? ORDER BY NumOrders;
?

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--Step1 ，首先对 FROM 后面的表进行笛卡尔乘积，生成虚表 STEP1

WITH STEP1

AS

(

SELECT C.CustomerID C_CustomerID,C.City C_City,O.OrderID O_OrderID,O.CustomerID O_CustomerID

? FROM Customers C,Orders O

)

SELECT * FROM STEP1
?

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--Step2 ，再次应用 ON 语句中的条件，如果没有外关联的话，这里的 ON 和 WHERE 实际上是没有什么差别的，生成虚表 STEP2

WITH STEP2

AS

(

SELECT C.CustomerID C_CustomerID,O.CustomerID O_CustomerID

? FROM Customers C

? JOIN Orders O

??? ON C.CustomerID=O.CustomerID

)

SELECT * FROM STEP2
?

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--Step3 ，如果指定了 OUTER JOIN ， SQL 会自动把 STEP2 表中未匹配的行作为外部行添加到 STEP3 中，此处找到了 CustomerID=FISSA,City=Madrid ，这个没有订单但又有相关名字的用户

WITH STEP3

AS

(

SELECT C.CustomerID C_CustomerID,O.CustomerID O_CustomerID

? FROM Customers C

? LEFT OUTER JOIN Orders O

??? ON C.CustomerID=O.CustomerID

)

SELECT * FROM STEP3
?

?

--Step4 ，应用 WHERE 条件，过滤不符合条件的记录

AS

(

SELECT C.CustomerID C_CustomerID,O.CustomerID O_CustomerID

? FROM Customers C

? LEFT OUTER JOIN Orders O

??? ON C.CustomerID=O.CustomerID

? WHERE C.City='Madrid'

)

SELECT * FROM STEP4
?

?

--Step5 ，对以上的结果集进行分组

WITH STEP5

AS

(

SELECT C.CustomerID,COUNT(O.OrderID) AS NumOrders

? FROM Customers C

? LEFT OUTER JOIN Orders O

??? ON C.CustomerID=O.CustomerID

? WHERE C.City='Madrid'

? GROUP BY C.CustomerID

)

SELECT * FROM STEP5
?

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--Step6 ，处理 CUBE 、 ROLLUP 之类的语句，此处无此需求

略

SELECT * FROM STEP6
?

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--Step7 处理 Having 筛选器，与 WHERE 条件有些类似

WITH STEP7

AS

(

SELECT C.CustomerID,COUNT(O.OrderID) AS NumOrders

? FROM Customers C

? LEFT OUTER JOIN Orders O

??? ON C.CustomerID=O.CustomerID

? WHERE C.City='Madrid'

? GROUP BY C.CustomerID

? HAVING COUNT(O.OrderID)<3

)

SELECT * FROM STEP7
?

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--Step8 ，处理 SELECT 列表，即别名转换把 COUNT(O.OrderID) 转换成 NumOrders

略

SELECT * FROM STEP8
?

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--Step9 ，应用 DISTINCT 语句，此处无此需求

略

SELECT * FROM STEP9
?

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--Step10 ，应用 ORDER BY 语句进行排序

SELECT C.CustomerID,COUNT(O.OrderID) AS NumOrders

? FROM Customers C

? LEFT OUTER JOIN Orders O

??? ON C.CustomerID=O.CustomerID

? WHERE C.City='Madrid'

? GROUP BY C.CustomerID

? HAVING COUNT(O.OrderID)<3

? ORDER BY NumOrders
?

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--Step11 ，执行 TOP 选项，此处无此需求

略

SELECT * FROM STEP11
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本文来自CSDN博客，转载请标明出处：http://blog.csdn.net/baoqiangwang/archive/2010/01/08/5155423.aspx

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