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。
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)
);
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);
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;
--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
--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,C.City C_City,O.OrderID O_OrderID,O.CustomerID O_CustomerID
FROM Customers C
JOIN Orders O
ON C.CustomerID=O.CustomerID
)
SELECT * FROM STEP2
--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,C.City C_City,O.OrderID O_OrderID,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,C.City C_City,O.OrderID O_OrderID,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
--Step6 ，处理 CUBE 、 ROLLUP 之类的语句，此处无此需求
略
SELECT * FROM STEP6
--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
--Step8 ，处理 SELECT 列表，即别名转换把 COUNT(O.OrderID) 转换成 NumOrders
略
SELECT * FROM STEP8
--Step9 ，应用 DISTINCT 语句，此处无此需求
略
SELECT * FROM STEP9
--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
--Step11 ，执行 TOP 选项，此处无此需求
略
SELECT * FROM STEP11
本文转自baoqiangwang51CTO博客，原文链接：http://blog.51cto.com/baoqiangwang/310463
，如需转载请自行联系原作者
