-- 1. cross join 联接两个表

select *  from TABLE_1 as T1
cross join TABLE_2 as T2【此句实际上是求笛卡尔积】

-- 2. cross join 联接表和表值函数，表值函数的参数是个“常量”select *from TABLE_1 T1
cross join FN_TableValue(100)-- 3. cross join  联接表和表值函数，表值函数的参数是“表T1中的字段”select *from TABLE_1 T1
cross join FN_TableValue(T1.column_a)Error：Msg 4104, Level 16, State 1, Line 1
The multi-part identifier "T1.column_a" could not be bound.最后的这个查询的语法有错误。在 cross join 时，表值函数的参数不能是表 T1 的字段， 为啥不能这样做呢？我猜可能微软当时没有加这个功能：），后来有客户抱怨后， 于是微软就增加了 cross apply 和 outer apply 来完善，请看 cross apply, outer apply 的例子： -- 4. cross applyselect *from TABLE_1 T1
cross apply FN_TableValue(T1.column_a)-- 5. outer applyselect *from TABLE_1 T1
outer apply FN_TableValue(T1.column_a)cross apply 和 outer apply 对于 T1 中的每一行都和派生表（表值函数根据T1当前行数据生成的动态结果集） 做了一个交叉联接。cross apply 和outer apply 的区别在于： 如果根据 T1 的某行数据生成的派生表为空，cross apply 后的结果集 就不包含 T1 中的这行数据，而 outer apply 仍会包含这行数据，并且派生表的所有字段值都为 NULL。 下面的例子摘自微软 SQL Server 2005 联机帮助，它很清楚的展现了 cross apply 和 outer apply 的不同之处： -- cross applyselect *from Departments as D
cross apply fn_getsubtree(D.deptmgrid) as STdeptid      deptname      deptmgrid   empid       empname       mgrid       lvl
----------- -----------   ----------- ----------- -----------   ----------- ------
1           HR            2           2           Andrew        1           0
1           HR            2           5           Steven        2           1
1           HR            2           6           Michael       2           1
2           Marketing     7           7           Robert        3           0
2           Marketing     7           11          David         7           1
2           Marketing     7           12          Ron           7           1
2           Marketing     7           13          Dan           7           1
2           Marketing     7           14          James         11          2
3           Finance       8           8           Laura         3           0
4           R&D           9           9           Ann           3           0
5           Training      4           4           Margaret      1           0
5           Training      4           10          Ina           4           1(12 row(s) affected)-- outer applyselect *from Departments as D
outer apply fn_getsubtree(D.deptmgrid) as STdeptid      deptname      deptmgrid   empid       empname       mgrid       lvl
----------- -----------   ----------- ----------- -----------   ----------- ------
1           HR            2           2           Andrew        1           0
1           HR            2           5           Steven        2           1
1           HR            2           6           Michael       2           1
2           Marketing     7           7           Robert        3           0
2           Marketing     7           11          David         7           1
2           Marketing     7           12          Ron           7           1
2           Marketing     7           13          Dan           7           1
2           Marketing     7           14          James         11          2
3           Finance       8           8           Laura         3           0
4           R&D           9           9           Ann           3           0
5           Training      4           4           Margaret      1           0
5           Training      4           10          Ina           4           1
6           Gardening     NULL        NULL        NULL          NULL        NULL(13 row(s) affected)注意 outer apply 结果集中多出的最后一行。 当 Departments 的最后一行在进行交叉联接时：deptmgrid 为 NULL，fn_getsubtree(D.deptmgrid) 生成的派生表中没有数据，但 outer apply 仍会包含这一行数据，这就是它和 cross join 的不同之处。 下面是完整的测试代码，你可以在 SQL Server 2005 联机帮助上找到： -- create Employees table and insert values
IF OBJECT_ID('Employees') IS NOT NULL
DROP TABLE Employees
GO
CREATE TABLE Employees
(
empid INT NOT NULL,
mgrid INT NULL,
empname VARCHAR(25) NOT NULL,
salary MONEY NOT NULL
)
GO
IF OBJECT_ID('Departments') IS NOT NULL
DROP TABLE Departments
GO
-- create Departments table and insert values
CREATE TABLE Departments
(
deptid INT NOT NULL PRIMARY KEY,
deptname VARCHAR(25) NOT NULL,
deptmgrid INT
)
GO-- fill datas
INSERT  INTO employees VALUES  (1,NULL,'Nancy',00.00)
INSERT  INTO employees VALUES  (2,1,'Andrew',00.00)
INSERT  INTO employees VALUES  (3,1,'Janet',00.00)
INSERT  INTO employees VALUES  (4,1,'Margaret',00.00)
INSERT  INTO employees VALUES  (5,2,'Steven',00.00)
INSERT  INTO employees VALUES  (6,2,'Michael',00.00)
INSERT  INTO employees VALUES  (7,3,'Robert',00.00)
INSERT  INTO employees VALUES  (8,3,'Laura',00.00)
INSERT  INTO employees VALUES  (9,3,'Ann',00.00)
INSERT  INTO employees VALUES  (10,4,'Ina',00.00)
INSERT  INTO employees VALUES  (11,7,'David',00.00)
INSERT  INTO employees VALUES  (12,7,'Ron',00.00)
INSERT  INTO employees VALUES  (13,7,'Dan',00.00)
INSERT  INTO employees VALUES  (14,11,'James',00.00)INSERT  INTO departments VALUES  (1,'HR',2)
INSERT  INTO departments VALUES  (2,'Marketing',7)
INSERT  INTO departments VALUES  (3,'Finance',8)
INSERT  INTO departments VALUES  (4,'R&D',9)
INSERT  INTO departments VALUES  (5,'Training',4)
INSERT  INTO departments VALUES  (6,'Gardening',NULL)
GO
--SELECT * FROM departments-- table-value function
IF OBJECT_ID('fn_getsubtree') IS NOT NULL
DROP FUNCTION  fn_getsubtree
GO
CREATE  FUNCTION dbo.fn_getsubtree(@empid AS INT)
RETURNS TABLE
AS
RETURN(WITH Employees_Subtree(empid, empname, mgrid, lvl)AS(-- Anchor Member (AM)SELECT empid, empname, mgrid, 0FROM employeesWHERE empid = @empid   UNION ALL-- Recursive Member (RM)SELECT e.empid, e.empname, e.mgrid, es.lvl+1FROM employees AS ejoin employees_subtree AS esON e.mgrid = es.empid)SELECT * FROM Employees_Subtree
)
GO-- cross apply query
SELECT  *
FROM Departments AS DCROSS APPLY fn_getsubtree(D.deptmgrid) AS ST-- outer apply query
SELECT  *
FROM Departments AS DOUTER APPLY fn_getsubtree(D.deptmgrid) AS ST-------------------中国风整理的两个小例子-------------------------------
create table #T(姓名 varchar(10))
insert into #T values('张三')
insert into #T values('李四')
insert into #T values(NULL )create table #T2(姓名 varchar(10) , 课程 varchar(10) , 分数 int)
insert into #T2 values('张三' , '语文' , 74)
insert into #T2 values('张三' , '数学' , 83)
insert into #T2 values('张三' , '物理' , 93)
insert into #T2 values(NULL , '数学' , 50)--drop table #t,#T2
goselect*
from#T a
cross apply(select 课程,分数 from #t2 where 姓名=a.姓名) b/*
姓名         课程         分数
---------- ---------- -----------
张三         语文         74
张三         数学         83
张三         物理         93(3 行受影响)*/
select*
from#T a
outer apply(select 课程,分数 from #t2 where 姓名=a.姓名) b
/*
姓名         课程         分数
---------- ---------- -----------
张三         语文         74
张三         数学         83
张三         物理         93
李四         NULL       NULL
NULL       NULL       NULL(5 行受影响)
*/