SQL Server 2005 中 Cross join & Cross Apply & Outer Apply 的区别

SQL Server 2005 新增 cross apply 和 outer apply 联接语句，增加这两个东东有啥作用呢？

我们知道有个 SQL Server 2000 中有个 cross join 是用于交叉联接的。实际上增加 cross apply 和 outer apply 是用于交叉联接表值函数（返回表结果集的函数）的， 更重要的是这个函数的参数是另一个表中的字段。这个解释可能有些含混不请，请看下面的例子：

-- 1. cross join 联接两个表
select *
  from TABLE_1 as T1
 cross join TABLE_2 as T2

-- 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)

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 apply
select *
  from TABLE_1 T1
 cross apply FN_TableValue(T1.column_a)

-- 5. outer apply
select *
  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 apply
select *
  from Departments as D
 cross apply fn_getsubtree(D.deptmgrid) as ST

deptid      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 apply
select *
  from Departments as D
 outer apply fn_getsubtree(D.deptmgrid) as ST

deptid      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
create table Employees
(
  empid   int         not null,
  mgrid   int         NULL,
  empname varchar(25) not null,
  salary  money       not null
)
go


-- create Departments table and insert values
create table Departments
(
  deptid    int not null primary key,
  deptname  varchar(25) not null
)
go

-- fill datas
insert into employees values(1 , NULL, 'Nancy'   , $10000.00)
insert into employees values(2 , 1   , 'Andrew'  , $5000.00)
insert into employees values(3 , 1   , 'Janet'   , $5000.00)
insert into employees values(4 , 1   , 'Margaret', $5000.00)
insert into employees values(5 , 2   , 'Steven'  , $2500.00)
insert into employees values(6 , 2   , 'Michael' , $2500.00)
insert into employees values(7 , 3   , 'Robert'  , $2500.00)
insert into employees values(8 , 3   , 'Laura'   , $2500.00)
insert into employees values(9 , 3   , 'Ann'     , $2500.00)
insert into employees values(10, 4   , 'Ina'     , $2500.00)
insert into employees values(11, 7   , 'David'   , $2000.00)
insert into employees values(12, 7   , 'Ron'     , $2000.00)
insert into employees values(13, 7   , 'Dan'     , $2000.00)
insert into employees values(14, 11  , 'James'   , $1500.00)

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


-- table-value function
create function dbo.fn_getsubtree(@empid AS INT) returns @TREE table
(
  empid   int not null,
  empname varchar(25) not null,
  mgrid   int null,
  lvl     int not null
)
as
begin
  with Employees_Subtree(empid, empname, mgrid, lvl)
  as
  (
    -- Anchor Member (AM)
    select empid, empname, mgrid, 0
      from employees
     where empid = @empid

    union all

    -- Recursive Member (RM)
    select e.empid, e.empname, e.mgrid, es.lvl+1
      from employees as e
              join employees_subtree as es
        on e.mgrid = es.empid
  )


  insert into @TREE
    select * from Employees_Subtree

  return
end
go

-- cross apply query
select *
  from Departments as D
 cross apply fn_getsubtree(D.deptmgrid) as ST


-- outer apply query
select *
  from Departments as D
 outer apply fn_getsubtree(D.deptmgrid) as ST

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