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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