SQL Server CROSS APPLY和OUTER APPLY的应用详解

SELECT
a
FROM
dbo.LargeTable AS LT --实际表
CROSS APPLY
dbo.Split(LT.Name,':') --自定义表值函数，完成字符串分隔
WHERE
a <> '' --去掉结果表中a字段为空的数据
-----------字符串分隔函数-------------
CREATE Function [dbo].[Split]
(
@Sql varchar(8000),
@Splits varchar(10)
)
returns @temp Table (a varchar(100))
As
Begin
Declare @i Int
Set @Sql = RTrim(LTrim(@Sql))
Set @i = CharIndex(@Splits,@Sql)

While @i >= 1
Begin
Insert @temp Values(Left(@Sql,@i-1))
Set @Sql = SubString(@Sql,@i+1,Len(@Sql)-@i)
Set @i = CharIndex(@Splits,@Sql)
End

If @Sql <> ''
Insert @temp Values (@Sql)

Return
End
------------调用示例-------------
SELECT
*
FROM
dbo.Split('581::579::519::279::406::361::560',':')

-- 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.

-- 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 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

-- 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, 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
)
SELECT * FROM Employees_Subtree
)
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