SQLServer2012多表连接查询功能实例代码

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废话不多说了,直接给大家贴代码了,具体代码如下所示:

-- 交叉连接产生笛卡尔值 (X*Y)
SELECT *
FROM  Student
    cross Join dbo.ClassInfo 
--另外一种写法
SELECT *
FROM  Student , ClassInfo 
-- 内连接 (Inner 可以省略)
SELECT *
FROM  Student
    JOIN dbo.ClassInfo ON dbo.Student.Class = dbo.ClassInfo.ID;
-- Inner Join
SELECT *
FROM  Student
    INNER JOIN dbo.ClassInfo ON dbo.Student.Class = dbo.ClassInfo.ID;
   -- on 条件,通常是 主外键,但是不限于主外键
   -- on 条件,允许有多个,而且可能是针对某个表的
SELECT *
FROM  Student
    INNER JOIN dbo.ClassInfo ON dbo.Student.Class = dbo.ClassInfo.ID
    AND dbo.Student.Class IN (1,2) -- 针对Student表增加查询条件
-- 不等于 (笛卡尔值减去 相等的值)
SELECT *
FROM  Student
    INNER JOIN dbo.ClassInfo ON dbo.Student.Class <> dbo.ClassInfo.ID;
--自连接 从Class表中查询到Class所在的系
SELECT T1.* ,T2.ClassName FROM dbo.ClassInfo AS T1
    INNER JOIN dbo.ClassInfo AS T2 ON T1.PID=T2.ID
-- Left Join
SELECT *
FROM  Student
    Left JOIN dbo.ClassInfo ON dbo.Student.Class = dbo.ClassInfo.ID;
-- Right Join
SELECT *
FROM  Student
    RIGHT  JOIN dbo.ClassInfo ON dbo.Student.Class = dbo.ClassInfo.ID;
-- 全外连接
SELECT *
FROM  Student
    FULL OUTER JOIN dbo.ClassInfo ON dbo.Student.Class = dbo.ClassInfo.ID;
-- 外连接和内连接的区别是:
-- 内连接,on条件不符合的话,会过滤掉数据
-- 外连接,以保留表为主,on条件成立显示数据,否则显示NULL
--Union 纵向连接
SELECT StuID,StuName,StuEnName,StuAge,StuBirthday
FROM  Student WHERE StuID<=2
UNION
SELECT StuID,StuName,StuEnName,StuAge,StuBirthday
FROM dbo.Student WHERE StuID>2
--Union 去重复
SELECT StuSex
FROM  Student WHERE StuID<=2
UNION
SELECT StuSex
FROM dbo.Student WHERE StuID>2
--Union 显示全部
SELECT StuSex
FROM  Student WHERE StuID<=2
UNION ALL
SELECT StuSex
FROM dbo.Student WHERE StuID>2
--Except 差集,排除
SELECT StuID,StuName,StuEnName,StuAge,StuBirthday
FROM  Student
Except
SELECT StuID,StuName,StuEnName,StuAge,StuBirthday
FROM dbo.Student WHERE StuID<=2
--Intersect 交集
SELECT StuID,StuName,StuEnName,StuAge,StuBirthday
FROM  Student WHERE StuID>=2
Except
SELECT StuID,StuName,StuEnName,StuAge,StuBirthday
FROM dbo.Student WHERE StuID<=3

以上所述是小编给大家介绍的SQL Server 2012 多表连接查询功能实例代码,希望对大家有所帮助,如果大家有任何疑问欢迎给我留言,小编会及时回复大家的!

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