发布于 2015-07-25 11:48:01 | 218 次阅读 | 评论: 0 | 来源: 网络整理
SQLite的UNION子句/运算符用于合并两个或多个SELECT语句的结果,不返回任何重复的行。
直接使用UNION,每个SELECT选择的列数必须具有相同的,相同数目的列表达式相同的数据类型,并让它们在相同的顺序,但它们不必具有相同的长度。
UNION的基本语法如下:
SELECT column1 [, column2 ]
FROM table1 [, table2 ]
[WHERE condition]
UNION
SELECT column1 [, column2 ]
FROM table1 [, table2 ]
[WHERE condition]
这里给定的条件可以是任何表达式,根据需要。
考虑以下两个表:COMPANY表如下:
sqlite> select * from COMPANY;
ID NAME AGE ADDRESS SALARY
---------- -------------------- ---------- ---------- ----------
1 Paul 32 California 20000.0
2 Allen 25 Texas 15000.0
3 Teddy 23 Norway 20000.0
4 Mark 25 Rich-Mond 65000.0
5 David 27 Texas 85000.0
6 Kim 22 South-Hall 45000.0
7 James 24 Houston 10000.0
另一表是DEPARTMENT如下:
ID DEPT EMP_ID
---------- -------------------- ----------
1 IT Billing 1
2 Engineering 2
3 Finance 7
4 Engineering 3
5 Finance 4
6 Engineering 5
7 Finance 6
现在,让我们加入这两个表使用SELECT语句一起UNION子句如下:
sqlite> SELECT EMP_ID, NAME, DEPT FROM COMPANY INNER JOIN DEPARTMENT
ON COMPANY.ID = DEPARTMENT.EMP_ID
UNION
SELECT EMP_ID, NAME, DEPT FROM COMPANY LEFT OUTER JOIN DEPARTMENT
ON COMPANY.ID = DEPARTMENT.EMP_ID;
这将产生以下结果:
EMP_ID NAME DEPT
---------- -------------------- ----------
1 Paul IT Billing
2 Allen Engineerin
3 Teddy Engineerin
4 Mark Finance
5 David Engineerin
6 Kim Finance
7 James Finance
UNION ALL运算符是用来结合两个SELECT语句,包括重复行的结果。
UNION 适用同样的规则适用于UNION 所有操作符。
UNION ALL的基本语法如下:
SELECT column1 [, column2 ]
FROM table1 [, table2 ]
[WHERE condition]
UNION ALL
SELECT column1 [, column2 ]
FROM table1 [, table2 ]
[WHERE condition]
这里给定的条件可以是任何表达式,根据需要。
现在,让我们连接上述两个表中的SELECT语句如下:
sqlite> SELECT EMP_ID, NAME, DEPT FROM COMPANY INNER JOIN DEPARTMENT
ON COMPANY.ID = DEPARTMENT.EMP_ID
UNION ALL
SELECT EMP_ID, NAME, DEPT FROM COMPANY LEFT OUTER JOIN DEPARTMENT
ON COMPANY.ID = DEPARTMENT.EMP_ID;
这将产生以下结果:
EMP_ID NAME DEPT
---------- -------------------- ----------
1 Paul IT Billing
2 Allen Engineerin
3 Teddy Engineerin
4 Mark Finance
5 David Engineerin
6 Kim Finance
7 James Finance
1 Paul IT Billing
2 Allen Engineerin
3 Teddy Engineerin
4 Mark Finance
5 David Engineerin
6 Kim Finance
7 James Finance