PostgreSQL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。 www.yiibai.com
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,但它们不必是相同的长度。
UNION的基本语法如下: yiibai.com
SELECT column1 [, column2 ] FROM table1 [, table2 ] [WHERE condition] UNION SELECT column1 [, column2 ] FROM table1 [, table2 ] [WHERE condition]yiibai.com
这里给定的条件可以是根据需要的任何表达式。
考虑以下两个表:COMPANY 表如下:
yiibai.com
testdb=# SELECT * from COMPANY; id | name | age | address | salary ----+-------+-----+-----------+-------- 1 | Paul | 32 | California| 20000 2 | Allen | 25 | Texas | 15000 3 | Teddy | 23 | Norway | 20000 4 | Mark | 25 | Rich-Mond | 65000 5 | David | 27 | Texas | 85000 6 | Kim | 22 | South-Hall| 45000 7 | James | 24 | Houston | 10000 (7 rows)yiibai.com
另一表是DEPARTMENT 如下: www.yiibai.com
testdb=# SELECT * from DEPARTMENT; 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 (7 rows)yiibai.com
现在,让我们加入这两个表使用SELECT语句连同UNION子句如下: www.yiibai.com
testdb=# 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; yiibai.com
This would produce following result:
emp_id | name | dept
--------+-------+--------------
5 | David | Engineering
6 | Kim | Finance
2 | Allen | Engineering
3 | Teddy | Engineering
4 | Mark | Finance
1 | Paul | IT Billing
7 | James | Finance
(7 rows)
yiibai.com
UNION ALL运算符是用来结合两个SELECT语句,包括重复行的结果。联盟适用同样的规则适用于UNION ALL运算符。 www.yiibai.com
UNION ALL的基本语法如下:
www.yiibai.com
SELECT column1 [, column2 ] FROM table1 [, table2 ] [WHERE condition] UNION ALL SELECT column1 [, column2 ] FROM table1 [, table2 ] [WHERE condition] www.yiibai.com
这里给定的条件可以是任何表达式,根据需要。
www.yiibai.com
现在,让我们携手上述两个表中的SELECT语句如下: www.yiibai.com
testdb=# 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; www.yiibai.com
这将产生以下结果: yiibai.com
emp_id | name | dept --------+-------+-------------- 1 | Paul | IT Billing 2 | Allen | Engineering 7 | James | Finance 3 | Teddy | Engineering 4 | Mark | Finance 5 | David | Engineering 6 | Kim | Finance 1 | Paul | IT Billing 2 | Allen | Engineering 7 | James | Finance 3 | Teddy | Engineering 4 | Mark | Finance 5 | David | Engineering 6 | Kim | Finance (14 rows) www.yiibai.com