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。
为了使用 UNION,每个 SELECT 被选择的列数必须是相同的,相同数目的列表达式,相同的数据类型,并确保它们有相同的顺序,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]
这里给定的条件根据需要可以是任何表达式。
假设有下面两个表,(1)COMPANY 表如下所示:
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
(2)另一个表是 DEPARTMENT,如下所示:
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,让我们使用 SELECT 语句及 UNION 子句来连接两个表,如下所示:
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 运算符用于结合两个 SELECT 语句的结果,包括重复行。
适用于 UNION 的规则同样适用于 UNION ALL 运算符。
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]
这里给定的条件根据需要可以是任何表达式。
现在,让我们使用 SELECT 语句及 UNION ALL 子句来连接两个表,如下所示:
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