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SQLite UNION子句/运算符用于合并两个或多个SELECT语句的结果,而不返回任何重复的行。
要使用UNION,每个SELECT必须具有相同数量的选定列,相同数量的列表达式,相同数据类型,并且具有相同的顺序,但是它们的长度不必相同。
以下是UNION的基本语法。
SELECT column1 [, column2 ] FROM table1 [, table2 ] [WHERE condition] UNION SELECT column1 [, column2 ] FROM table1 [, table2 ] [WHERE condition]
在此,给定条件可以是根据您的要求的任何给定表达式。
考虑以下两个表,(a)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
(b)另一个表是部门(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
现在,让我们使用SELECT语句和UNION子句将这两个表连接起来,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 Engineering 3 Teddy Engineering 4 Mark Finance 5 David Engineering 6 Kim Finance 7 James Finance
UNION ALL运算符用于合并两个SELECT语句的结果,包括重复的行。
适用于UNION的相同规则也适用于UNION ALL运算符。
以下是的基本语法UNION ALL。
SELECT column1 [, column2 ] FROM table1 [, table2 ] [WHERE condition] UNION ALL SELECT column1 [, column2 ] FROM table1 [, table2 ] [WHERE condition]
在此,给定条件可以是根据您的要求的任何给定表达式。
现在,让我们在SELECT语句中将上述两个表连接如下:
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 Engineering 3 Teddy Engineering 4 Mark Finance 5 David Engineering 6 Kim Finance 7 James Finance 1 Paul IT Billing 2 Allen Engineering 3 Teddy Engineering 4 Mark Finance 5 David Engineering 6 Kim Finance 7 James Finance