例子
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
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,以下是使用 TABLE ALIAS 我们分别使用 C 和 D 作为 COMPANY 和 DEPARTMENT 表的别名 -
sqlite> SELECT C.ID, C.NAME, C.AGE, D.DEPT
FROM COMPANY AS C, DEPARTMENT AS D
WHERE C.ID = D.EMP_ID;
上面的 SQLite 语句将产生以下结果 -
ID NAME AGE DEPT
---------- ---------- ---------- ----------
1 Paul 32 IT Billing
2 Allen 25 Engineering
3 Teddy 23 Engineering
4 Mark 25 Finance
5 David 27 Engineering
6 Kim 22 Finance
7 James 24 Finance
考虑一个使用的例子 COLUMN ALIAS 其中 COMPANY_ID 是 ID 列的别名,COMPANY_NAME 是 name 列的别名。
sqlite> SELECT C.ID AS COMPANY_ID, C.NAME AS COMPANY_NAME, C.AGE, D.DEPT
FROM COMPANY AS C, DEPARTMENT AS D
WHERE C.ID = D.EMP_ID;
上面的 SQLite 语句将产生以下结果 -
COMPANY_ID COMPANY_NAME AGE DEPT
---------- ------------ ---------- ----------
1 Paul 32 IT Billing
2 Allen 25 Engineering
3 Teddy 23 Engineering
4 Mark 25 Finance
5 David 27 Engineering
6 Kim 22 Finance
7 James 24 Finance