oracle distinct 的使用方法

distinct这个关键字来过滤掉多余的重复记录只保留一条，但往往只用它来返回不重复记录的条数，而不是用它来返回不重记录的所有值。其原因是distinct只有用二重循环查询来解决，而这样对于一个数据量非常大的站来说，无疑是会直接影响到效率的。

SQL> -- create demo table
SQL> create table Employee(
2    ID                 VARCHAR2(4 BYTE)         NOT NULL,
3    First_Name         VARCHAR2(10 BYTE),
4    Last_Name          VARCHAR2(10 BYTE),
5    Start_Date         DATE,
6    End_Date           DATE,
7    Salary             Number(8,2),
8    City               VARCHAR2(10 BYTE),
9    Description        VARCHAR2(15 BYTE)
10 )
11 /

Table created.

SQL>
SQL> -- prepare data
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,       Description)
2               values ('01','Jason',    'Martin', to_date('19960725','YYYYMMDD'), to_date('20060725','YYYYMMDD'), 1234.56, 'Toronto', 'Programmer')
3 /

1 row created.

SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,       Description)
2                values('02','Alison',   'Mathews', to_date('19760321','YYYYMMDD'), to_date('19860221','YYYYMMDD'), 2334.78, 'Vancouver','Tester')
3 /

1 row created.

SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,       Description)
2                values('03','James',    'Smith',   to_date('19781212','YYYYMMDD'), to_date('19900315','YYYYMMDD'), 2334.78, 'Vancouver','Tester')
3 /

1 row created.

SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,       Description)
2                values('04','Celia',    'Rice',    to_date('19821024','YYYYMMDD'), to_date('19990421','YYYYMMDD'), 2334.78, 'Vancouver','Manager')
3 /

1 row created.

SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,       Description)
2                values('05','Robert',   'Black',   to_date('19840115','YYYYMMDD'), to_date('19980808','YYYYMMDD'), 2334.78, 'Vancouver','Tester')
3 /

1 row created.

SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,        Description)
2                values('06','Linda',    'Green',   to_date('19870730','YYYYMMDD'), to_date('19960104','YYYYMMDD'), 2334.78,'New York', 'Tester')
3 /

1 row created.

SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,        Description)
2                values('07','David',    'Larry',   to_date('19901231','YYYYMMDD'), to_date('19980212','YYYYMMDD'), 2334.78,'New York', 'Manager')
3 /

1 row created.

SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,        Description)
2                values('08','James',    'Cat',     to_date('19960917','YYYYMMDD'), to_date('20020415','YYYYMMDD'), 2334.78,'Vancouver', 'Tester')
3 /

1 row created.

SQL>
SQL>
SQL>
SQL> -- display data in the table
SQL> select * from Employee
2 /

ID   FIRST_NAME LAST_NAME START_DAT END_DATE      SALARY CITY       DESCRIPTION
---- ---------- ---------- --------- --------- ---------- ---------- ---------------
01   Jason      Martin     25-JUL-96 25-JUL-06    1234.56 Toronto    Programmer
02   Alison     Mathews    21-MAR-76 21-FEB-86    2334.78 Vancouver Tester
03   James      Smith      12-DEC-78 15-MAR-90    2334.78 Vancouver Tester
04   Celia      Rice       24-OCT-82 21-APR-99    2334.78 Vancouver Manager
05   Robert     Black      15-JAN-84 08-AUG-98    2334.78 Vancouver Tester
06   Linda      Green      30-JUL-87 04-JAN-96    2334.78 New York   Tester
07   David      Larry      31-DEC-90 12-FEB-98    2334.78 New York   Manager
08   James      Cat        17-SEP-96 15-APR-02    2334.78 Vancouver Tester

8 rows selected.

SQL>
SQL>
SQL>
SQL>
SQL>
SQL> -- Remember that the DISTINCT operator applies to the entire select list.
SQL>
SQL> SELECT DISTINCT City, Description FROM Employee;

CITY       DESCRIPTION
---------- ---------------
New York   Manager
Vancouver Tester
Toronto    Programmer
Vancouver Manager
New York   Tester

同时与groupy count 使用的用法

SQL> select Coder
2 ,      count(distinct course)
3 ,      count(*)
4 from   offerings
5 group by Coder;

     CODER COUNT(DISTINCTCOURSE)   COUNT(*)
---------- --------------------- ----------
         1                     2          3
         4                     2          2
         8                     2          2
        11                     1          1
        13                     2          2
                               3          3

6 rows selected.

补充：数据库,Oracle教程