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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.

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