| 标题 | oracle distinct 的使用方法 |
| 范文 | 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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