Oracle PL/SQL/Select Query/Derive values

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Use Arithmetic operators with literal values to derive values

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SQL> SQL> 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> SQL> SQL> SQL> SQL> --Use Arithmetic operators with literal values to derive values: * SQL> SQL> SELECT ID, First_Name, Salary * 5 FROM Employee; ID FIRST_NAME SALARY*5


---------- ----------

01 Jason 6172.8 02 Alison 11673.9 03 James 11673.9 04 Celia 11673.9 05 Robert 11673.9 06 Linda 11673.9 07 David 11673.9 08 James 11673.9 8 rows selected. SQL> SQL> SQL> SQL> SQL> SQL> SQL> SQL> -- clean the table SQL> drop table Employee

 2  /

Table dropped. SQL> SQL> SQL>

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