Oracle PL/SQL Tutorial/Query Select/ALL
Версия от 13:45, 26 мая 2010; (обсуждение)
Greater than all subquery
SQL>
SQL> create table emp
2 ( empno NUMBER(4) constraint E_PK primary key
3 , ename VARCHAR2(8)
4 , init VARCHAR2(5)
5 , job VARCHAR2(8)
6 , mgr NUMBER(4)
7 , bdate DATE
8 , sal NUMBER(6,2)
9 , comm NUMBER(6,2)
10 , deptno NUMBER(2) default 10
11 ) ;
Table created.
SQL> insert into emp values(1,"Tom","N", "TRAINER", 13,date "1965-12-17", 800 , NULL, 20);
1 row created.
SQL> insert into emp values(2,"Jack","JAM", "Tester",6,date "1961-02-20", 1600, 300, 30);
1 row created.
SQL> insert into emp values(3,"Wil","TF" , "Tester",6,date "1962-02-22", 1250, 500, 30);
1 row created.
SQL> insert into emp values(4,"Jane","JM", "Designer", 9,date "1967-04-02", 2975, NULL, 20);
1 row created.
SQL> insert into emp values(5,"Mary","P", "Tester",6,date "1956-09-28", 1250, 1400, 30);
1 row created.
SQL> insert into emp values(6,"Black","R", "Designer", 9,date "1963-11-01", 2850, NULL, 30);
1 row created.
SQL> insert into emp values(7,"Chris","AB", "Designer", 9,date "1965-06-09", 2450, NULL, 10);
1 row created.
SQL> insert into emp values(8,"Smart","SCJ", "TRAINER", 4,date "1959-11-26", 3000, NULL, 20);
1 row created.
SQL> insert into emp values(9,"Peter","CC", "Designer",NULL,date "1952-11-17", 5000, NULL, 10);
1 row created.
SQL> insert into emp values(10,"Take","JJ", "Tester",6,date "1968-09-28", 1500, 0, 30);
1 row created.
SQL> insert into emp values(11,"Ana","AA", "TRAINER", 8,date "1966-12-30", 1100, NULL, 20);
1 row created.
SQL> insert into emp values(12,"Jane","R", "Manager", 6,date "1969-12-03", 800 , NULL, 30);
1 row created.
SQL> insert into emp values(13,"Fake","MG", "TRAINER", 4,date "1959-02-13", 3000, NULL, 20);
1 row created.
SQL> insert into emp values(14,"Mike","TJA","Manager", 7,date "1962-01-23", 1300, NULL, 10);
1 row created.
SQL>
SQL>
SQL> select e.empno, e.ename, e.job, e.sal
2 from emp e
3 where e.sal > ALL (select x.sal
4 from emp x
5 where x.job = "Designer");
no rows selected
SQL>
SQL> drop table emp;
Table dropped.
Use the ANY operator in a WHERE clause to compare a value with any of the values in a list
You must place an =, <>, <, >, <=, or >= operator before ANY.
The following SELECT statement uses the ANY operator to retrieve rows from the employee table where the value in the id column is greater than any of the values 2, 3, or 4:
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"), 6661.78, "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"), 6544.78, "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"), 2344.78, "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"), 4322.78,"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"), 7897.78,"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"), 1232.78,"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 6661.78 Vancouver Tester
03 James Smith 12-DEC-78 15-MAR-90 6544.78 Vancouver Tester
04 Celia Rice 24-OCT-82 21-APR-99 2344.78 Vancouver Manager
05 Robert Black 15-JAN-84 08-AUG-98 2334.78 Vancouver Tester
06 Linda Green 30-JUL-87 04-JAN-96 4322.78 New York Tester
07 David Larry 31-DEC-90 12-FEB-98 7897.78 New York Manager
08 James Cat 17-SEP-96 15-APR-02 1232.78 Vancouver Tester
8 rows selected.
SQL>
SQL>
SQL> SELECT * FROM employee WHERE id > ANY (2, 3, 4);
ID FIRST_NAME LAST_NAME START_DAT END_DATE SALARY CITY DESCRIPTION
---- ---------- ---------- --------- --------- ---------- ---------- ---------------
03 James Smith 12-DEC-78 15-MAR-90 6544.78 Vancouver Tester
04 Celia Rice 24-OCT-82 21-APR-99 2344.78 Vancouver Manager
05 Robert Black 15-JAN-84 08-AUG-98 2334.78 Vancouver Tester
06 Linda Green 30-JUL-87 04-JAN-96 4322.78 New York Tester
07 David Larry 31-DEC-90 12-FEB-98 7897.78 New York Manager
08 James Cat 17-SEP-96 15-APR-02 1232.78 Vancouver Tester
6 rows selected.
SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee
2 /
Table dropped.
SQL>
Using the ALL Operator with subquery
SQL>
SQL>
SQL>
SQL> CREATE TABLE dept (
2 deptID INT NOT NULL PRIMARY KEY,
3 empID INT NOT NULL,
4 ClassID INT NOT NULL,
5 EnrolledOn DATE,
6 Grade INT);
Table created.
SQL>
SQL> INSERT INTO dept (deptID,empID,ClassID,EnrolledOn,Grade) VALUES (1,1,1,DATE "2002-09-23",62);
1 row created.
SQL> INSERT INTO dept (deptID,empID,ClassID,EnrolledOn,Grade) VALUES (2,1,2,DATE "2002-09-30",70);
1 row created.
SQL> INSERT INTO dept (deptID,empID,ClassID,EnrolledOn,Grade) VALUES (3,2,3,DATE "2003-09-23",51);
1 row created.
SQL> INSERT INTO dept (deptID,empID,ClassID,EnrolledOn,Grade) VALUES (4,2,5,DATE "2003-09-23",41);
1 row created.
SQL> INSERT INTO dept (deptID,empID,ClassID,EnrolledOn,Grade) VALUES (5,2,6,DATE "2003-09-23",68);
1 row created.
SQL> INSERT INTO dept (deptID,empID,ClassID,EnrolledOn,Grade) VALUES (6,3,4,DATE "2002-09-30",78);
1 row created.
SQL> INSERT INTO dept (deptID,empID,ClassID,EnrolledOn,Grade) VALUES (7,3,7,DATE "2002-09-30",80);
1 row created.
SQL> INSERT INTO dept (deptID,empID,ClassID,EnrolledOn,Grade) VALUES (8,4,8,DATE "2002-09-20",70);
1 row created.
SQL>
SQL>
SQL> CREATE TABLE empExam (
2 empID INT NOT NULL,
3 ExamID INT NOT NULL,
4 Mark INT,
5 Taken SMALLINT,
6 Comments VARCHAR(255),
7 CONSTRAINT PK_empExam PRIMARY KEY (empID, ExamID));
SQL>
SQL>
SQL>
SQL> SELECT empID, Grade FROM dept e
2 WHERE Grade > ALL (
3 SELECT Mark FROM empExam s
4 WHERE s.empID = e.empID);
EMPID GRADE
---------- ----------
2 68
3 78
3 80
4 70
4 rows selected.
SQL>
SQL>
SQL> drop table dept;
Table dropped.
SQL> drop table empExam;
Table dropped.