Oracle PL/SQL Tutorial/Character String Functions/LENGTH
Содержание
LENGTH(x) get the number of characters in x
The following example displays the length of the strings in the first_name column.
<source lang="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"), 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> SELECT first_name, LENGTH(first_name) FROM employee; FIRST_NAME LENGTH(FIRST_NAME)
------------------
Jason 5 Alison 6 James 5 Celia 5 Robert 6 Linda 5 David 5 James 5 8 rows selected. SQL> SQL> SQL> SQL> SQL> -- clean the table SQL> drop table Employee
2 /
Table dropped. SQL></source>
Order by the length of employee name
<source lang="sql">
SQL> SQL> SQL> create table employees(
2 empno NUMBER(4) 3 , ename VARCHAR2(8) 4 , init VARCHAR2(5) 5 , job VARCHAR2(8) 6 , mgr NUMBER(4) 7 , bdate DATE 8 , msal NUMBER(6,2) 9 , comm NUMBER(6,2) 10 , deptno NUMBER(2) ) ;
Table created. SQL> SQL> SQL> insert into employees values(1,"Jason", "N", "TRAINER", 2, date "1965-12-18", 800 , NULL, 10); 1 row created. SQL> insert into employees values(2,"Jerry", "J", "SALESREP",3, date "1966-11-19", 1600, 300, 10); 1 row created. SQL> insert into employees values(3,"Jord", "T" , "SALESREP",4, date "1967-10-21", 1700, 500, 20); 1 row created. SQL> insert into employees values(4,"Mary", "J", "MANAGER", 5, date "1968-09-22", 1800, NULL, 20); 1 row created. SQL> insert into employees values(5,"Joe", "P", "SALESREP",6, date "1969-08-23", 1900, 1400, 30); 1 row created. SQL> insert into employees values(6,"Black", "R", "MANAGER", 7, date "1970-07-24", 2000, NULL, 30); 1 row created. SQL> insert into employees values(7,"Red", "A", "MANAGER", 8, date "1971-06-25", 2100, NULL, 40); 1 row created. SQL> insert into employees values(8,"White", "S", "TRAINER", 9, date "1972-05-26", 2200, NULL, 40); 1 row created. SQL> insert into employees values(9,"Yellow", "C", "DIRECTOR",10, date "1973-04-27", 2300, NULL, 20); 1 row created. SQL> insert into employees values(10,"Pink", "J", "SALESREP",null,date "1974-03-28", 2400, 0, 30); 1 row created. SQL> SQL> select lower(job), initcap(ename)
2 from employees 3 where upper(job) = "SALESREP" 4 order by length(ename);
LOWER(JO INITCAP(
--------
salesrep Joe salesrep Pink salesrep Jord salesrep Jerry SQL> SQL> SQL> drop table employees; Table dropped.</source>
Use length() function three times in a select statement
<source lang="sql">
SQL> SQL> CREATE TABLE book(
2 title_id CHAR(3) NOT NULL, 3 title_name VARCHAR(40) NOT NULL, 4 type VARCHAR(10) NULL , 5 pub_id CHAR(3) NOT NULL, 6 pages INTEGER NULL , 7 price DECIMAL(5,2) NULL , 8 sales INTEGER NULL , 9 pubdate DATE NULL , 10 contract SMALLINT NOT NULL 11 );
Table created. SQL> SQL> SQL> SQL> INSERT INTO book VALUES("T01","Java","history","P01",111,21.99,566,DATE "2000-08-01",1); 1 row created. SQL> INSERT INTO book VALUES("T02","Oracle","history","P03", 114,19.95,9566,DATE "1998-04-01",1); 1 row created. SQL> INSERT INTO book VALUES("T03","SQL","computer","P02", 122,39.95,25667,DATE "2000-09-01",1); 1 row created. SQL> INSERT INTO book VALUES("T04","C++","psychology","P04", 511,12.99,13001,DATE "1999-05-31",1); 1 row created. SQL> INSERT INTO book VALUES("T05","Python","psychology","P04", 101,6.95,201440,DATE "2001-01-01",1); 1 row created. SQL> INSERT INTO book VALUES("T06","JavaScript","biography","P01", 173,19.95,11320,DATE "2000-07-31",1); 1 row created. SQL> INSERT INTO book VALUES("T07","LINQ","biography","P03", 331,23.95,1500200,DATE "1999-10-01",1); 1 row created. SQL> INSERT INTO book VALUES("T08","C#","children","P04", 861,10.00,4095,DATE "2001-06-01",1); 1 row created. SQL> INSERT INTO book VALUES("T09","SQL Server","children","P04", 212,13.95,5000,DATE "2002-05-31",1); 1 row created. SQL> INSERT INTO book VALUES("T10","AJAX","biography","P01", NULL,NULL,NULL,NULL,0); 1 row created. SQL> INSERT INTO book VALUES("T11","VB","psychology","P04", 821,7.99,94123,DATE "2000-11-30",1); 1 row created. SQL> INSERT INTO book VALUES("T12","Office","biography","P01", 507,12.99,100001,DATE "2000-08-31",1); 1 row created. SQL> INSERT INTO book VALUES("T13","VBA","history","P03", 812,29.99,10467,DATE "1999-05-31",1); 1 row created. SQL> SQL> SQL> SELECT title_name,
2 LENGTH(title_name) AS "Len" 3 FROM book 4 WHERE LENGTH(title_name) < 30 5 ORDER BY LENGTH(title_name) ASC;
TITLE_NAME Len
----------
VB 2 C# 2 VBA 3 SQL 3 C++ 3 Java 4 LINQ 4 AJAX 4 Office 6 Oracle 6 Python 6 TITLE_NAME Len
----------
SQL Server 10 JavaScript 10 13 rows selected. SQL> SQL> drop table book; Table dropped. SQL> SQL> SQL></source>
Use Length() to deal with number column
The following example displays the total number of characters that make up the salary using LENGTH().
The decimal point (.) is counted in the number of salary characters.
<source lang="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"), 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 salary, LENGTH(salary) FROM employee;
SALARY LENGTH(SALARY)
--------------
1234.56 7 6661.78 7 6544.78 7 2344.78 7 2334.78 7 4322.78 7 7897.78 7 1232.78 7
8 rows selected. SQL> SQL> SQL> SQL> -- clean the table SQL> drop table Employee
2 /
Table dropped. SQL></source>