Oracle PL/SQL Tutorial/Character String Functions/LENGTH

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LENGTH(x) get the number of characters in x

The following example displays the length of the strings in the first_name column.



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>


Order by the length of employee name

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.


Use length() function three times in a select statement

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>


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.



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>