Oracle PL/SQL Tutorial/Regular Expressions Functions/REGEXP SUBSTR

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Backreference

The backslash may be followed by a number.
In the backreference version, "\1" says to match the same string as was matched by the nth subexpression.
As a first example, we can use the backreference in a manner similar to the repeat operator.

SQL>
SQL> SELECT REGEXP_SUBSTR("Yababa dababa do","(ab)")
  2  FROM dual;
RE
--
ab
SQL>
SQL>
SQL> SELECT REGEXP_SUBSTR("Yababa dababa do","(ab)\1")
  2  FROM dual;
REGE
----
abab
SQL>
SQL>
SQL> SELECT REGEXP_SUBSTR("Yababa dababa do","(ab){2}") from dual;
REGE
----
abab
SQL>


Grouping

SQL>
SQL> -- create demo table
SQL> create table myTable(
  2    id           NUMBER(2),
  3    value        VARCHAR(50)
  4  );
Table created.
SQL>
SQL> insert into myTable(id, value)values(1,"1234 4th St. Vancouver");
1 row created.
SQL> insert into myTable(id, value)values(2,"4 Maple Ct. New York");
1 row created.
SQL> insert into myTable(id, value)values(3,"4321 Green Blvd. London");
1 row created.
SQL> insert into myTable(id, value)values(4,"33 Third St. Toronto");
1 row created.
SQL> insert into myTable(id, value)values(5,"One First Drive. Queen");
1 row created.
SQL> insert into myTable(id, value)values(6,"1664 1/2 Springhill Ave");
1 row created.
SQL> insert into myTable(id, value)values(7,"665 Fall Ave. Linken");
1 row created.
SQL>
SQL> select * from mytable;
        ID VALUE
---------- --------------------------------------------------
         1 1234 4th St. Vancouver
         2 4 Maple Ct. New York
         3 4321 Green Blvd. London
         4 33 Third St. Toronto
         5 One First Drive. Queen
         6 1664 1/2 Springhill Ave
         7 665 Fall Ave. Linken
7 rows selected.
SQL>
SQL> SELECT value, REGEXP_SUBSTR(value,"ird|irs")
  2  FROM myTable;
VALUE                                              REGEXP_SUBSTR(VALUE,"IRD|IRS")
-------------------------------------------------- -------------------------------------------------
-
1234 4th St. Vancouver
4 Maple Ct. New York
4321 Green Blvd. London
33 Third St. Toronto                               ird
One First Drive. Queen                             irs
1664 1/2 Springhill Ave
665 Fall Ave. Linken
7 rows selected.
SQL>
SQL>
SQL>
SQL>
SQL> drop table myTable;
Table dropped.
SQL>
SQL>


Group the letters "ir" together by putting them in parentheses and then parenthesizing the suffix using alternation

SQL>
SQL> -- create demo table
SQL> create table myTable(
  2    id           NUMBER(2),
  3    value        VARCHAR(50)
  4  );
Table created.
SQL>
SQL> insert into myTable(id, value)values(1,"1234 4th St. Vancouver");
1 row created.
SQL> insert into myTable(id, value)values(2,"4 Maple Ct. New York");
1 row created.
SQL> insert into myTable(id, value)values(3,"4321 Green Blvd. London");
1 row created.
SQL> insert into myTable(id, value)values(4,"33 Third St. Toronto");
1 row created.
SQL> insert into myTable(id, value)values(5,"One First Drive. Queen");
1 row created.
SQL> insert into myTable(id, value)values(6,"1664 1/2 Springhill Ave");
1 row created.
SQL> insert into myTable(id, value)values(7,"665 Fall Ave. Linken");
1 row created.
SQL>
SQL> select * from mytable;
        ID VALUE
---------- --------------------------------------------------
         1 1234 4th St. Vancouver
         2 4 Maple Ct. New York
         3 4321 Green Blvd. London
         4 33 Third St. Toronto
         5 One First Drive. Queen
         6 1664 1/2 Springhill Ave
         7 665 Fall Ave. Linken
7 rows selected.
SQL>
SQL>
SQL> SELECT value, REGEXP_SUBSTR(value,"(ir)(d|s)")
  2  FROM myTable;
VALUE                                              REGEXP_SUBSTR(VALUE,"(IR)(D|S)")
-------------------------------------------------- -------------------------------------------------
-
1234 4th St. Vancouver
4 Maple Ct. New York
4321 Green Blvd. London
33 Third St. Toronto                               ird
One First Drive. Queen                             irs
1664 1/2 Springhill Ave
665 Fall Ave. Linken
7 rows selected.
SQL>
SQL>
SQL>
SQL> drop table myTable;
Table dropped.
SQL>


Regexp_Substr returns a string of data type VARCHAR2 or CLOB

REGEXP_SUBSTR uses regular expressions to specify the beginning and ending points of the returned string.

The simplest format for this function is:

REGEXP_SUBSTR(source_string, pattern_to_find)

The general format for the REGEXP_SUBSTR function with all the options is:

REGEXP_SUBSTR(source_string, pattern_to_find [, position,occurrence, match_parameter])



SQL>
SQL> SELECT REGEXP_SUBSTR("Mississippi", "si", 1, 2, "i") FROM dual;
RE
--
si
SQL>
SQL>


REGEXP_SUBSTR(x, pattern [, start [, occurrence [, match_option]]]) gets a substring of x that matches pattern, which begins at the position specified by start.

The following example returns the substring that matches the regular expression lalpha: {4} using REGEXP_SUBSTR():



SQL>
SQL> SELECT REGEXP_SUBSTR("abcedfghijklumnoprstuvwxyzabcedfghijklumnoprstuvwxyzabcedfghijklumnoprstuvwxyz", "l[[:alpha:]]{4}") AS result
  2  FROM dual;
RESUL
-----
lumno
SQL>
SQL>


Search for (

SQL>
SQL> create table history
  2  ( empno      NUMBER(4)
  3  , beginyear  NUMBER(4)
  4  , begindate  DATE
  5  , enddate    DATE
  6  , deptno     NUMBER(2)
  7  , msal       NUMBER(6,2)
  8  , comments   VARCHAR2(60)
  9  ) ;
Table created.
SQL>
SQL>
SQL>
SQL> insert into history values (9,2000,date "2000-01-01",date "2002-01-02",40, 950,"history for 9");
1 row created.
SQL> insert into history values (8,2000,date "2000-01-02", NULL       ,20, 800,"");
1 row created.
SQL> insert into history values (7,1988,date "2000-01-06",date "2002-01-07",30,1000,"");
1 row created.
SQL> insert into history values (6,1989,date "2000-01-07",date "2002-01-12",30,1300,"");
1 row created.
SQL> insert into history values (5,1993,date "2000-01-12",date "2002-01-10",30,1500,"history for 5");
1 row created.
SQL> insert into history values (4,1995,date "2000-01-10",date "2002-01-11",30,1700,"");
1 row created.
SQL> insert into history values (3,1999,date "2000-01-11", NULL       ,30,1600,"");
1 row created.
SQL> insert into history values (2,1986,date "2000-01-10",date "2002-01-08",20,1000,"history for 2");
1 row created.
SQL> insert into history values (1,1987,date "2000-01-08",date "2002-01-01",30,1000,"history for 1");
1 row created.
SQL> insert into history values (7,1989,date "2000-01-01",date "2002-05-12",30,1150,"history for 7");
1 row created.
SQL>
SQL>
SQL>
SQL> select comments
  2  ,      regexp_substr(comments, "\([^\)]+\)") as substring
  3  from   history
  4  where  comments like "%(%";
no rows selected
SQL>
SQL>
SQL> drop table history;
Table dropped.
Источник — «http://sqle.ru/index.php?title=Oracle_PL/SQL_Tutorial/Regular_Expressions_Functions/REGEXP_SUBSTR&oldid=4102»