Oracle PL/SQL Tutorial/Regular Expressions Functions/REGEXP SUBSTR
Содержание
- 1 Backreference
- 2 Grouping
- 3 Group the letters "ir" together by putting them in parentheses and then parenthesizing the suffix using alternation
- 4 Regexp_Substr returns a string of data type VARCHAR2 or CLOB
- 5 REGEXP_SUBSTR(x, pattern [, start [, occurrence [, match_option]]]) gets a substring of x that matches pattern, which begins at the position specified by start.
- 6 Search for (
Backreference
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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></source>
Grouping
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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></source>
Group the letters "ir" together by putting them in parentheses and then parenthesizing the suffix using alternation
<source lang="sql">
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></source>
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])
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SQL> SQL> SELECT REGEXP_SUBSTR("Mississippi", "si", 1, 2, "i") FROM dual; RE -- si SQL> SQL></source>
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():
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SQL> SQL> SELECT REGEXP_SUBSTR("abcedfghijklumnoprstuvwxyzabcedfghijklumnoprstuvwxyzabcedfghijklumnoprstuvwxyz", "lalpha:{4}") AS result
2 FROM dual;
RESUL
lumno SQL> SQL></source>
Search for (
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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.</source>