oracle正则表达式

--The following query returns the first and last names for those employees with a first name of Steven or Stephen 
--(where first_name begins with Ste and ends with en and in between is either v or ph):
--查询返回结果为first_name以ste开头，en结尾，中间为v或ph
SQL> SELECT first_name, last_name
  2  FROM employees
  3  WHERE REGEXP_LIKE (first_name, ‘^Ste(v|ph)en$‘)
  4  ORDER BY first_name, last_name;
FIRST_NAME           LAST_NAME
-------------------- -------------------------
Stephen              Stiles
Steven               King
Steven               Markle

--The following query returns the last name for those employees with a double vowel in their last name 
--(where last_name contains two adjacent occurrences of either a, e, i, o, or u, regardless of case):
--last_name中包含aeiou中任何一个，不区分大小写
SELECT last_name
FROM employees
WHERE REGEXP_LIKE (last_name, ‘([aeiou])\1‘, ‘i‘)
ORDER BY last_name;

LAST_NAME
-------------------------
De Haan
Greenberg
Khoo
Gee
Greene
Lee
Bloom
Feeney

--用到的实例 判断客户信息中手机号格式是否正确
   REGEXP_LIKE(CUS_MOBILE1, ‘\d*([0-9]{1}+[0-9]{1}+)\1{4,}\d*‘) --类似1313131……
OR REGEXP_LIKE(CUS_MOBILE1, ‘\d*(\d)\1{6,}\d*‘) --某一位数连续重复5次
OR  REGEXP_LIKE(CUS_MOBILE1,‘0123|1234|2345|3456|4567|5678|6789|7890|01234|12345|23456|34567|45678|56789|67890
                            |012345|123456|234567|345678|456789|567890
                            |0123456|1234567|2345678|3456789|4567890|01234567|12345678|23456789|34567890
                            |012345678|123456789|234567890|0123456789|1234567890|‘) --连续增长7位以上

--The following example examines phone_number,
-- looking for the pattern xxx.xxx.xxxx. Oracle reformats this pattern with (xxx) xxx-xxxx.
SELECT
  REGEXP_REPLACE(phone_number,
                 ‘([[:digit:]]{3})\.([[:digit:]]{3})\.([[:digit:]]{4})‘,
                 ‘(\1) \2-\3‘) "REGEXP_REPLACE"
  FROM employees
  ORDER BY "REGEXP_REPLACE";

REGEXP_REPLACE
--------------------------------------------------------------------------------
(515) 123-4444
(515) 123-4567
(515) 123-4568
(515) 123-4569
(515) 123-5555

--The following example examines the string, looking for two or more spaces. 
--Oracle replaces each occurrence of two or more spaces with a single space.
SELECT
  REGEXP_REPLACE(‘500   Oracle     Parkway,    Redwood  Shores, CA‘,
                 ‘( ){2,}‘, ‘ ‘) "REGEXP_REPLACE"
  FROM DUAL;

REGEXP_REPLACE
--------------------------------------
500 Oracle Parkway, Redwood Shores, CA

--The following example examines the string, looking for the first substring bounded by commas. 
--Oracle Database searches for a comma followed by one or more occurrences of non-comma characters followed by a comma. 
--Oracle returns the substring, including the leading and trailing commas.
SELECT
  REGEXP_SUBSTR(‘500 Oracle Parkway, Redwood Shores, CA‘,
                ‘,[^,]+,‘) "REGEXPR_SUBSTR"
  FROM DUAL;

REGEXPR_SUBSTR
-----------------
, Redwood Shores,

--The following example examines the string, 
--looking for http:// followed by a substring of one or more alphanumeric characters and optionally, 
--a period (.). Oracle searches for a minimum of three and a maximum of four occurrences of this substring between 
--http:// and either a slash (/) or the end of the string.
SELECT
  REGEXP_SUBSTR(‘http://www.example.com/products‘,
                ‘http://([[:alnum:]]+\.?){3,4}/?‘) "REGEXP_SUBSTR"
  FROM DUAL;

REGEXP_SUBSTR
----------------------
http://www.example.com/

--The next two examples use the subexpr argument to return a specific subexpression of pattern.
--The first statement returns the first subexpression in pattern:
SELECT REGEXP_SUBSTR(‘1234567890‘, ‘(123)(4(56)(78))‘, 1, 1, ‘i‘, 1) 
"REGEXP_SUBSTR" FROM DUAL;

REGEXP_SUBSTR
-------------------
123

--The next statement returns the fourth subexpression in pattern:
SELECT REGEXP_SUBSTR(‘1234567890‘, ‘(123)(4(56)(78))‘, 1, 1, ‘i‘, 4) 
"REGEXP_SUBSTR" FROM DUAL;

REGEXP_SUBSTR
-------------------
78

--The following example examines the string, looking for occurrences of one or more non-blank characters. 
--Oracle begins searching at the first character in the string and 
--returns the starting position (default) of the sixth occurrence of one or more non-blank characters.
SELECT
  REGEXP_INSTR(‘500 Oracle Parkway, Redwood Shores, CA‘,
               ‘[^ ]+‘, 1, 6) "REGEXP_INSTR"
  FROM DUAL;

REGEXP_INSTR
------------
          37

--The following example examines the string, looking for occurrences of words beginning with s, r, or p, regardless of case,
--followed by any six alphabetic characters. Oracle begins searching at the third character in the string and 
--returns the position in the string of the character following the second occurrence of a seven-letter word beginning with s, r, or p, 
--regardless of case.
SELECT
  REGEXP_INSTR(‘500 Oracle Parkway, Redwood Shores, CA‘,
               ‘[s|r|p][[:alpha:]]{6}‘, 3, 2, 1, ‘i‘) "REGEXP_INSTR"
  FROM DUAL;

REGEXP_INSTR
------------
          28

--The following examples use the subexpr argument to search for a particular subexpression in pattern. 
--The first statement returns the position in the source string of the first character in the first subexpression, which is ‘123‘:
SELECT REGEXP_INSTR(‘1234567890‘, ‘(123)(4(56)(78))‘, 1, 1, 0, ‘i‘, 1) 
"REGEXP_INSTR" FROM DUAL;

REGEXP_INSTR
-------------------
1

--The next statement returns the position in the source string of the first character in the second subexpression, which is ‘45678‘:
SELECT REGEXP_INSTR(‘1234567890‘, ‘(123)(4(56)(78))‘, 1, 1, 0, ‘i‘, 2) 
"REGEXP_INSTR" FROM DUAL;

REGEXP_INSTR
-------------------
4

--The next statement returns the position in the source string of the first character in the fourth subexpression, which is ‘78‘:
SELECT REGEXP_INSTR(‘1234567890‘, ‘(123)(4(56)(78))‘, 1, 1, 0, ‘i‘, 4) 
"REGEXP_INSTR" FROM DUAL;

REGEXP_INSTR
-------------------
7