Split string by space and character as delimiter in Oracle with regexp_substr

随声附和 提交于 2019-11-26 17:19:08

问题


I'm trying to split a string with regexp_subtr, but i can't make it work.

So, first, i have this query

select regexp_substr('Helloworld - test!' ,'[[:space:]]-[[:space:]]') from dual

which very nicely extracts my delimiter - blank-blank

But then, when i try to split the string with this option, it just doesn't work.

select regexp_substr('Helloworld - test!' ,'[^[[:space:]]-[[:space:]]]+')from dual

The query returns nothing.

Help will be much appreciated! Thanks


回答1:


SQL Fiddle

Oracle 11g R2 Schema Setup:

CREATE TABLE TEST( str ) AS
          SELECT 'Hello world - test-test! - test' FROM DUAL
UNION ALL SELECT 'Hello world2 - test2 - test-test2' FROM DUAL;

Query 1:

SELECT Str,
       COLUMN_VALUE AS Occurrence,
       REGEXP_SUBSTR( str ,'(.*?)([[:space:]]-[[:space:]]|$)', 1, COLUMN_VALUE, NULL, 1 ) AS split_value
FROM   TEST,
       TABLE(
         CAST(
           MULTISET(
             SELECT LEVEL
             FROM   DUAL
             CONNECT BY LEVEL < REGEXP_COUNT( str ,'(.*?)([[:space:]]-[[:space:]]|$)' )
           )
           AS SYS.ODCINUMBERLIST
         )
       )

Results:

|                               STR | OCCURRENCE |  SPLIT_VALUE |
|-----------------------------------|------------|--------------|
|   Hello world - test-test! - test |          1 |  Hello world |
|   Hello world - test-test! - test |          2 |   test-test! |
|   Hello world - test-test! - test |          3 |         test |
| Hello world2 - test2 - test-test2 |          1 | Hello world2 |
| Hello world2 - test2 - test-test2 |          2 |        test2 |
| Hello world2 - test2 - test-test2 |          3 |   test-test2 |



回答2:


If i understood correctly, this will help you. Currently you are getting output as Helloworld(with space at the end). So i assume u don't want to have space at the end. If so you can simply use the space in the delimiter also like.

select regexp_substr('Helloworld - test!' ,'[^ - ]+',1,1)from dual;

OUTPUT
Helloworld(No space at the end)

As u mentioned in ur comment if u want two columns output with Helloworld and test!. you can do the following.

select regexp_substr('Helloworld - test!' ,'[^ - ]+',1,1),
       regexp_substr('Helloworld - test!' ,'[^ - ]+',1,3) from dual;

OUTPUT
col1         col2
Helloworld   test!



回答3:


Trying to negate the match string '[[:space:]]-[[:space:]]' by putting it in a character class with a circumflex (^) to negate it will not work. Everything between a pair of square brackets is treated as a list of optional single characters except for named named character classes which expand out to a list of optional characters, however, due to the way character classes nest, it's very likely that your outer brackets are being interpreted as follows:

  • [^[[:space:]] A single non space non left square bracket character
  • - followed by a single hyphen
  • [[:space:]] followed by a single space character
  • ]+ followed by 1 or more closing square brackets.

It may be easier to convert your multi-character separator to a single character with regexp_replace, then use regex_substr to find you individual pieces:

select regexp_substr(regexp_replace('Helloworld - test!'
                                   ,'[[:space:]]-[[:space:]]'
                                   ,chr(11))
                    ,'([^'||chr(11)||']*)('||chr(11)||'|$)'
                    ,1 -- Start here
                    ,2 -- return 1st, 2nd, 3rd, etc. match
                    ,null
                    ,1 -- return 1st sub exp
                    )
  from dual;

In this code I first changed - to chr(11). That's the ASCII vertical tab (VT) character which is unlikely to appear in most text strings. Then the match expression of the regexp_substr matches all non VT characters followed by either a VT character or the end of line. Only the non VT characters are returned (the first subexpression).




回答4:


Slight improvement on MT0's answer. Dynamic count using regexp_count and proves it handles nulls where the format of [^delimiter]+ as a pattern does NOT handle NULL list elements. More info on that here: Split comma seperated values to columns

SQL> with tbl(str) as (
  2    select ' - Hello world - test-test! -  - test - ' from dual
  3  )
  4  SELECT LEVEL AS Occurrence,
  5         REGEXP_SUBSTR( str ,'(.*?)([[:space:]]-[[:space:]]|$)', 1, LEVEL, NULL, 1 ) AS split_value
  6  FROM   tbl
  7  CONNECT BY LEVEL <= regexp_count(str, '[[:space:]]-[[:space:]]')+1;

OCCURRENCE SPLIT_VALUE
---------- ----------------------------------------
         1
         2 Hello world
         3 test-test!
         4
         5 test
         6

6 rows selected.

SQL>



回答5:


CREATE OR REPLACE FUNCTION field(i_string            VARCHAR2
                                ,i_delimiter         VARCHAR2
                                ,i_occurance         NUMBER
                                ,i_return_number     NUMBER DEFAULT 0
                                ,i_replace_delimiter VARCHAR2) RETURN VARCHAR2     IS
  -----------------------------------------------------------------------
  -- Function Name.......: FIELD
  -- Author..............: Dan Simson
  -- Date................: 05/06/2016 
  -- Description.........: This function is similar to the one I used from 
  --                       long ago by Prime Computer.  You can easily
  --                       parse a delimited string.
  -- Example.............: 
  --  String.............: This is a cool function
  --  Delimiter..........: ' '
  --  Occurance..........: 2
  --  Return Number......: 3
  --  Replace Delimiter..: '/'
  --  Return Value.......: is/a/cool
  --------------------------------------------------------------------------    ---                                    
  v_return_string  VARCHAR2(32767);
  n_start          NUMBER := i_occurance;
  v_delimiter      VARCHAR2(1);
  n_return_number  NUMBER := i_return_number;
  n_max_delimiters NUMBER := regexp_count(i_string, i_delimiter);
BEGIN
  IF i_return_number > n_max_delimiters THEN
    n_return_number := n_max_delimiters + 1;
  END IF;
  FOR a IN 1 .. n_return_number LOOP
    v_return_string := v_return_string || v_delimiter || regexp_substr    (i_string, '[^' || i_delimiter || ']+', 1, n_start);
    n_start         := n_start + 1;
    v_delimiter     := nvl(i_replace_delimiter, i_delimiter);
  END LOOP;
  RETURN(v_return_string);
END field;


SELECT field('This is a cool function',' ',2,3,'/') FROM dual;

SELECT regexp_substr('This is a cool function', '[^ ]+', 1, 1) Word1
      ,regexp_substr('This is a cool function', '[^ ]+', 1, 2) Word2
      ,regexp_substr('This is a cool function', '[^ ]+', 1, 3) Word3
      ,regexp_substr('This is a cool function', '[^ ]+', 1, 4) Word4
      ,regexp_substr('This is a cool function', '[^ ]+', 1, 5) Word5
  FROM dual;


来源:https://stackoverflow.com/questions/31654411/split-string-by-space-and-character-as-delimiter-in-oracle-with-regexp-substr

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