Oracle - need to extract text between given strings - sql

Question

Example - need to extract everything between "Begin begin" and "End end". I tried this way:
with phrases as (
select 'stackoverflow is awesome. Begin beginHello, World!End end It has everything!' as phrase
from dual
)
select regexp_replace(phrase
, '([[:print:]]+Begin begin)([[:print:]]+)(End end[[:print:]]+)', '\2')
from phrases
;
Result: Hello, World!
However it fails if my text contains new line characters. Any tip how to fix this to allow extracting text containing also new lines?
[edit]How does it fail:
with phrases as (
select 'stackoverflow is awesome. Begin beginHello,
World!End end It has everything!' as phrase
from dual
)
select regexp_replace(phrase
, '([[:print:]]+Begin begin)([[:print:]]+)(End end[[:print:]]+)', '\2')
from phrases
;
Result:
stackoverflow is awesome. Begin beginHello, World!End end It has
everything!
Should be:
Hello,
World!
[edit]
Another issue. Let's see to this sample:
WITH phrases AS (
SELECT 'stackoverflow is awesome. Begin beginHello,
World!End end It has everything!End endTESTESTESTES' AS phrase
FROM dual
)
SELECT REGEXP_REPLACE(phrase, '.+Begin begin(.+)End end.+', '\1', 1, 1, 'n')
FROM phrases;
Result:
Hello,
World!End end It has everything!
So it matches last occurence of end string and this is not what I want. Subsgtring should be extreacted to first occurence of my label, so result should be:
Hello,
World!
Everything after first occurence of label string should be ignored. Any ideas?

Answer 1

I'm not that familiar with the POSIX [[:print:]] character class but I got your query functioning using the wildcard .. You need to specify the n match parameter in REGEXP_REPLACE() so that . can match the newline character:
WITH phrases AS (
SELECT 'stackoverflow is awesome. Begin beginHello,
World!End end It has everything!' AS phrase
FROM dual
)
SELECT REGEXP_REPLACE(phrase, '.+Begin begin(.+)End end.+', '\1', 1, 1, 'n')
FROM phrases;
I used the \1 backreference as I didn't see the need to capture the other groups from the regular expression. It might also be a good idea to use the * quantifier (instead of +) in case there is nothing preceding or following the delimiters. If you want to capture all of the groups then you can use the following:
WITH phrases AS (
SELECT 'stackoverflow is awesome. Begin beginHello,
World!End end It has everything!' AS phrase
FROM dual
)
SELECT REGEXP_REPLACE(phrase, '(.+Begin begin)(.+)(End end.+)', '\2', 1, 1, 'n')
FROM phrases;
UPDATE - FYI, I tested with [[:print:]] and it doesn't work. This is not surprising since [[:print:]] is supposed to match printable characters. It doesn't match anything with an ASCII value below 32 (a space). You need to use ..
UPDATE #2 - per update to question - I don't think a regex will work the way you want it to. Adding the lazy quantifier to (.+) has no effect and Oracle regular expressions don't have lookahead. There are a couple of things you might do, one is to use INSTR() and SUBSTR():
WITH phrases AS (
SELECT 'stackoverflow is awesome. Begin beginHello,
World!End end It has everything!End endTESTTESTTEST' AS phrase
FROM dual
)
SELECT SUBSTR(phrase, str_start, str_end - str_start) FROM (
SELECT INSTR(phrase, 'Begin begin') + LENGTH('Begin begin') AS str_start
, INSTR(phrase, 'End end') AS str_end, phrase
FROM phrases
);
Another is to combine INSTR() and SUBSTR() with a regular expression:
WITH phrases AS (
SELECT 'stackoverflow is awesome. Begin beginHello,
World!End end It has everything!End endTESTTESTTEST' AS phrase
FROM dual
)
SELECT REGEXP_REPLACE(SUBSTR(phrase, 1, INSTR(phrase, 'End end') + LENGTH('End end')), '.+Begin begin(.+)End end.+', '\1', 1, 1, 'n')
FROM phrases;

Answer 2

Try this regex:
([[:print:]]+Begin begin)(.+?)(End end[[:print:]]+)
Sample usage:
SELECT regexp_replace(
phrase ,
'([[:print:]]+Begin begin)(.+?)(End end[[:print:]]+)',
'\2',
1, -- Start at the beginning of the phrase
0, -- Replace ALL occurences
'n' -- Let dot meta character matches new line character
)
FROM
(SELECT 'stackoverflow is awesome. Begin beginHello, '
|| chr(10)
|| ' World!End end It has everything!' AS phrase
FROM DUAL
)
The dot meta character (.) matches any character in the database character set and the new line character. However, when regexp_replace is called, the match_parameter must contain n switch for dot matches new lines.

Answer 3

In order to get your second option to work you need to add [[:space:][:print:]]* as follows:
with phrases as (
select 'stackoverflow is awesome. Begin beginHello,
World!End end It has everything!' as phrase
from dual
)
select regexp_replace(phrase
, '([[:print:]]+Begin begin)([[:print:]]+[[:space:][:print:]]*)(End end[[:print:]]+)', '\2')
from phrases
;
But still it will break if you have more \n, for instance it won't work for
with phrases as (
select 'stackoverflow is awesome. Begin beginHello,
World!End end
It has everything!' as phrase
from dual
)
select regexp_replace(phrase
, '([[:print:]]+Begin begin)([[:print:]]+[[:space:][:print:]]*)(End end[[:print:]]+)', '\2')
from phrases
;
Then you need to add
with phrases as (
select 'stackoverflow is awesome. Begin beginHello,
World!End end
It has everything!' as phrase
from dual
)
select regexp_replace(phrase
, '([[:print:]]+Begin begin)([[:print:]]+[[:space:][:print:]]*)(End end[[:print:]]+[[:space:][:print:]]*)', '\2')
from phrases
;
The problem of regex is that you might have to scope the variations and create a rule that match all of them. If something falls out of your scope, you'll have to visit the regex and add the new exception.
You can find extra info here.

Answer 4

Description.........: This is a function similar to the one that was available from PRIME Computers
back in the late 80/90's. This function will parse out a segment of a string
based on a supplied delimiter. The delimiters can be anything.
Usage:
Field(i_string =>'This.is.a.cool.function'
,i_deliiter => '.'
,i_start_pos => 2
,i_occurrence => 2)
Return value = is.a
FUNCTION field(i_string VARCHAR2
,i_delimiter VARCHAR2
,i_occurance NUMBER DEFAULT 1
,i_return_instances NUMBER DEFAULT 1) RETURN VARCHAR2 IS
--
v_delimiter VARCHAR2(1);
n_end_pos NUMBER;
n_start_pos NUMBER := 1;
n_delimiter_pos NUMBER;
n_seek_pos NUMBER := 1;
n_tbl_index PLS_INTEGER := 0;
n_return_counter NUMBER := 0;
v_return_string VARCHAR2(32767);
TYPE tbl_type IS TABLE OF VARCHAR2(4000) INDEX BY PLS_INTEGER;
tbl tbl_type;
e_no_delimiters EXCEPTION;
v_string VARCHAR2(32767) := i_string || i_delimiter;
BEGIN
BEGIN
LOOP
----------------------------------------
-- Search for the delimiter in the
-- string
----------------------------------------
n_delimiter_pos := instr(v_string, i_delimiter, n_seek_pos);
--
IF n_delimiter_pos = length(v_string) AND n_tbl_index = 0 THEN
------------------------------------------
-- The delimiter you are looking for is
-- not in this string.
------------------------------------------
RAISE e_no_delimiters;
END IF;
--
EXIT WHEN n_delimiter_pos = 0;
n_start_pos := n_seek_pos;
n_end_pos := n_delimiter_pos - n_seek_pos;
n_seek_pos := n_delimiter_pos + 1;
--
n_tbl_index := n_tbl_index + 1;
-----------------------------------------------
-- Store the segments of the string in a tbl
-----------------------------------------------
tbl(n_tbl_index) := substr(i_string, n_start_pos, n_end_pos);
END LOOP;
----------------------------------------------
-- Prepare the results for return voyage
----------------------------------------------
v_delimiter := NULL;
FOR a IN tbl.first .. tbl.last LOOP
IF a >= i_occurance AND n_return_counter < i_return_instances THEN
v_return_string := v_return_string || v_delimiter || tbl(a);
v_delimiter := i_delimiter;
n_return_counter := n_return_counter + 1;
END IF;
END LOOP;
--
EXCEPTION
WHEN e_no_delimiters THEN
v_return_string := i_string;
END;
RETURN TRIM(v_return_string);
END;