I have the following string:
aaa'dd?'d'xxx'
The delimiter is
'
but if it has
?
in front of it, it should not be consider delimiter but just a literal (the ? is a escape character for delimiter).
The result I want to display is:
aaa
dd'd
xxx
In this moment I am using [^']+ which does not take into consideration the escaping character(?).
Can you help me, please?
A simple option is to replace offending string with something else; for example, I used #. For the final result, replace it with a single quote, '.
SQL> with test (col) as
2 (select q'[aaa'dd?'d'xxx']' from dual),
3 inter as
4 (select replace(col, '?''', '#') icol
5 from test
6 )
7 select replace(regexp_substr(icol, '[^'']+', 1, level), '#', '''') result
8 from inter
9 connect by level <= regexp_count(icol, '''');
RESULT
-------------
aaa
dd'd
xxx
If you want to this without replacing the ?' pattern with a fixed dummy character - whether that's a '#' or anything else you are sure will never actually appear - then you can use a regular expression pattern like this:
-- bind variable for sample value
var str varchar2(20);
exec :str := q'[aaa'dd?'d'xxx']';
select regexp_substr(:str, '((.*?[^?])*?)(''|$)', 1, level, null, 1) as result
from dual
connect by level < regexp_count(:str, '((.*?[^?])*?)(''|$)');
RESULT
--------------------
aaa
dd?'d
xxx
and you can then just apply a simple replace afterwards:
select replace(
regexp_substr(:str, '((.*?[^?])*?)(''|$)', 1, level, null, 1),
'?''',
'''') as result
from dual
connect by level < regexp_count(:str, '((.*?[^?])*?)(''|$)');
RESULT
--------------------
aaa
dd'd
xxx
If you have two adjacent unescaped delimiters you get a null element back from that position (this didn't happen with an earlier version of the regex pattern):
exec :str := q'[aaa''dd?'d'xxx']';
-- just to make them more visible...
set null (null)
select replace(
regexp_substr(:str, '((.*?[^?])*?)(''|$)', 1, level, null, 1),
'?''',
'''') as result
from dual
connect by level < regexp_count(:str, '((.*?[^?])*?)(''|$)');
RESULT
--------------------
aaa
(null)
dd'd
xxx
Related
I need to replace all characters with nothing before the . character and also replace all [ and ] with nothing.
Please see examples below:
from
to
[PINWHEEL_ASSET].[MX5530]
MX5530
[PINWHEEL_TRADE].[AR5403]
AR5403
The parts before and after the . dot are variables.
with
sample_data (my_string) as (
select '[PINWHEEL_ASSET].[MX5530]' from dual
)
select rtrim(substr(my_string, instr(my_string, '.') + 2), ']') as second_part
from sample_data
;
SECOND_PART
-----------
MX5530
This assumes that the input string looks exactly like this: [first].[second], where "first" and "second" are (possibly empty) strings that do not contain periods or closing brackets.
Yet another option is to use regular expressions (see line #6).
Sample data:
SQL> with test (col) as
2 (select '[PINWHEEL_ASSET].[MX5530]' from dual union all
3 select '[PINWHEEL_TRADE].[AR5403]' from dual
4 )
Query begins here:
5 select col,
6 regexp_substr(col, '\w+', 1, 2) result
7 from test;
COL RESULT
------------------------- --------------------
[PINWHEEL_ASSET].[MX5530] MX5530
[PINWHEEL_TRADE].[AR5403] AR5403
SQL>
I used the SUBSTR function for the similar purposes, but I encountered the following issue:
I am extracting 6 characters from the right, but the data in column is inconsistent and for some rows it has characters less than 6, i.e. 5 or 4. So for such rows, the function returns blanks. How can I fix this?
Example Scenario 1:
SUBSTR('0000123456',-6,6)
Output: 123456
Scenario 2 (how do I fix this?, I need it to return '23456'):
SUBSTR('23456',-6,6)
Output: ""
You can use a case expression: if the string length is strictly greater than 6 then return just the last 6 characters; otherwise return the string itself. This way you don't need to call substr unless it is really needed.
Alternatively, if speed is not the biggest issue and you are allowed to use regular expressions, you can write this more compactly - select between 0 and 6 characters - as many as possible - at the end of the string.
Finally, if you don't mind using undocumented functions, you can use reverse and standard substr (starting from character 1 and extracting the first 6 characters; that will work as expected even if the string has length less than 6). So: reverse the string, extract first (up to) 6 characters, and then reverse again to restore the order. WARNING: This is shown only for fun; DO NOT USE THIS METHOD!
with
test_data (str) as (
select '0123449389' from dual union all
select '00000000' from dual union all
select null from dual union all
select 'abcd' from dual
)
select str,
case when length(str) > 6 then substr(str, -6) else str end as case_substr,
regexp_substr(str, '.{0,6}$') as regexp_substr,
reverse(substr(reverse(str), 1, 6)) as rev_substr
from test_data
;
STR CASE_SUBSTR REGEXP_SUBSTR REV_SUBSTR
---------- ------------- ------------- --------------
0123449389 449389 449389 449389
00000000 000000 000000 000000
abcd abcd abcd abcd
One method uses coalesce():
select coalesce(substr('23456', -6, 6), '23456')
Another tweaks the length:
select substr('23456', greatest(- length('23456'), -6), 6)
I have a problem with understanding a regex.
I have the following string:
aaa'dd?'d'xxx'
In this string, the
'
is substring delimiter and
?
is a escape character for the
'
.
In Oracle SQL, I have a sentence which splits my string in substrings, based on substring delimiter:
select replace(
regexp_substr(q'[aaa'dd?'d'xxx']', '(.*?[^?])(''|$)', 1, level, null, 1),
'?''',
'''') as result
FROM dual
connect by level <= regexp_count(q'[aaa'dd?'d'xxx']', '(.*?[^?])(''|$)');
In this case, the result is:
aaa
dd'd
xxx
... which is correct.
My problem comes from the fact that I want to change the sub-string delimiter from
'
into
+
.
In this case, the main string becomes
aaa+dd?+d+xxx+
I modified the SQL statement in:
SELECT REPLACE(
regexp_substr(q'[aaa+dd?+d+xxx+]', '(.*?[^?])(+|$)', 1, level, null, 1),
'?''',
'''') as result
FROM dual
connect by level <= regexp_count(q'[aaa+dd?+d+xxx+]', '(.*?[^?])(+|$)');
... and the result is different:
a
a
a
+
d
d
?+
d
+
x
x
x
+
Can you point me what am I doing wrong in my modified script in order to get same result, please?
In regexp + means 1 or more of the preceding pattern. Try escaping the + with \ making your regexp '(.*?[^?])(\+|$)'
I have a string as follows: first, last (123456) the expected result should be 123456. Could someone help me in which direction should I proceed using Oracle?
It will depend on the actual pattern you care about (I assume "first" and "last" aren't literal hard-coded strings), but you will probably want to use regexp_substr.
For example, this matches anything between two brackets (which will work for your example), but you might need more sophisticated criteria if your actual examples have multiple brackets or something.
SELECT regexp_substr(COLUMN_NAME, '\(([^\)]*)\)', 1, 1, 'i', 1)
FROM TABLE_NAME
Your question is ambiguous and needs clarification. Based on your comment it appears you want to select the six digits after the left bracket. You can use the Oracle instr function to find the position of a character in a string, and then feed that into the substr to select your text.
select substr(mycol, instr(mycol, '(') + 1, 6) from mytable
Or if there are a varying number of digits between the brackets:
select substr(mycol, instr(mycol, '(') + 1, instr(mycol, ')') - instr(mycol, '(') - 1) from mytable
Find the last ( and get the sub-string after without the trailing ) and convert that to a number:
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE test ( str ) AS
SELECT 'first, last (123456)' FROM DUAL UNION ALL
SELECT 'john, doe (jr) (987654321)' FROM DUAL;
Query 1:
SELECT TO_NUMBER(
TRIM(
TRAILING ')' FROM
SUBSTR(
str,
INSTR( str, '(', -1 ) + 1
)
)
) AS value
FROM test
Results:
| VALUE |
|-----------|
| 123456 |
| 987654321 |
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
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 |
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!
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)||'|$)'
,1 -- Start here
,2 -- return 1st, 2nd, 3rd, etc. match
,null
,1 -- return 1st sub exp
)
from dual;
In this code I first changed - to chr(11). That's the ASCII vertical tab (VT) character which is unlikely to appear in most text strings. Then the match expression of the regexp_substr matches all non VT characters followed by either a VT character or the end of line. Only the non VT characters are returned (the first subexpression).
Slight improvement on MT0's answer. Dynamic count using regexp_count and proves it handles nulls where the format of [^delimiter]+ as a pattern does NOT handle NULL list elements. More info on that here: Split comma seperated values to columns
SQL> with tbl(str) as (
2 select ' - Hello world - test-test! - - test - ' from dual
3 )
4 SELECT LEVEL AS Occurrence,
5 REGEXP_SUBSTR( str ,'(.*?)([[:space:]]-[[:space:]]|$)', 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>
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;