How can we convert a string of any length into a comma separated string with comma after every n characters. I am using Oracle 10g and above. I tried with REGEXP_SUBSTR but couldn't get desired result.
e.g.: for below string comma after every 5 characters.
input:
aaaaabbbbbcccccdddddeeeeefffff
output:
aaaaa,bbbbb,ccccc,ddddd,eeeee,fffff,
or
aaaaa,bbbbb,ccccc,ddddd,eeeee,fffff
Thanks in advance.
This can be done with regexp_replace, like so:
WITH sample_data AS (SELECT 'aaaaabbbbbcccccdddddeeeeefffff' str FROM dual UNION ALL
SELECT 'aaaa' str FROM dual UNION ALL
SELECT 'aaaaabb' str FROM dual)
SELECT str,
regexp_replace(str, '(.{5})', '\1,')
FROM sample_data;
STR REGEXP_REPLACE(STR,'(.{5})','\
------------------------------ --------------------------------------------------------------------------------
aaaaabbbbbcccccdddddeeeeefffff aaaaa,bbbbb,ccccc,ddddd,eeeee,fffff,
aaaa aaaa
aaaaabb aaaaa,bb
The regexp_replace simply looks for any 5 characters (.{5}), and then replaces them with the same 5 characters plus a comma. The brackets around the .{5} turn it into a labelled subexpression - \1, since it's the first set of brackets - which we can then use to represent our 5 characters in the replacement section.
You would then need to trim the extra comma off the resultant string, if necessary.
SELECT RTRIM ( REGEXP_REPLACE('aaaaabbbbbcccccdddddeeeeefffff', '(.{5})' ,'\1,') ,',') replaced
FROM DUAL;
This worked for me:
WITH strlen AS
(
SELECT 'aaaaabbbbbcccccdddddeeeeefffffggggg' AS input,
LENGTH('aaaaabbbbbcccccdddddeeeeefffffggggg') AS LEN,
5 AS part
FROM dual
)
,
pattern AS
(
SELECT regexp_substr(strlen.input, '[[:alnum:]]{5}', 1, LEVEL)
||',' AS line
FROM strlen,
dual
CONNECT BY LEVEL <= strlen.len / strlen.part
)
SELECT rtrim(listagg(line, '') WITHIN GROUP (
ORDER BY 1), ',') AS big_bang$
FROM pattern ;
Related
Like if I have a string "123456,852369,7852159,1596357"
The out put looking for "1234,8523,7852,1596"
Requirement is....we want to collect 4 char after every ',' separator
like split, substring and again concat
select
REGEXP_REPLACE('MEDA,MEDA,MEDA,MEDA,MEDA,MEDA,MEDA,MEDA,MDCB,MDCB,MDCB,MDCB,MDCB,MDCB', '([^,]+)(,\1)+', '\1')
from dual;
we want to collect 4 char after every ',' separator
Here is an approach using regexp_replace:
select regexp_replace(
'123456,852369,7852159,1596357',
'([^,]{4})[^,]*(,|$)',
'\1\2'
)
from dual
Regexp breakdown:
([^,]{4}) 4 characters others than "," (capture that group as \1)
[^,]* 0 to n characters other than "," (no capture)
(,|$) either character "," or the end of string (capture this as \2)
The function replaces each match with capture 1 (the 4 characters we want) followed by capture 2 (the separator, if there is one).
Demo:
RESULT
1234,8523,7852,1596
One option might be to split the string, extract 4 characters and aggregate them back:
SQL> with test (col) as
2 (select '123456,852369,7852159,1596357' from dual)
3 select listagg(regexp_substr(col, '[^,]{4}', 1, level), ',')
4 within group (order by level) result
5 from test
6 connect by level <= regexp_count(col, ',') + 1;
RESULT
--------------------------------------------------------------------------------
1234,8523,7852,1596
SQL>
With REGEX_REPLACE:
select regexp_replace(the_string, '(^|,)([^,]{4})[^,]*', '\1\2')
from mytable;
This looks for
the beginning of the string or the comma
then four characters that are not a comma
then any number of trailing characters that are not a comma
And only keeps
the beginning or the comma
the four characters that follow
Demo: https://dbfiddle.uk/efUFvKyO
anyone can help me to build proper syntax for regexp_replace to remove any multiplicated non-digits and non-letters from string ? If digit/letter is multiplicated - it is not changed
eg.
source and expected result:
'ABBC000001223, ABC00000212,,, '
'ABBC000001223, ABC00000212, '
(removed second occurance of space after comma and second and third comma )
Use this REGEXP_REPLACE to match any non alphanumeric character in the first group
([^[:alnum:]])
followed by one or more same charcters (group 1)
([^[:alnum:]])(\1)+
and replace it with the original character (group 1)
I added some other data to demonstrate the result
with dta as (
select 'ABBC000001223, ABC00000212,,, ' txt from dual union all
select ',.,;,;;;;,,,,,,,,,,,,#''++`´' txt from dual union all
select 'ABBC000001223ABC00000212' txt from dual)
select txt,
regexp_replace(txt,'([^[:alnum:]])(\1)+', '\1') result
from dta
TXT
-------------------------------
RESULT
--------------------------------
ABBC000001223, ABC00000212,,,
ABBC000001223, ABC00000212,
,.,;,;;;;,,,,,,,,,,,,#'++`´
,.,;,;,#'+`´
ABBC000001223ABC00000212
ABBC000001223ABC00000212
I'm trying to get first string after a character.
Example is like
ABCDEF||GHJ||WERT
I need only
GHJ
I tried to use REGEXP but i couldnt do it.
Can anyone help me with please?
Thank you
Somewhat simpler:
SQL> select regexp_substr('ABCDEF||GHJ||WERT', '\w+', 1, 2) result from dual;
^
RES |
--- give me the 2nd "word"
GHJ
SQL>
which reads as: give me the 2nd word out of that string. Won't work properly if GHJ consists of several words (but that's not what your example suggests).
Something like I interpret with a separator in place, In this case it is || or | example is with oracle database
-- pattern -- > [^] represents non-matching character and + for says one or more character followed by ||
-- 3rd parameter --> starting position
-- 4th parameter --> nth occurrence
WITH tbl(str) AS
(SELECT 'ABCDEF||GHJ||WERT' str FROM dual)
SELECT regexp_substr(str
,'[^||]+'
,1
,2) output
FROM tbl;
I think the most general solution is:
WITH tbl(str) AS (
SELECT 'ABCDEF||GHJ||WERT' str FROM dual UNION ALL
SELECT 'ABC|DEF||GHJ||WERT' str FROM dual UNION ALL
SELECT 'ABClDEF||GHJ||WERT' str FROM dual
)
SELECT regexp_replace(str, '^.*\|\|(.*)\|\|.*', '\1')
FROM tbl;
Note that this works even if the individual elements contain punctuation or a single vertical bar -- which the other solutions do not. Here is a comparison.
Presumably, the double vertical bar is being used for maximum flexibility.
You should use regexp_substr function
select regexp_substr('ABCDEF||GHJ||WERT ', '\|{2}([^|]+)', 1, 1, 'i', 1) str
from dual;
STR
---
GHJ
The input column has comma separated integer values like
Sample input 1)
1,2,200000,2345323,1200000
Sample Output 1)
1,2,2345323
Sample input 2)
546^515,400000,657180,3
Sample Output 2)
546^515, 657180,3
The output string should filter out all integers which have "5" trailing zeros.
Excising the five-zero numbers is straightforward with regexp_replace(). But it seems we also need to tidy up the commas left behind. This solution has another regexp_replace() to catch the ,, when the excised number is inside the string, and a simple trim() for when the number is the first or last in the string:
with cte as (
select '1,2,200000,2345323,1200000' as str from dual union all
select '546^515,400000,657180,3' as str from dual union all
select '546^515,400000,1200000,657180,3' as str from dual union all
select '54000000,400000,1200000,657180,300000000' as str from dual
)
select trim(both ',' from regexp_replace(regexp_replace(str, '([0-9]+)00000', null),',(,+)',',')) as new_str
from cte
/
I need to insert character string after each character in Oracle SQL.
Example:
ABC will A,B,C
DEFG will be D,E,F,G
This question gives only one character in string
Oracle insert character into a string
Edit: As some fellows have mentioned, Oracle does not admit this regex. So my approach would be to do a regex to match all characters, add them a comma after the character and then removing the last comma.
WITH regex AS (SELECT REGEXP_REPLACE('ABC', '(.)', '\1,') as reg FROM dual) SELECT SUBSTR(reg, 1, length(reg)-1) FROM regex;
Note that with the solution of rtrim there could be errors if the string you want to parse has a final ending comma and you don't want to remove it.
Previous solution: (Not working on Oracle)
Check if this does the trick:
SELECT REGEXP_REPLACE('ABC', '(.)(?!$)', '\1,') FROM dual;
It does a regexp_replace of every character, but the last one for the same character followed by a ,
To see how regexp_replace works I recommend you: https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions130.htm
SELECT rtrim(REGEXP_REPLACE('ABC', '(.)', '\1,'),',') "REGEXP_REPLACE" FROM dual;
You could do it using:
REGEXP_REPLACE
RTRIM
For example,
SQL> WITH sample_data AS(
2 SELECT 'ABC' str FROM dual UNION ALL
3 SELECT 'DEFG' str FROM dual UNION ALL
4 SELECT 'XYZ' str FROM dual
5 )
6 -- end of sample_data mimicking a real table
7 SELECT str,
8 rtrim(regexp_replace(str, '(\w?)', '\1,'),',') new_str
9 FROM sample_data;
STR NEW_STR
---- ----------
ABC A,B,C
DEFG D,E,F,G
XYZ X,Y,Z
Since there is no way to negate the end of string in an Oracle regex (that does not support lookarounds), you may use
SELECT REGEXP_REPLACE(
REGEXP_REPLACE('ABC', '([^,])([^,])','\1,\2'),
'([^,])([^,])',
'\1,\2')
AS Result from dual
See the DB Fiddle. The point here is to use REGEXP_REPLACE with ([^,])([^,]) pattern twice to cater for consecutive matches.
The ([^,])([^,]) pattern matches any non-comma char into Group 1 (\1) and then any non-comma char into Group 2 (\2), and inserts a comma in between them.