How to replace multiple whole characters, except those in combinations...?
The below code replaces multiple characters, but it also disturbing those in combinations.
SELECT regexp_replace('a,ca,va,ea,r,y,q,b,g','(a|y|q|g)','X') RESULT FROM dual;
Current output:
RESULT
--------------------
X,cX,vX,eX,r,X,X,b,X
Expected output:
RESULT
------------------------
'X,ca,va,ea,r,X,X,b,X
I just want to replace only separate whole characters('a','y','q','g'), but not the 1 in combinations('ca','va','ea')...
Because you are delimiting with a comma ',' you can combine that like ',a,'
and this will replace only single a's.
you can try follows:
with t as
(
select 'a,ca,va,ea,r,y,q,b,g' str
from dual
)
select substr(sys_connect_by_path(regexp_replace(regexp_substr(str, '[^,]+', 1, level), '^(a|y|q|g)$', 'X'), ','), 2) as str
from t
where connect_by_isleaf = 1
connect by level <= length(regexp_replace(str, '[^,]*')) + 1;
Sadly oracle doesn´t support lookahead and lookbehind. But this is a solution i came up with.
SELECT regexp_replace
(regexp_replace
('a,ca,va,ea,r,y,q,b,g',
'^[ayqg](,)|(,)[ayqg](,)|(,)[ayqg]$',
'\2\4X\1\3'),'(,)[ayqg](,)','\1X\2')
RESULT FROM dual;
I had to use the regexp twice sadly, since it doesn´t find two similar values following after each other and replacing it. ..,a,y,.. is getting replaced as ..,X,y,... So the second call replaces the missing [ayqg] with the exact values. In the first inner regexp call replaces the first and last values.
Maybe this could be simplified into one expression, but i am not that conform with the regex from oracle.
As a explanation i am grouping the commata and basicly replace every ,[ayqg], with ,X, by backreferencing the commata
You would look for word boundaries, which is \b, and which is unfortunately not supported by Oracle's regexp_replace.
So let's look for a non-word character \W or the beginning ^ or ending $ of the text.
select
regexp_replace('a,ca,va,ea,r,y,q,b,g','(^|$|\W)(a|y|q|g)(^|$|\W)','\1X\3') as result
from dual;
In order to not remove the non-word characters, we must have them in the replace string: \1 for the expression in the first parenteses, \3 for the ones in the third. Thus we only change the expression in the second parentheses, which is a, y, q or g, with X.
Unfortunately above gives
X,ca,va,ea,r,X,q,b,X
The q was not replaced, because we recognize ',y,' thus being positioned a 'g,' whereas we'd need to be positioned at ',g,' to recognize g as a word, too.
So we need to replace in iterations (i.e. recursively):
with results(txt, num) as
(
select 'a,ca,va,ea,r,y,q,b,g' as txt, 0 as num from dual
union all
select regexp_replace(txt, '(^|$|\W)(a|y|q|g)(^|$|\W)','\1X\3'), num + 1 as num
from results
where txt <> regexp_replace(txt, '(^|$|\W)(a|y|q|g)(^|$|\W)','\1X\3')
)
select max(txt) keep (dense_rank last order by num) as result
from results;
EDIT: Kevin Esche is right; of course one has to do it only twice. Hence you can also do:
select
regexp_replace(txt, search_str, replace_str) as result
from
(
select
regexp_replace(txt, search_str, replace_str) as txt, search_str, replace_str
from
(
select
'a,ca,va,ea,r,y,q,y,q,b,g' as txt,
'(^|$|\W)(a|y|q|g)(^|$|\W)' as search_str,
'\1X\3' as replace_str
from dual
)
);
with replaced_values as (
SELECT case when length(val)=1 then regexp_replace(val,'(a|y|q|g)','X') else val end new_val, lvl
from (
SELECT regexp_substr('a,ca,va,ea,r,y,q,b,g','[^,]+', 1, LEVEL) val, level lvl FROM dual
connect by regexp_substr('a,ca,va,ea,r,y,q,b,g','[^,]+',1, LEVEL) is not null
) all_values
)
select lISTAGG(new_val, ',') WITHIN GROUP (ORDER BY lvl) RESULT
from replaced_values
This statement pivots data into rows and replaces only lines wich contains one character.
Data are then unpivoted in one rows
This sql works also with empty entries like 'a,,,b,c' and more complex regular expressions:
with t as
(select ',a,,ca,va,ea,bbb,ba,r,y,q,b,g,,,' as str,
',' as delimiter,
'(a|y|q|g|ea|[b]*)' as regexp_expr,
'X' as replace_expr
from dual)
(select substr (sys_connect_by_path(regexp_replace(substr(str,
decode(level - 1, 0, 0, instr(str, ',', 1, level - 1)) + 1,
decode(instr(str, ',', 1, level),
0,
length(str),
instr(str, ',', 1, level) - 1) -
decode(level - 1, 0, 0, instr(str, ',', 1, level - 1))),
'^' || regexp_expr || '$',
replace_expr), ','), 2)
from t
where connect_by_isleaf = 1
connect by level <= length(regexp_replace(str, '[^'|| delimiter||']')) + 1)
Result
,X,,ca,va,X,X,ba,r,X,X,X,X,,,
Don't Know much Oracle, but I would have thought something like this could work. Assuming the delimiter is always a comma.
SELECT
regexp_replace(regexp_replace(regexp_replace(regexp_replace(regexp_replace('a,ca,va,ea,r,y,q,b,g','(,a,|,y,|,q,|,g,)',',X,') ,'(,a,|,y,|,q,|,g,)',',X,'), '(^a,|^y,|^q,|^g,)','X,'), '(,a$|,y$|,q$|,g$)',',X'), '(^a$|^y$|^q$|^g$)','X')
RESULT FROM test;
The first two parts replaces a single character in commas in the middle, the third part gets those at the start of the string, the fourth is for the end of the string and the fifth is for when then string has just one character.
This answer might will be simplifiable by advanced Regexp use.
How i can replace words?
RS & OS ===> D, LS & IS ==== >
SECTION_ID Output required
1-LS-1991 1-P-1991
1-IS-1991 1-P-1991
1-RS-1991 1- D- 1991
1-OS-1991 1-D-1991
Related
/abc/required_string/2/ should return abc with regexp_substr
SELECT REGEXP_SUBSTR ('/abc/blah/blah/', '/([a-zA-Z0-9]+)/', 1, 1, NULL, 1) first_val
from dual;
You might try the following:
SELECT TRIM('/' FROM REGEXP_SUBSTR(mycolumn, '^\/([^\/]+)'))
FROM mytable;
This regular expression will match the first occurrence of a pattern starting with / (I habitually escape /s in regular expressions, hence \/ which won't hurt anything) and including any non-/ characters that follow. If there are no such characters then it will return NULL.
Hope this helps.
You can search for /([^/]+)/, which says:
/ forward slash
( start of subexpression (usually called "group" in other languages)
[^/] any character other than forward slash
+ match the preceding expression one or more times
) end of subexpression
/ forward slash
You can use the 6th argument to regexp_substr to select a subexpression.
Here we pass 1 to match only the characters between the /s:
select regexp_substr(txt, '/([^/]+)/', 1, 1, null, 1)
from t1
See it working at SQL Fiddle.
Classic SUBSTR + INSTR offer a simple solution; I know you specified regular expressions, but - consider this too, might work better for a large data volume.
SQL> with test (col) as
2 (select '/abc/required_string/2/' from dual)
3 select substr(col, 2, instr(col, '/', 1, 2) - 2) result
4 from test;
RES
---
abc
SQL>
Here's another way to get the 2nd occurrence of a string of characters followed by a forward slash. It handles the problem if that element happens to be NULL as well. Always expect the unexpected!
Note: If you use the regex form of [^/]+, and that element is NULL it will return "required string" which is NOT what you expect! That form does NOT handle NULL elements. See here for more info: [https://stackoverflow.com/a/31464699/2543416]
with tbl(str) as (
select '/abc/required_string/2/' from dual union all
select '//required_string1/3/' from dual
)
select regexp_substr(str, '(.*?)(/)', 1, 2, null, 1)
from tbl;
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 have a sample String as below,
A|SDFR|RESTA|PRET|PRUQA
B|121|BBCTRI|9ALFA|DEV|5AS
I want to extract the part that is coming after send delimiter,
Expected,
RESTA|PRET|PRUQA
BBCTRI|9ALFA|DEV|5AS
What i got is just extracting single characters regexp_substr
Assuming you mean after the second delimiter, you don't need to use regular expressions for this; you can use the basic ]substr()](http://docs.oracle.com/database/121/SQLRF/functions196.htm) function, getting the starting position with instr():
substr(<your_string>, instr(<your_string>, '|', 1, 2) + 1)
The third argument to instr() says you want the second occurrence; the second argument says you're starting from position 1. That then points to the second delimiter, and you want to start at the next character after that delimiter, so have to add 1.
Demo:
with t (str) as (
select 'A|SDFR|RESTA|PRET|PRUQA' from dual
union all select 'B|121|BBCTRI|9ALFA|DEV|5AS' from dual
)
select substr(str, instr(str, '|', 1, 2) + 1)
from t;
SUBSTR(STR,INSTR(STR,'|',1
--------------------------
RESTA|PRET|PRUQA
BBCTRI|9ALFA|DEV|5AS
try this:
substr(string, instr(string, '|', 1, 2)+1)
Maybe like this;
with a as (select 'B|121|BBCTRI|9ALFA|DEV|5AS' test from dual)
select substr(test,instr(test,'|',3)+1) from a
I have a string and I would like to split that string by delimiter at a certain position.
For example, my String is F/P/O and the result I am looking for is:
Therefore, I would like to separate the string by the furthest delimiter.
Note: some of my strings are F/O also for which my SQL below works fine and returns desired result.
The SQL I wrote is as follows:
SELECT Substr('F/P/O', 1, Instr('F/P/O', '/') - 1) part1,
Substr('F/P/O', Instr('F/P/O', '/') + 1) part2
FROM dual
and the result is:
Why is this happening and how can I fix it?
Therefore, I would like to separate the string by the furthest delimiter.
I know this is an old question, but this is a simple requirement for which SUBSTR and INSTR would suffice. REGEXP are still slower and CPU intensive operations than the old subtsr and instr functions.
SQL> WITH DATA AS
2 ( SELECT 'F/P/O' str FROM dual
3 )
4 SELECT SUBSTR(str, 1, Instr(str, '/', -1, 1) -1) part1,
5 SUBSTR(str, Instr(str, '/', -1, 1) +1) part2
6 FROM DATA
7 /
PART1 PART2
----- -----
F/P O
As you said you want the furthest delimiter, it would mean the first delimiter from the reverse.
You approach was fine, but you were missing the start_position in INSTR. If the start_position is negative, the INSTR function counts back start_position number of characters from the end of string and then searches towards the beginning of string.
You want to use regexp_substr() for this. This should work for your example:
select regexp_substr(val, '[^/]+/[^/]+', 1, 1) as part1,
regexp_substr(val, '[^/]+$', 1, 1) as part2
from (select 'F/P/O' as val from dual) t
Here, by the way, is the SQL Fiddle.
Oops. I missed the part of the question where it says the last delimiter. For that, we can use regex_replace() for the first part:
select regexp_replace(val, '/[^/]+$', '', 1, 1) as part1,
regexp_substr(val, '[^/]+$', 1, 1) as part2
from (select 'F/P/O' as val from dual) t
And here is this corresponding SQL Fiddle.
I have below string:
ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence
So I want to select Sentence since it is the string after the last period. How can I do this?
Just for completeness' sake, here's a solution using regular expressions (not very complicated IMHO :-) ):
select regexp_substr(
'ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence',
'[^.]+$')
from dual
The regex
uses a negated character class to match anything except for a dot [^.]
adds a quantifier + to match one or more of these
uses an anchor $ to restrict matches to the end of the string
You can probably do this with complicated regular expressions. I like the following method:
select substr(str, - instr(reverse(str), '.') + 1)
Nothing like testing to see that this doesn't work when the string is at the end. Something about - 0 = 0. Here is an improvement:
select (case when str like '%.' then ''
else substr(str, - instr(reverse(str), ';') + 1)
end)
EDIT:
Your example works, both when I run it on my local Oracle and in SQL Fiddle.
I am running this code:
select (case when str like '%.' then ''
else substr(str, - instr(reverse(str), '.') + 1)
end)
from (select 'ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence' as str from dual) t
And yet another way.
Not sure from a performance standpoint which would be best...
The difference here is that we use -1 to count backwards to find the last . when doing the instr.
With CTE as
(Select 'ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence' str, length('ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence') len from dual)
Select substr(str,instr(str,'.',-1)+1,len-instr(str,'.',-1)+1) from cte;
select
substr(
'ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence',
INSTR(
'ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence',
'.',
-1
)+1
)
from dual;
The INSTR function accepts a third parameter, the occurrence. It defaults to 1 (the first occurrence), but also accepts negative numbers (meaning counting from the last occurrence backwards).
select substr(str, instr(str, '.', -1) + 1)
from (
select 'ThisSentence.ShouldBe.SplitAfterLastPeriod.Sentence'
as str
from dual);
Sentence
how many dots in a string?
select length(str) - length(replace(str, '.', '') number_of_dots from ...
get substring after last dot:
select substr(str, instr(str, '.', 1, number_of_dots)+1) from ...