I have a string with double quotes inside.
EG:
<cosmtio :ff "intermit"ksks>
I need the data between the ""
I have tried the regexp_substr but still couldn't get the value between double-quotes.
We could try using REGEXP_REPLACE here:
SELECT
string,
REGEXP_REPLACE(string, '.*"([^"]+)".*', '\1') AS quoted_term
FROM yourTable;
Data:
WITH yourTable AS (
SELECT '<cosmtio :ff "intermit"ksks>' AS string FROM dual
)
Demo
Another option, using REGEXP_SUBSTR:
SELECT
string,
TRIM(BOTH '"' FROM REGEXP_SUBSTR(string, '".*"'))
FROM yourTable;
But this approach requires nesting two function calls, which means it might not outperform the REGEXP_REPLACE version.
You need to use REGEXP_SUBSTR:
SELECT REGEXP_SUBSTR('<cosmtio :ff "intermit"ksks>', '"([^"]+)"', 1, 1, NULL, 1) AS Result FROM DUAL
See the online demo.
The regex is simple: "([^"]+)" matches ", then captures any 1+ chars other than " into Group 1 and then matches ". The last argument is 1 telling Oracle REGEXP_SUBSTR to return the Group 1 values. The first (position) and the second (occurrence) arguments are default, 1. NULL means no specific options need to be passed to the regex engine.
You can try the following:
SELECT REGEXP_REPLACE('<cosmtio :ff "intermit"ksks>', '^[^"]*("([^"]*)")?.*', '\2') FROM dual
It is possible with regexp_substr as following:
Select
regexp_substr('<cosmtio :ff "intermit"ksks>', '[^"]+', 1, 2)
from dual;
Cheers!!
Related
How to get rest of string after specific char?
I have a string 'a|b|c|2|:x80|3|rr|' and I would like to get result after 3rd occurance of |. So the result should be like 2|:x80|3|rr|
The query
select REGEXP_SUBSTR('a|b|c|2|:x80|3|rr|','[^|]+$',1,4)
from dual
Returned me NULL
Use SUBSTR / INSTR combination
WITH t ( s ) AS (
SELECT 'a|b|c|2|:x80|3|rr|'
FROM dual
) SELECT substr(s,instr(s,'|',1,3) + 1)
FROM t;
Demo
REGEXP_REPLACE() will do the trick. Skip 3 groups of anything followed by a pipe, then replace with the 2nd group, which is the rest of the line (anchored to the end).
SQL> select regexp_replace('a|b|c|2|:x80|3|rr|', '(.*?\|){3}(.*)$', '\2') trimmed
2 from dual;
TRIMMED
------------
2|:x80|3|rr|
SQL>
I suggest a nice by long way by using regexp_substr, regexp_count and listagg together as :
select listagg(str) within group (order by lvl)
as "Result String"
from
(
with t(str) as
(
select 'a|b|c|2|:x80|3|rr|' from dual
)
select level-1 as lvl,
regexp_substr(str,'(.*?)(\||$)',1,level) as str
from dual
cross join t
connect by level <= regexp_count('a|b|c|2|:x80|3|rr|','\|')
)
where lvl >= 3;
Rextester Demo
If you use oracle 11g and above you can specify a subexpression to return like this:
select REGEXP_SUBSTR('a|b|c|2|:x80|3|rr|','([^|]+\|){3}(.+)$',1,1,null,2) from dual
Erkko,
You need to use the combination of SUBSTR and REGEXP_INSTR OR INSTR.
Your query will look like this. (Without Regex)
SELECT SUBSTR('a|b|c|2|:x80|3|rr|',INSTR('a|b|c|2|:x80|3|rr|','|',1,3)+1) from dual;
Your query will look like this. (With Regex as you want to use)
SELECT SUBSTR('a|b|c|2|:x80|3|rr|',REGEXP_INSTR('a|b|c|2|:x80|3|rr|','\|',1,3)+1) from dual;
Explanation:
First, you will need to find the place of the string you want as you mentioned. So in your case | comes at place 6. So that +1 would be your position to start to substring.
Second, from the original string, substring from that position+1 to unlimited.(Where your string ends)
Example:
https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=6fd782db95f575201eded084493232ee
I have String like a123bcd-e2343fg-hij-dfgh
and I want OUTPUT e2343fg-hij-dfgh using Regular_expression in oracle.
select regexp_substr('abcd-efg-hij','-[^-]+'1) from dual;
You may apply regexp_substr with [^-]+[-^] pattern and then ltrim as :
select ltrim('a123bcd-e2343fg-hij-dfgh',
regexp_substr('a123bcd-e2343fg-hij-dfgh','[^-]+[-^]')) as output_string
from dual;
or better to call with bind variable :
select ltrim('&str', regexp_substr('&str','[^-]+[-^]')) as output_string
from dual;
where &str may be replaced with a123bcd-e2343fg-hij-dfgh after prompted.
Rextester Demo
Why regular expression, when a trivial SUBSTR + INSTR does the job nicely & quickly? True, it will look smarter, but I can't see any other benefit.
SQL> with test (col) as
2 (select 'a123bcd-e2343fg-hij-dfgh' from dual)
3 select substr(col, instr(col, '-') + 1) result
4 from test;
RESULT
----------------
e2343fg-hij-dfgh
SQL>
select substr('abcd-efg-hij',
regexp_instr('abcd-efg-hij','-[^-]+')+1,length('abcd-efg-hij'))
from dual;
try this
For the sake of argument, regexp_replace works too. This regex matches anything up to and including the first dash, and remembers the rest which it returns.
with tbl(str) as (
select 'a123bcd-e2343fg-hij-dfgh' from dual
)
select regexp_replace(str, '^.*?-(.*)', '\1')
from tbl;
Keep in mind if regexp_substr() does not find a match, it returns NULL but if regexp_replace() does not find a match it return the original string.
/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;
Getting Examples from similar Stack Overflow threads,
Remove all characters after a specific character in PL/SQL
and
How to Select a substring in Oracle SQL up to a specific character?
I would want to retrieve only the first characters before the occurrence of a string.
Example:
STRING_EXAMPLE
TREE_OF_APPLES
The Resulting Data set should only show only STRING_EXAM and TREE_OF_AP because PLE is my delimiter
Whenever i use the below REGEXP_SUBSTR, It gets only STRING_ because REGEXP_SUBSTR treats PLE as separate expressions (P, L and E), not as a single expression (PLE).
SELECT REGEXP_SUBSTR('STRING_EXAMPLE','[^PLE]+',1,1) from dual;
How can i do this without using numerous INSTRs and SUBSTRs?
Thank you.
The problem with your query is that if you use [^PLE] it would match any characters other than P or L or E. You are looking for an occurence of PLE consecutively. So, use
select REGEXP_SUBSTR(colname,'(.+)PLE',1,1,null,1)
from tablename
This returns the substring up to the last occurrence of PLE in the string.
If the string contains multiple instances of PLE and only the substring up to the first occurrence needs to be extracted, use
select REGEXP_SUBSTR(colname,'(.+?)PLE',1,1,null,1)
from tablename
Why use regular expressions for this?
select substr(colname, 1, instr(colname, 'PLE')-1) from...
would be more efficient.
with
inputs( colname ) as (
select 'FIRST_EXAMPLE' from dual union all
select 'IMPLEMENTATION' from dual union all
select 'PARIS' from dual union all
select 'PLEONASM' from dual
)
select colname, substr(colname, 1, instr(colname, 'PLE')-1) as result
from inputs
;
COLNAME RESULT
-------------- ----------
FIRST_EXAMPLE FIRST_EXAM
IMPLEMENTATION IM
PARIS
PLEONASM
I've got a string that ends with a certain number of '=' characters at the end. It's basically a base 64 string.
How can I get this count of '=' characters at the end? A built-in SQL function or regex would be preferred.
I know about the instr function, but it doesn't seem like it could be applied here. I'm not sure if a regex would apply here either.
Use length and replace
select length(some_column) - length(replace(some_column, '=', ''))
from your_table
SELECT REGEXP_COUNT('hello world==', '=') cnt FROM dual
/
SELECT count(distinct(Instr('hello world==','=', LEVEL))) cnt
FROM dual
CONNECT BY LEVEL <= Length('hello world==')
ORDER BY 1
/