I have a field with following values, now i want to extract only those rows with "xyz" in the field value mentioned below, can you please help?
Mydata_xyz_aug21
Mydata2_zzz_aug22
Mydata3_xyz_aug33
One more requirement
I want to extract only "aIBM_MyProjectFile" from following string below, can you please help me with this?
finaldata/mydata/aIBM_MyProjectFile.exe.ld
I've tried this but it didn't work.
select
regexp_substr('FinalProject/MyProject/aIBM_MyProjectFile.exe.ld','([^/]*)[\.]') exp
from dual;
To extract substrings between the first pair of underscores, you need to use
regexp_substr('Mydata_xyz_aug21','_([^_]+)_', 1, 1, NULL, 1)
To get the file name without the extension, you need
regexp_substr('FinalProject/MyProject/aIBM_MyProjectFile.exe.ld','.*/([^.]+)', 1, 1, NULL, 1)
Note that each regex contains a capturing group (a pattern inside (...)) and this value is accessed with the last 1 argument to the regexp_substr function.
The _([^_]+)_ pattern finds the first _, then places 1 or more chars other than _ into Group 1 and then matches another _.
The .*/([^.]+) pattern matches the whole text up to the last /, then captures 1 or more chars other than . into Group 1 using ([^.]+).
For the first requirement, it would suffice to use LIKE, as posted in answer above:
SELECT column
FROM table
WHERE column LIKE '%xyz%';
For your second requirement (extraction) you will have to use REGEXP_SUBSTR function:
SELECT REGEXP_SUBSTR ('FinalProject/MyProject/aIBM_MyProjectFile.exe.ld', '.*/([^.]+)', 1, 1, NULL, 1)
FROM DUAL
I hope it helped!
Another way to do this is to skip regexp completely:
WITH
aset AS
(SELECT 'with_extension.txt' txt FROM DUAL
UNION ALL
SELECT 'without_extension' FROM DUAL)
SELECT CASE
WHEN INSTR (txt, '.', -1) > 0
THEN
SUBSTR (txt, 1, INSTR (txt, '.', -1) - 1)
ELSE
txt
END
txt
FROM aset
The result of this is
with_extension
without_extension
A BIG Caveat where the regexp is better:
My method doesn't handle this case correctly:
\this\is.a\test
So after I have gone to all this effort, stay with the regexp solutions. I'll leave this here so that others may learn from it.
/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 need to replace the entire word with 0 if the word has any non-digit character. For example, if digital_word='22B4' then replace with 0, else if digital_word='224' then do not replace.
SELECT replace_funtion(digital_word,'has non numeric character pattern',0,digital_word)
FROM dual;
I tried decode, regexp_instr, regexp_replace but could not come up with the right solution.
Please advise.
Thank you.
the idea is simple - you need check if the value is numeric or not
script:
with nums as
(
select '123' as num from dual union all
select '456' as num from dual union all
select '7A9' as num from dual union all
select '098' as num from dual
)
select n.*
,nvl2(LENGTH(TRIM(TRANSLATE(num, ' +-.0123456789', ' '))),'0',num)
from nums n
result
1 123 123
2 456 456
3 7A9 0
4 098 098
see more articles below to see which way is better to you
How can I determine if a string is numeric in SQL?
https://asktom.oracle.com/pls/asktom/f?p=100:11:0::::P11_QUESTION_ID:15321803936685
How to tell if a value is not numeric in Oracle?
You might try the following:
SELECT CASE WHEN REGEXP_LIKE(digital_word, '\D') THEN '0' ELSE digital_word END
FROM dual;
The regular expression class \D matches any non-digit character. You could also use [^0-9] to the same effect:
SELECT CASE WHEN REGEXP_LIKE(digital_word, '\D') THEN '0' ELSE digital_word END
FROM dual;
Alternately you could see if the value of digital_word is made up of nothing but digits:
SELECT CASE WHEN REGEXP_LIKE(digital_word, '^\d+$') THEN digital_word ELSE '0' END
FROM dual;
Hope this helps.
The fastest way is to replace all digits with null (to simply delete them) and see if anything is left. You don't need regular expressions (slow!) for this, you just need the standard string function TRANSLATE().
Unfortunately, Oracle has to work around their own inconsistent treatment of NULL - sometimes as empty string, sometimes not. In the case of the TRANSLATE() function, you can't simply translate every digit to nothing; you must also translate a non-digit character to itself, so that the third argument is not an empty string (which is treated as a real NULL, as in relational theory). See the Oracle documentation for the TRANSLATE() function. https://docs.oracle.com/cd/E11882_01/server.112/e41084/functions216.htm#SQLRF06145
Then, the result can be obtained with a CASE expression (or various forms of NULL handling functions; I prefer CASE, which is SQL Standard):
with
nums ( num ) as (
select '123' from dual union all
select '-56' from dual union all
select '7A9' from dual union all
select '0.9' from dual
)
-- End of simulated inputs (for testing only, not part of the solution).
-- SQL query begins BELOW THIS LINE. Use your own table and column names.
select num,
case when translate(num, 'z0123456789', 'z') is null
then num
else '0'
end as result
from nums
;
NUM RESULT
--- ------
123 123
-56 0
7A9 0
0.9 0
Note: everything here is in varchar2 data type (or some other kind of string data type). If the results should be converted to number, wrap the entire case expression within TO_NUMBER(). Note also that the strings '-56' and '0.9' are not all-digits (they contain non-digits), so the result is '0' for both. If this is not what you needed, you must correct the problem statement in the original post.
Something like the following update query will help you:
update [table] set [col] = '0'
where REGEXP_LIKE([col], '.*\D.*', 'i')
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
If the key word is "Find", is it possible to extract a string that is between the "Find"?
stackoverflow is awesome. FindHello, World!Find It has everything!
The result should be 'Hello, World!' because the string is between "Find"
My initial idea was to use Instr to locate two "Find", then locate what's between "Find".
Is there any better way to do this?
You can use either regular expressions or instr() to achieve what you're after.
I actually prefer regular expressions, if you're using version 10g or later, because I find doing multiple contortions with instr() fairly unwieldy, but it's up to you.
with phrases as (
select 'stackoverflow is awesome. FindHello, World!Find It has everything!' as phrase
from dual
)
select substr( phrase
, instr(phrase,'Find',1,1) + 4
, instr(phrase,'Find',1,2)
- instr(phrase,'Find',1,1)
- 4
)
from phrases
This gets the first and second occurrences of the string Find, starting from the first character, then uses these to work out the positions that you should be doing the sub-string on.
Alternatively, using regular expressions:
with phrases as (
select 'stackoverflow is awesome. FindHello, World!Find It has everything!' as phrase
from dual
)
select regexp_replace(phrase
, '([[:print:]]+Find)([[:print:]]+)(Find[[:print:]]+)', '\2')
from phrases
;
This takes any printable character multiple times, followed by the string Find etc. But, the main bit is the grouping (), which separates each part of the phrase. The \2 means that of the original matched string only the second group, i.e. that between the Find's is returned.
Here's a little SQL Fiddle to demonstrate.
This query suppose to handle more than two 'Find's
with SourceString as(
select 'Find123Find45345Find76876234Find87687Find' s_string
, 'Find' delimiter
from dual
)
select substr(s_string, f_f - s_f + length(delimiter), s_f-Length(delimiter ) )
from (select f_f
, s_f
from(select f_f
, f_f - lag(f_f, 1, f_f) over(order by 1) s_f
from (select Instr(s_string, delimiter , 1, level) f_f
from SourceString
connect by level <= Length(s_string))
)
where s_f > 0)
, SourceString