I have the following example
05.04.2018 at 11:10:37 AEST
My goal is to remove all alpha chars from the string.
The expected result is to remove the ' at' and ' AEST' sub-strings:
Note: There should be only one space between the date and the time. There should be no space at the end of the string
05.04.2018 11:10:37
The 'AEST' sub-string is a timezone and can change.
This is my current SQL query:
select SUBSTR(REGEXP_REPLACE(REGEXP_REPLACE('05.04.2018 at 11:10:37 AEST',' at',''), ' EST| AEDT| AEST', ''),1) from dual;
I'm looking to enhance my query (preferably using regex) so I will not have to specify explicitly all potential values for timezone (as currently being done in the query)
Thanks
You may use \s*[a-zA-Z]+ / \s*[[:alpha:]]+ regex:
select REGEXP_REPLACE('05.04.2018 at 11:10:37 AEST','\s*[a-zA-Z]+','') as Result from dual
The pattern matches
\s* - 0+ whitespace chars
[a-zA-Z]+ - 1+ ASCII letters ([[:alpha:]]+ will match any letters).
See an online Oracle demo. Output:
Something like this?
SQL> with test as (select '05.04.2018 at 11:10:37 AEST' col from dual)
2 select regexp_replace(col, '\s*[[:alpha:]]+') result
3 from test;
RESULT
-------------------
05.04.2018 11:10:37
SQL>
You can use:
select trim(regexp_replace(col, '[a-zA-Z]', ''))
I assume you want to remove the final space as well.
Keep it simple! Why not without regexp? The date part and the time part are always at the same position.
select substr(col,1,10) -- the date part
||' '|| -- the blank
substr(col,15,8) -- the time part
from tab;
e.g.
SQL> select substr(col,1,10)
||' '||
substr(col,15,8) "date+time"
from (
select '05.04.2018 at 11:10:37 AEST' col
from dual) tab;
date+time
-------------------
05.04.2018 11:10:37
Related
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
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.
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.
I have the following problem.
There is a String:
There is something 2015.06.06. in the air 1234567 242424 2015.06.07. 12125235
I need to show only just the last date from this string: 2015.06.07.
I tried with regexp_substr with insrt but it doesn't work.
So this is just test, and if I can solve this after it with this solution I should use it for a CLOB query where there are multiple date, and I need only the last one. I know there is regexp_count, and it is help to solve this, but the database what I use is Oracle 10g so it wont work.
Can somebody help me?
The key to find the solution of this problem is the idea of reversing the words in the string presented in this answer.
Here is the possible solution:
WITH words AS
(
SELECT regexp_substr(str, '[^[:space:]]+', 1, LEVEL) word,
rownum rn
FROM (SELECT 'There is something 2015.06.06. in the air 1234567 242424 2015.06.07. 2015.06.08 2015.06.17. 2015.07.01. 12345678999 12125235' str
FROM dual) tab
CONNECT BY LEVEL <= LENGTH(str) - LENGTH(REPLACE(str, ' ')) + 1
)
, words_reversed AS
(
SELECT *
FROM words
ORDER BY rn DESC
)
SELECT regexp_substr(word, '\d{4}\.\d{2}\.\d{2}', 1, 1)
FROM words_reversed
WHERE regexp_like(word, '\d{4}\.\d{2}\.\d{2}')
AND rownum = 1;
From the documentation on regexp_substr, I see one problem immediately:
The . (period) matches any character. You need to escape those with a backslash: \. in order to match only a period character.
For reference, I am linking this post which appears to be the approach you are taking with substr and instr.
Relevant documentation from Oracle:
INSTR(string , substring [, position [, occurrence]])
When position is negative, then INSTR counts and searches backward from the end of string. The default value of position is 1, which means that the function begins searching at the beginning of string.
The problem here is that your regular expression only returns a single value, as explained here, so you will be giving the instr function the appropriate match in the case of multiple dates.
Now, because of this limitation, I recommend using the approach that was proposed in this question, namely reverse the entire string (and your regular expression, i.e. \d{2}\.\d{2}\.\d{4}) and then the first match will be the 'last match'. Then, perform another string reversal to get the original date format.
Maybe this isn't the best solution, but it should work.
There are three different PL/SQL functions that will get you there.
The INSTR function will identify where the first "period" in the date string appears.
SUBSTR applied to the entire string using the value from (1) as the start point
TO_DATE for a specific date mask: YYYY.MM.DD will convert the result from (2) into a Oracle date time type.
To make this work in procedural code, the standard blocks apply:
DECLARE
v_position pls_integer;
... other variables
BEGIN
sql code and function calls;
END
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE finddate
(column1 varchar2(11), column2 varchar2(39))
;
INSERT ALL
INTO finddate (column1, column2)
VALUES ('row1', '1234567 242424 2015.06.07. 12125235')
INTO finddate (column1, column2)
VALUES ('string2', '1234567 242424 2015.06.07. 12125235')
SELECT * FROM dual
;
Query 1:
select instr(column2,'.',1) from finddate
where column1 = 'string2'
select substr(column2,(20-4),10) from finddate
select to_date('2015.06.07','YYYY.MM.DD') from finddate
Results:
| TO_DATE('2015.06.07','YYYY.MM.DD') |
|------------------------------------|
| June, 07 2015 00:00:00 |
| June, 07 2015 00:00:00 |
Here's a way using regexp_replace() that should work with 10g, assuming the format of the lines will be the same:
with tbl(col_string) as
(
select 'There is something 2015.06.06. in the air 1234567 242424 2015.06.07. 12125235'
from dual
)
select regexp_replace(col_string, '^.*(\d{4}\.\d{2}\.\d{2})\. \d*$', '\1')
from tbl;
The regex can be read as:
^ - Match the start of the line
. - followed by any character
* - followed by 0 or more of the previous character (which is any character)
( - Start a remembered group
\d{4}\.\d{2}\.\d{2} - 4 digits followed by a literal period followed by 2 digits, etc
) - End the first remembered group
\. - followed by a literal period
- followed by a space
\d* - followed by any number of digits
$ - followed by the end of the line
regexp_replace then replaces all that with the first remembered group (\1).
Basically describe the whole line as a regular expression, group around what you want to return. You will most likely need to tweak the regex for the end of the line if it could be other characters than digits but this should give you an idea.
For the sake of argument this works too ONLY IF there are 2 occurrences of the date pattern:
with tbl(col_string) as
(
select 'There is something 2015.06.06. in the air 1234567 242424 2015.06.07. 12125235' from dual
)
select regexp_substr(col_string, '\d{4}\.\d{2}\.\d{2}', 1, 2)
from tbl;
returns the second occurrence of the pattern. I expect the above regexp_replace more accurately describes the solution.