I have to remove all whitespaces in String '5 000 000,5' to '5000000,5'.
I tried 3 below but it did not work
select replace('5 000 000,5',' ','') from dual;
select regexp_replace('5 000 000,5', '[[:space:]]*','') from dual;
select regexp_replace('5 000 000,5', ' ','') from dual;
Or anyone know how to convert this String '5 000 000,5' to number because TO_NUMBER failed.
Thanks
Using REGEXP_REPLACE and SPACE class.
Select regexp_replace('your_value', '[[:space:]]+', '') from dual:
Using REPLACE
Select REPLACE('your_value', chr(32), '') from dual:
I think the proplem is your NLS_NUMERIC_CHARACTERS, that should work
select to_number('5 000 000,5', '9G999G999D0', 'NLS_NUMERIC_CHARACTERS = '', ''')
from dual
You can try this and remove any non-numeric chars like comma (,)
SELECT to_number(regexp_replace('5 000 000,5', '[^0-9]', '')) FROM dual;
Related
I have following Oracle SQL code:
SELECT TO_NUMBER(TRIM(REGEXP_REPLACE(per_growth, '(%)(\s)')),
'FM99999999999999999990D099999999999999999',
'NLS_NUMERIC_CHARACTERS = '', ''') AS per_growth
FROM sometable;
This code supposed to look for percentage sign first then space and exclude them from the result. However, it is showing
ORA-01722: invalid number
error. I am learning sql yet and do not know exact cause. Is it something went wrong with (%)(\s)? The value in the table is 50%
You can use TRANSLATE to get rid of all instances of unwanted characters:
SELECT TO_NUMBER(
TRANSLATE(per_growth, '0% ', '0'),
'FM99999999999999999990D099999999999999999',
'NLS_NUMERIC_CHARACTERS = '', '''
) as per_growth
FROM sometable;
Note: TRANSLATE(expr, from_string, to_string) works by swapping all instances of the characters in from_string with the corresponding characters in to_string and if there are more characters in from_string than to_string then the remaining characters are removed. It is faster than using regular expressions and on a par with using REPLACE but it can handle multiple replacements at once, which REPLACE cannot.
If you did want to use the slower REGEXP_REPLACE then you can replace all whitespace characters and all percent characters, whether together or not, using:
SELECT TO_NUMBER(
REGEXP_REPLACE(per_growth, '[%[:space:]]'),
'FM99999999999999999990D099999999999999999',
'NLS_NUMERIC_CHARACTERS = '', '''
) as per_growth
FROM sometable;
Which, for the sample data:
CREATE TABLE sometable (per_growth) AS
SELECT '1%' FROM DUAL UNION ALL
SELECT '%2' FROM DUAL UNION ALL
SELECT '3 %' FROM DUAL UNION ALL
SELECT '4% ' FROM DUAL UNION ALL
SELECT '5,0%' FROM DUAL UNION ALL
SELECT '%%% 123 456 789,0123456 %' FROM DUAL;
Both output:
PER_GROWTH
1
2
3
4
5
123456789.0123456
db<>fiddle here
Did you try good, old REPLACE?
select replace(replace(per_growth, '%', ''), ' ', '') as result
from your_table
You can use
REGEXP_REPLACE(per_growth, '( )(%)')
in order to get rid of % sign and whitespace(s) together
or
TRIM(REPLACE(per_growth,'%'))
to get rid of % sign first, and then leading and trailing spaces next,
before numerical conversion.
I am using PostgreSQL. Data in the table looks like below there is ''. I would like to remove '' with the empty value.
Query:
select 'Patriots Colony Family Monthly'
Actual Result Screenshot:
Expected result:
Abcd
You can use regex_replace to remove non alphanumeric chars
select regexp_replace(yourColumn, '[^[:alnum:]]', ' ', 'g')
from yourTable;
You can also do like this
select regexp_replace(yourColumn, '[^a-zA-Z0-9]+',' ','g')
from yourTable;
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 want to replace all the dots before # in an email with empty string in oracle query
like:
anurag.mart#hotmail.com >> anuragmart#hotmail.com
Instr - To identify the position(#)
Substr - To extract data between start(1) and end(#) position
Replace - To replace . with ''
|| - To concatenate two strings
Try this
SELECT Replace(Substr('anurag.mart#hotmail.com', 1,
Instr('anurag.mart#hotmail.com', '#', 1)), '.', '')
|| Substr('anurag.mart#hotmail.com', Instr('anurag.mart#hotmail.com','#')+1)
FROM dual
Result:
anuragmart#hotmail.com
SqlFiddle Demo
The easiest way is to use REGEXP_REPLACE to identify the pattern and replace it with required pattern.
regexp_replace('anurag.mart#hotmail.com', '(\w+)\.(\w+)(#+)', '\1\2\3')
For example,
SQL> SELECT 'anurag.mart#hotmail.com' email_id,
2 regexp_replace('anurag.mart#hotmail.com', '(\w+)\.(\w+)(#+)', '\1\2\3') new_email_id
3 FROM dual;
EMAIL_ID NEW_EMAIL_ID
----------------------- ----------------------
anurag.mart#hotmail.com anuragmart#hotmail.com
I came on this page while looking for solutions for SQL servers, I converted the above for SQL server for my project, Here is SQL if anybody else needs it.
SELECT
CONCAT(
REPLACE(
SUBSTRING(EmailAddress, 1, CHARINDEX('#', EmailAddress)-1),
'.',
''
),
SUBSTRING(EmailAddress, CHARINDEX('#', EmailAddress), LEN(EmailAddress))
)
FROM [Applicant]
I'm trying to write a SQL query that will remove all spaces so that if a string only has spaces the final string is just ''.
I've tried this code but apparently it isn't working for more than one space:
regexp_replace(:P14_search_text, '( ){1,}', '')
Being :P14_search_text the string I want to modify.
Any help?
how about:
regexp_replace(:P14_search_text, '[[:space:]]*', '');
try this:
Select Replace(:P14_search_text, ' ', '');
Hope this helps you,
SELECT REGEXP_REPLACE(' Any String ','( ){1,}','') "REGEXP_REPLACE" FROM DUAL;
SELECT REGEXP_REPLACE(' ','( ){1,}','') "REGEXP_REPLACE" FROM DUAL;
I tried the same method as #Don suggested and it works in oracle 10 xe.
select replace(' lkjds d s adkj ', ' ', '') from dual
result
lkjdsdsadkj
the following query works for me in oracle:
select
:tst_val AS INPUT,
regexp_replace(:tst_val, '[[:space:]]*', '') AS MODIFIED
from
dual
if this query does not work for you, would you show us what results you're getting?