My string is - ABC Corp., NY., ("Lender") As Agency
I need to replace comma with ~ and double quotes with ^.
Required Output - ABC Corp.~ NY.~ (^Lender^) As Agency
How do I do it in Oracle 11g SQL using regexp_replace()?? Or is there any other way?
The "other way" is to just do two normal replaces
select
REPLACE(REPLACE('ABC Corp., NY., ("Lender")', ',', '~'), '"', '^')
from dual
Or a TRANSLATE, which is easiest if you only need to switch single characters.
select
TRANSLATE('ABC Corp., NY., ("Lender")', ',"', '~^')
from dual
this will work:
select regexp_replace(regexp_replace(yourcolumn,',','~'),'"','^') from yourtable;
Related
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!!
I have names in my dataset and they include parentheses. But, I am trying to clean up the names to exclude those parentheses.
Example: ABC Company (Somewhere, WY)
What I want to turn it into is: ABC Company
I'm using standard SQL with google big query.
I've done some research and I know big query has left(), but I do not know the equivalent of find(). My plan was to do something that finds the ( and then gives me everything to the left of -1 characters from the (.
My plan was to do something that finds the ( and then gives me everything to the left of -1 characters from the (.
Good plan! In BigQuery Standard SQL - equivalent of LEFT is SUBSTR(value, position[, length]) and equivalent of FIND is STRPOS(value1, value2)
With this in mind your query can look like (which is exactly as you planned)
#standardSQL
WITH names AS (
SELECT 'ABC Company (Somewhere, WY)' AS name
)
SELECT SUBSTR(name, 1, STRPOS(name, '(') - 1) AS clean_name
FROM names
Usually, string functions are less expensive than regular expression functions, so if you have pattern as in your example - you should go with above version
But in more generic cases, when pattern to clean is more dynamic like in Graham's answer - you should go with solution in Graham's answer
Just use REGEXP_REPLACE + TRIM. This will work with all variants (just not nested parentheses):
#standardSQL
WITH
names AS (
SELECT
'ABC Company (Somewhere, WY)' AS name
UNION ALL
SELECT
'(Somewhere, WY) ABC Company' AS name
UNION ALL
SELECT
'ABC (Somewhere, WY) Company' AS name)
SELECT
TRIM(REGEXP_REPLACE(name,r'\(.*?\)',''), ' ') AS cleaned
FROM
names
Use REGEXP_EXTRACT:
SELECT
RTRIM(REGEXP_EXTRACT(names, r'([^(]*)')) AS new_name
FROM yourTable
The regex used here will greedily consume and match everything up until hitting an opening parenthesis. I used RTRIM to remove any unwanted whitespace picked up by the regex.
Note that this approach is robust with respect to the edge case of an address record not having any term with parentheses. In this case, the above query would just return the entire original value.
I can't test this solution at the moment, but you can combine SUBSTR and INSTR. Like this:
SELECT CASE WHEN INSTR(name, '(') > 0 THEN SUBSTR( name, 1, INSTR(name, '(') ) ELSE name END as name FROM table;
My query is not working as expected :
SELECT REGEXP_REPLACE('This is testing data SIL(TM)T for once ',
'SIL(TM)T', 'SIL<REFERENCE ID="8208" TYPE="trademark"/>T')as
Newdescriptiontext from dual
output should be:
This is testing data SIL<REFERENCE ID="8208" TYPE="trademark"/>T for once
which is not the case .Need guidance .
try replace instead of regexp_replace
SELECT REPLACE('This is testing data SIL(TM)T for once ',
'SIL(TM)T', 'SIL<REFERENCE ID="8208" TYPE="trademark"/>T')as
Newdescriptiontext from dual;
You simply have to escape the parentheses:
SELECT REGEXP_REPLACE('This is testing data SIL(TM)T for once ',
'SIL\(TM\)T', 'SIL<REFERENCE ID="8208" TYPE="trademark"/>T')as Newdescriptiontext
from dual
In a regexp, they are used to delimit a "subexpression", so '(TM)' matches 'TM'; if you escape them, they'll be interpreted as plain characters, thus having '\(TM\)' matching '(TM)'
I have sql query and want to replace all characters except hyphen(-) , alphabets and numbers.
How can I do that in sql query?
You can represent non hyphen or alphanumeric characters by the class:
[^\-a-zA-Z0-9]
Then use REGEXP_REPLACE to remove these characters from your column:
SELECT REGEXP_REPLACE (col, '[^\-a-zA-Z0-9]', '')
FROM dual;
This would remove all the alphabets and numbers from the input string and will leave '-'.
SELECT 'Rajkakla-53535-' As Strng, REGEXP_REPLACE(REGEXP_REPLACE ('Rajkakla-53535-', '[A-Za-z]',''), '[0-9]','') As No_Alphnum
FROM dual;
Or you can use:
SELECT REGEXP_REPLACE ('Rajkakla-53535-', '[a-zA-Z0-9]', '')
FROM dual;
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.