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oracle regular expression issue


By : ghtse
Date : October 18 2020, 03:08 PM
it helps some times You cannot match Unicode hard space with [:space:], nor [^[:print:]], nor \s patterns.
You may trim the strings with the following solution:
code :
select regexp_replace(' some stuff  ', '^[[:space:] ]+|[[:space:] ]+$', '') as result from dual
select regexp_replace(' some stuff  ', '^([^[:print:]]| )+|([^[:print:]]| )+$', '') as result from dual


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Javascript regular expression issue - run-time error JS5017: Syntax error in regular expression


By : Julio Azevedo
Date : March 29 2020, 07:55 AM
will be helpful for those in need If you really wanted that first backslash in those expressions, they must be escaped:
code :
$htmlPattern =  [
                    /\\<br>(.*?)/ig,
                    /\\<br\/>(.*?)/ig
                ];
$htmlPattern =  [
                    /<br>(.*?)/ig,
                    /<br\/>(.*?)/ig
                ];

Regular Expression with oracle


By : user3590798
Date : March 29 2020, 07:55 AM
wish of those help Im looking to find a pattern for a regular expression that start with 3 letters and have 10 numbers after to use it with REGEXP_LIKE in oracle example for String: ABC1236547890 i need to find the pattern ,
start with 3 letters and have 10 numbers after
code :
SQL> WITH DATA AS(
  2  SELECT 'ABC1234567890' STR FROM DUAL UNION ALL
  3  SELECT 'AB1234567890123' STR FROM DUAL UNION ALL
  4  SELECT 'ABCD123456789' STR FROM DUAL
  5  )
  6  SELECT * FROM DATA
  7  WHERE REGEXP_LIKE(STR, '^[[:alpha:]]{3}[[:digit:]]{10}$');

STR
---------------
ABC1234567890

SQL>

Oracle extract json fields using regular expression with oracle regexp_substr


By : Zachary Huffman
Date : March 29 2020, 07:55 AM
wish helps you Allowing for the different data format. This results in {"name":"osama"} which I hope is what you want:
code :
select regexp_replace(regexp_substr('{"code":"001","message":"success","transactionId":437,"results":{"name":"osama"}}','"results":\s*{"(.*| )*"\}', 1, level), '"results":\s*(\{.*\})', '\1', 1, 1) results
 from dual
connect by regexp_substr('{"code":"001","message":"success","transactionId":437,"results":{"name":"osama"}}', '"results":\{(.*)"\}', 1, level) is not null;

oracle regular expression sub string and regular expression replace


By : 4waqar4
Date : March 29 2020, 07:55 AM
around this issue I have Phone_number column in my table where multiple numbers inserted with character values(eg.(123).254,5674). I need to compare each values without character string and needs to remove duplicate values. Column value= '(245)289.4321,(897)201-7210,(897)2017210,8964253712' I need to remove duplicates and only need distinct values. In advance for your help , This should work:
code :
with tmp_tbl as
(select '(245)289.4321,(897)201-7210,(897)2017210,8964253712' phone_numbers  from dual)
select distinct
  regexp_replace((regexp_substr(tmp_tbl.phone_numbers,'[^,]+', 1, level)),'\D','') phone_number
from 
  tmp_tbl
connect by
  regexp_substr(tmp_tbl.phone_numbers,'[^,]+', 1, level) is not null

Regular expression in Oracle


By : van long
Date : March 29 2020, 07:55 AM
Any of those help I want to fetch a string pattern in column of table. I am using a query like , Are you looking for the following?
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