I'm trying to all rows where a column contains any control charters with the exception of the line feed character (hex value of A). I've tried the following, but this only returns results that have a control character and don't have a line feed. I really want the set of characters that are control characters, LESS the line feed character. Is there a 'minus' operation for character sets, where you can exclude particular ones from it?
SELECT *
FROM MyTable
WHERE REGEXP_LIKE(MyColumn, '[:cntrl: &&[^' || UTL_RAW.CAST_TO_VARCHAR2(HEXTORAW('A')) || ']]{1,}');
Any thoughts?
Thanks!
Well here's a first try that will work but I'm sure this can be made more elegant and efficient:
SELECT *
FROM MyTable
WHERE regexp_like(MyColumn, '[[:cntrl:]]')
AND MyColumn NOT like '%' || chr(10) || '%';
Related
I am trying to come up with an equivalent of the below Oracle statement in Snowflake. This would check if the different parts of the string separated by '.' matches the number of characters in the REGEXP_LIKE expression. I have come up with a rudimentary version to perform the check in Snowflake but I am sure there's a better and cleaner way to do it. I am looking to come up with a one-liner regular expression check in Snowflake similar to Oracle. Appreciate your help!
-- Oracle
SELECT -- would return True
CASE
WHEN REGEXP_LIKE('AB.XYX.12.34.5670.89', '^\w{2}\.\w{3}\.\w{2}') THEN 'True'
ELSE NULL
END AS abc
FROM DUAL
-- Snowflake
SELECT -- would return True
REGEXP_LIKE(SPLIT_PART('AB.XYX.12.34.5670.89', '.', 1), '[A-Z0-9]{2}') AND
REGEXP_LIKE(SPLIT_PART('AB.XYX.12.34.5670.89', '.', 2), '[A-Z0-9]{3}') AND
REGEXP_LIKE(SPLIT_PART('AB.XYX.12.34.5670.89', '.', 3), '[A-Z0-9]{2}') AS abc
You need to add a .* at the end as the REGEXP_LIKE adds explicit ^ && $ to string:
The function implicitly anchors a pattern at both ends (i.e. '' automatically becomes '^$', and 'ABC' automatically becomes '^ABC$'). To match any string starting with ABC, the pattern would be 'ABC.*'.
select
column1 as str,
REGEXP_LIKE(str, '\\w{2}\\.\\w{3}\\.\\w{2}.*') as oracle_way
FROM VALUES
('AB.XYX.12.34.5670.89')
;
gives:
STR
ORACLE_WAY
AB.XYX.12.34.5670.89
TRUE
Or in the context of your question:
SELECT IFF(REGEXP_LIKE('AB.XYX.12.34.5670.89', '\\w{2}\\.\\w{3}\\.\\w{2}.*'), 'True', null) AS abc;
Your use of \w seems to suggest you don't need delimited strings to be strictly [A-Z0-9] since word characters allow underscore and period. If all bets were off and the only requirement was to have . at 3rd, 7th and 10th position, you could have used like this way.
select 'AB.XGH.12.34.5670.89' like '__.___.__.%' ;
I would like to get:
82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
from the following expression
LASTNAME_FIRSTNAME_82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
Does someone know how I can get this using regexp_substr ?
EDIT
Basically I have a field which has 7 sets each separated by _ . The string I gave is just one example. I wanted to retrieve everything after the second _ . There is no fixed character length so I can not use a substr function. Hence I was using regexp_substr. I was able to get away by using a simplified version
Select FILE_NAME, ( (REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,3)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,4)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,5)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,6)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+',1,7)) ) as RegExp
from tbl
Here is some more data from the FILE_NAME field
LAST_FIRST_82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
SMITH_JOHN_82961_0130BPQX9QZN9G4P5RDTPA9HR4R_MR_1_1of1
LASTNAME_FIRSTNAME_99999_01V0MU4XUQK0Y24Y9RYTFA7W1CM_MR_3_1of1
To get everything after the second underscore, you do not need regular expressions, but can use something like the following:
select substr(FILE_NAME, instr(FILE_NAME, '_', 1, 2) +1 ) from tbl
The instr returns the position of the second occurrence of '_', starting by the first character; the substr simply gets everything starting from the position given by instr + 1
From your Requirement, you can just go ahead and use the simple SUBSTRfunction. Its faster, and it addresses the simple need to remove the String LASTNAME_FIRSTNAME.
select substr('LASTNAME_FIRSTNAME_82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1', 20) data_string
from dual;
Output:
data_string
-----------------
82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
Unless you have another underlying logic you need to address?
Kindly clarify so i can edit the answer accordingly.
I had a file that I was processing and it was constantly giving me errors. After some checking I realized that it had some special characters that were hidden.
I have just manually found the hidden characters and made a simple replace like
REPLACE( String,'', '')
How could I prevent this from happening in the future?
I have tried to make a table that stores these hidden ascii characters which are in the range from 125-255 but the database does not store them accordingly.
For example chr(168) is not the same as ascii 168.
select chr('168'),
convert(chr('168'),
'US7ASCII',
'WE8ISO8859P1')
from dual;
What else can I try?
Easy to write, but not the fastest to execute - using regexp_replace:
update table_name
set column_name =
regexp_replace(column_name, '[' || chr(125) || '-' || chr(255) || ']', '')
;
More efficient, but a pain in the neck to write and maintain (you can write some code to generate the code though, to save some typing):
...
set column_name =
translate(column_name, '~' || chr(125) || chr(126) || ..... || chr(255), '~')
;
Yes, with translate() you will have to spell them all out, no "range" concept... {:-(
That's to change the data already in the tables. It would be even better to fix the data as it is coming in, if you can - using a similar transformation.
Got stuck
select *
from table a
where ID like 'delr' || trim(substr(busns_asct_id,5)) || '%';
based on the above query, I get records like "delr112591-8218-1" but in that I need to eliminate ID like "delr60201-T,delr99999-A" and "delr1 -5"
How to eliminate this?
This how to deal with LIKE in Oracle:
An underscore _ in the pattern matches exactly one character
A percent sign % in the pattern can match zero or more characters
ESCAPE '\' to escape the _ or % with any character you want, here I used \
To match delr112591-8218-1 you can use that
like 'delr' || trim(substr(busns_asct_id,5)) || '-____-_';
any other string obtained by replacing _ by any other character will match, so "delr60201-T","delr99999-A" and "delr1 -5" will not match, if you want something more accurate, you have to use REGEX.
LIKE condition in Oracle
REGEX in Oracle
select *
from table a
where ID like 'delr' || trim(substr(busns_asct_id,5)) || '-_' or ID like 'delr_-_;
I am trying to select a column from a table that contains newline (NL) characters (and possibly others \n, \r, \t). I would like to use the REGEXP to select the data and replace (only these three) characters with a space, " ".
No need for regex. This can be done easily with the ASCII codes and boring old TRANSLATE()
select translate(your_column, chr(10)||chr(11)||chr(13), ' ')
from your_table;
This replaces newline, tab and carriage return with space.
TRANSLATE() is much more efficient than its regex equivalent. However, if your heart is set on that approach, you should know that we can reference ASCII codes in regex. So this statement is the regex version of the above.
select regexp_replace(your_column, '([\x0A|\x0B|`\x0D])', ' ')
from your_table;
The tweak is to reference the ASCII code in hexadecimal rather than base 10.
select translate(your_column, chr(10)||chr(11)||chr(13), ' ') from your_table;
to clean it is essential to serve non-null value as params ...
(oracle function basically will return null once 1 param is null, there are few excpetions like replace-functions)
select translate(your_column, ' '||chr(10)||chr(11)||chr(13), ' ') from your_table;
this examples uses ' '->' ' translation as dummy-value to prevent Null-Value in parameter 3