How to ignore special characters and get only number with the below input as string.
Input: '33-01-616-000'
Output should be 3301616000
Use the REPLACE() function to remove the - characters.
REPLACE(columnname, '-', '')
Or if there can be other non-numeric characters, you can use REGEXP_REPLACE() to remove anything that isn't a number.
REGEXP_REPLACE(columnname, '\D', '')
Standard string functions (like REPLACE, TRANSLATE etc.) are often much faster (one order of magnitude faster) than their regular expression counterparts. Of course, this is only important if you have a lot of data to process, and/or if you don't have that much data but you must process it very frequently.
Here is one way to use TRANSLATE for this problem even if you don't know ahead of time what other characters there may be in the string - besides digits:
TRANSLATE(columnname, '0123456789' || columnname, '0123456789')
This will map 0 to 0, 1 to 1, etc. - and all other characters in the input string columnname to nothing (so they will be simply removed). Note that in the TRANSLATE mapping, only the first occurrence of a character in the second argument matters - any additional mapping (due to the appearance of the same character in the second argument more than once) is ignored.
You can also use REGEXP_REPLACE function. Try code below,
SELECT REGEXP_REPLACE('33-01-61ASDF6-0**(98)00[],./123', '([^[:digit:]])', NULL)
FROM DUAL;
SELECT regexp_replace('33-01-616-000','[^0-9]') digits_only FROM dual;
/
Related
I am using Snowflake SQL. I would like to remove characters from a string after a special character ~. How can I do that?
here is the whole scenario. Let me explain. I do have a string like 'CK#123456~fndkjfgdjkg'. Now, i want only the number after #.And not anything after ~. This is number length varies for that field value. It might be 1 or 5 or 3. And i want to add the condition in where class where this number is equal to check_num from other table after joining. I am trying REGEXP_SUBSTR(A.SRC_TXT, '(?<=CK#)(.+?\b)') = C.CHK_NUM in the where condition. I am getting the error as 'No repititive argument after ?'
You can use a regex for this
-- To remove just the character after a ~
select regexp_replace('fo~o bar','~.', '');
-- returns 'fo bar'
--If you want to keep the ~
select regexp_replace('fo~o bar','~.', '~');
-- returns 'fo~ bar'
--If you want to remove everything after the ~
select regexp_replace('fo~o bar','~.*', '');
--returns 'fo'
If you need to remove other specific character sets after a ~, you can probably do this with a slightly more complicated regex, but I'd need examples of your desired input/output to help with that.
EDIT for updated question
This regex replace should get what you need.
select regexp_replace('CK#123456~fndkjfgdjkg','CK#(\\d*)~.*', '\\1');
-- returns 123456
(\\d*) gets ANY number of digits in a row, and the \\1 causes it to replace the match with what was in the first set of parenthesis, which is your list of digits. the CK# and ~.* are there to make sure the whole string gets matched and replaced.
If the CK# can vary as well, you can use .*? like this.
select regexp_replace('ABCD123HI#123456~fndkjfgdjkg','.*?#(\\d*)~.*', '\\1')
-- returns 123456
I'd probably do something like the following, easy enough but not as cool as RegEx type of functions.
set my_string='fooo~12345';
set search_for_me = '~';
SELECT SUBSTR($my_string, 1, DECODE(position($search_for_me, $my_string), 0, length($my_string), position($search_for_me, $my_string)));
I hope this helps...Rich
It looks like lookahead and lookbehinds are not supported in REGEXP functions, they seem to work in the PATTERN clause of a LIST command. Snowflake documentation makes no mention either way of lookahead or lookbehinds.
In your example:
It seems that the query engine is looking for that repeating argument, where you are attempting a lookbehind
You have not specified what you wanted extracted. You have two capture groups, but in this scenario everything would be returned
Since you are looking to remove everything after ~ you have a delimiter, why not use it in your REGEXP_SUBSTR function?
Try the following:
SELECT $1,REGEXP_SUBSTR($1,'\\w+#(.+?)~',1,1,'is',1)
FROM VALUES
('CK#123456~fndkjfgdjkg')
,('QH#128fklj924~fndkjfgdjkg')
;
This looks for:
One or more word characters
Followed by #
Capturing one or more characters upto and not including ~
Returns the characters within the capture group
You can change the .+? to \\d+? to make sure the pattern is only digits. Backslashes must be escaped with a backslash.
The descriptions for each argument of the function can be found here:
https://docs.snowflake.net/manuals/sql-reference/functions/regexp_substr.html
You could check this!!
select substr('CK#123456~fndkjfgdjkg',4,6) from dual;
OUTPUT
123456
https://docs.snowflake.net/manuals/sql-reference/functions/substr.html
I want to extract alphanumeric text of varied length from a string between the second occurrence of a specific characters.
I have tried various forms of substr and regexp_substr but can't seem to get the syntax right. This is for use in Teradata SQL assistant. In the past I would have to create a temp table and use substr twice before trimming down the string to what I need. I want to do it all in one go.
SELECT regexp_substr('Channel:DF GB, Order Num:12345T6, Order Date:01/01/2019, Charge Codes:TAXES,,GBRAX', 'Num\\:+(\\:+)',1,2, ':') as RESULTING_STRING
My desired result is to return ONLY what is between "Num:" and the next "," in this case "12345T6". The length of the order number can vary so it is not a fixed length. When I run my code the actual output is a '?' returned by Teradata. What am I doing wrong?
This seems to work:
SELECT regexp_substr('Channel:DF GB, Order Num:12345T6, Order Date:01/01/2019, Charge Codes:TAXES,,GBRAX', 'Num:(\w*)', 1, 1, NULL, 1) as RESULTING_STRING from dual
Finds Num: and then captures as many word characters (, is not a word char) as there are available. The last parameter - subexpr - specifies which subexpression (aka capture group) you want, without it the whole thing will be matched (Num:12345T6).
Assuming you use Teradata SQL Assistant to query a Teradata system (but why do you tag Oracle then) the RegEx syntax is slightly different (both use a different RegEx dialects):
Teradata's RegExp_Substr doesn't support the subexpression parameter, you can either switch to the (I really don't know why) undocumented RegExp_Substr_gpl
RegExp_Substr_gpl(x, 'Num:([^,]*)', 1, 1, 'i', 1)
or tell the RegEx to forget the previous match using \K:
RegExp_Substr(x, 'Num:\K[^,]*', 1,1, 'i')
You can give a try to the below pattern search !
SELECT REGEXP_REPLACE ((REGEXP_SUBSTR('Channel:DF GB, Order Num:12345T6, Order Date:01/01/2019, Charge Codes:TAXES,,GBRAX', 'Num:[A-Za-z0-9]*',1,1, 'i')),'Num:','',1,1,'i') AS RESULTING_STRING
Regexp_substr pattern search ['Num:[A-Za-z0-9]*'], will first filter out the alphanumeric characters that follow the pattern 'Num:',astriek, helps to find out zero or more occurrences of the specified pattern.
For eg:, in this 'Num:12345T6' will be filtered out of the string provided, also note the last parameter in the regexp_substr is 'i', which ensures case in-specific search.
Lastly, Regexp_replace will replace the pattern 'Num:' from the output of the regexp_substr with an empty string,resulting in a final string as '12345T6'.
I want to extract a specific part of column values.
The target column and its values look like
TEMP_COL
---------------
DESCOL 10MG
TEGRAL 200MG 50S
COLOSPAS 135MG 30S
The resultant column should look like
RESULT_COL
---------------
10MG
200MG
135MG
This can be done using a regular expression:
SELECT regexp_substr(TEMP_COL, '[0-9]+MG')
FROM the_table;
Note that this is case sensitive and it always returns the first match.
I would probably approach this using REGEXP_SUBSTR() rather than base functions, because the structure of the prescription text varies from record to record.
SELECT TRIM(REGEXP_SUBSTR(TEMP_COL, '(\s)(\S*)', 1, 1))
FROM yourTable
The pattern (\s)(\S*) will match a single space followed by any number of non-space characters. This should match the second term in all cases. We use TRIM() to remove a leading space which is matched and returned.
how do you know what is the part you want to extract? how do you know where it begins and where it ends? using the white-spaces?
if so, you can use substr for cutting the data and instr for finding the white-spaces.
example:
select substr(tempcol, -- string
instr(tempcol, ' ', 1), -- location of first white-space
instr(tempcol, ' ', 1, 2) - instr(tempcol, ' ', 1)) -- length until next space
from dual
another solution is using regexp_substr (but it might be harder on performance if you have a lot of rows):
SELECT REGEXP_SUBSTR (tempcol, '(\S*)(\s*)', 1, 2)
FROM dual;
edit: fixed the regular expression to include expressions that don't have space after the parsed text. sorry about that.. ;)
I have a questions regarding below data.
You clearly can see each EMP_IDENTIFIER has connected with EMP_ID.
So I need to pull only identifier which is 10 characters that will insert another column.
How would I do that?
I did some traditional way, using INSTR, SUBSTR.
I just want to know is there any other way to do it but not using INSTR, SUBSTR.
EMP_ID(VARCHAR2)EMP_IDENTIFIER(VARCHAR2)
62049 62049-2162400111
6394 6394-1368000222
64473 64473-1814702333
61598 61598-0876000444
57452 57452-0336503555
5842 5842-0000070666
75778 75778-0955501777
76021 76021-0546004888
76274 76274-0000454999
73910 73910-0574500122
I am using Oracle 11g.
If you want the second part of the identifier and it is always 10 characters:
select t.*, substr(emp_identifier, -10) as secondpart
from t;
Here is one way:
REGEXP_SUBSTR (EMP_IDENTIFIER, '-(.{10})',1,1,null,1)
That will give the 1st 10 character string that follows a dash ("-") in your string. Thanks to mathguy for the improvement.
Beyond that, you'll have to provide more details on the exact logic for picking out the identifier you want.
Since apparently this is for learning purposes... let's say the assignment was more complicated. Let's say you had a longer input string, and it had several groups separated by -, and the groups could include letters and digits. You know there are at least two groups that are "digits only" and you need to grab the second such "purely numeric" group. Then something like this will work (and there will not be an instr/substr solution):
select regexp_substr(input_str, '(-|^)(\d+)(-|$)', 1, 2, null, 2) from ....
This searches the input string for one or more digits ( \d means any digit, + means one or more occurrences) between a - or the beginning of the string (^ means beginning of the string; (a|b) means match a OR b) and a - or the end of the string ($ means end of the string). It starts searching at the first character (the second argument of the function is 1); it looks for the second occurrence (the argument 2); it doesn't do any special matching such as ignore case (the argument "null" to the function), and when the match is found, return the fragment of the match pattern included in the second set of parentheses (the last argument, 2, to the regexp function). The second fragment is the \d+ - the sequence of digits, without the leading and/or trailing dash -.
This solution will work in your example too, it's just overkill. It will find the right "digits-only" group in something like AS23302-ATX-20032-33900293-CWV20-3499-RA; it will return the second numeric group, 33900293.
I have strings like 'keepme:cutme' or 'string-without-separator' which should become respectively 'keepme' and 'string-without-separator'. Can this be done in PostgreSQL? I tried:
select substring('first:last' from '.+:')
But this leaves the : in and won't work if there is no : in the string.
Use split_part():
SELECT split_part('first:last', ':', 1) AS first_part
Returns the whole string if the delimiter is not there. And it's simple to get the 2nd or 3rd part etc.
Substantially faster than functions using regular expression matching. And since we have a fixed delimiter we don't need the magic of regular expressions.
Related:
Split comma separated column data into additional columns
regexp_replace() may be overload for what you need, but it also gives the additional benefit of regex. For instance, if strings use multiple delimiters.
Example use:
select regexp_replace( 'first:last', E':.*', '');
SQL Select to pick everything after the last occurrence of a character
select right('first:last', charindex(':', reverse('first:last')) - 1)