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'.
Related
I am having following string in my query
.\ABC\ABC\2021\02\24\ABC__123_123_123_ABC123.txt
beginning with a period from which I need to extract the segment between the final \ and the file extension period, meaning following expected result
ABC__123_123_123_ABC123
Am fairly new to using REGEXP and couldn't help myself to an elegant (or workable) solution with what Q&A here or else. In all queries the pattern is the same in quantity and order but for my growth of knowledge I'd prefer to not just count and cut.
You can use REGEXP_REPLACE function such as
REGEXP_REPLACE(col,'(.*\\)(.*)\.(.*)','\2')
in order to extract the piece starting from the last slash upto the dot. Preceding slashes in \\ and \. are used as escape characters to distinguish the special characters and our intended \ and . characters.
Demo
You need just regexp_substr and simple regexp ([^\]+)\.[^.]*$
select
regexp_substr(
'.\ABC\ABC\2021\02\24\ABC__123_123_123_ABC123.txt',
'([^\]+)\.[^.]*$',
1, -- position
1, -- occurence
null, -- match_parameter
1 -- subexpr
) substring
from dual;
([^\]+)\.[^.]*$ means:
([^\]+) - find one or more(+) any characters except slash([] - set, ^ - negative, ie except) and name it as group \1(subexpression #1)
\. - then simple dot (. is a special character which means any character, so we need to "escape" it using \ which is an escape character)
[^.]* - zero or more any characters except .
$ - end of line
So this regexp means: find a substring which consist from: one or more any characters except slash followed by dot followed by zero or more any characters except dot and it should be in the end of string. And subexpr parameter = 1, says oracle to return first subexpression (ie first matched group in (...))
Other parameters you can find in the doc.
Here is my simple full compatible example with Oracle 11g R2, PCRE2 and some other languages.
Oracle 11g R2 using function substr (Reference documentation)
select
regexp_substr(
'.\ABC\ABC\2021\02\24\ABC__123_123_123_ABC123.txt',
'((\w)+(_){2}(((\d){3}(_)){3}){1}((\w)+(\d)+){1}){1}',
1,
1
) substring
from dual;
Pattern: ((\w)+(_){2}(((\d){3}(_)){3}){1}((\w)+(\d)+){1}){1}
Result: ABC__123_123_123_ABC123
Just as simple as it can be, regular expressions always follow a minimal standard, as you can see portability also provided, just for the case someone else is interested in going the simplest way.
Hopefully, this will help you out!
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)
I got this query and want to extract the value between the brackets.
select de_desc, regexp_substr(de_desc, '\[(.+)\]', 1)
from DATABASE
where col_name like '[%]';
It however gives me the value with the brackets such as "[TEST]". I just want "TEST". How do I modify the query to get it?
The third parameter of the REGEXP_SUBSTR function indicates the position in the target string (de_desc in your example) where you want to start searching. Assuming a match is found in the given portion of the string, it doesn't affect what is returned.
In Oracle 11g, there is a sixth parameter to the function, that I think is what you are trying to use, which indicates the capture group that you want returned. An example of proper use would be:
SELECT regexp_substr('abc[def]ghi', '\[(.+)\]', 1,1,NULL,1) from dual;
Where the last parameter 1 indicate the number of the capture group you want returned. Here is a link to the documentation that describes the parameter.
10g does not appear to have this option, but in your case you can achieve the same result with:
select substr( match, 2, length(match)-2 ) from (
SELECT regexp_substr('abc[def]ghi', '\[(.+)\]') match FROM dual
);
since you know that a match will have exactly one excess character at the beginning and end. (Alternatively, you could use RTRIM and LTRIM to remove brackets from both ends of the result.)
You need to do a replace and use a regex pattern that matches the whole string.
select regexp_replace(de_desc, '.*\[(.+)\].*', '\1') from DATABASE;