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.. ;)
Related
I have a column as varchar2 datatype, the data in it is in format:
100323.3819823.222
100.323123.443422
1001010100.233888
LOL12333.DDD33.44
I need to remove the whole part after the first occurrence of '.'
In the end it should look like this:
100323
100
1001010100
LOL12333
I cant seem to find the exact substring expression due to the fact that there is not any fix length of the first part.
One way is to use REGEXP_SUBSTR:
SELECT REGEXP_SUBSTR(column_name,'^[^.]*') FROM table
The other way is to combine SUBSTR with INSTR, which is a bit faster, but will result in NULL if the data doesn't contain a dot, so you'll have to add a switch if needed:
SELECT SUBSTR(column_name, 1, INSTR(column_name,'.') - 1) FROM table
For oracle you can try this:
select substr (i,1,Instr(i,'.',i)-1) from Table name.
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'.
Input string: ["1189-13627273","89-13706681","118-13708388"]
Expected Output: ["14013627273","14013706681","14013708388"]
What I am trying to achieve is to replace any numbers till the '-' for each item with hard coded text like '140'
SELECT replace(value_to_replace, '-', '140')
FROM (
VALUES ('1189-13627273-77'), ('89-13706681'), ('118-13708388')
) t(value_to_replace);
check this
I found the right way to achieve that using the below regular expression.
SELECT REGEXP_REPLACE (string_to_change, '\\"[0-9]+\\-', '140')
You don't need a regexp for this, it's as easy as concatenation of 140 and the substring from - (or the second part when you split by -)
select '140'||substring('89-13706681' from position('-' in '89-13706681')+1 for 1000)
select '140'||split_part('89-13706681','-',2)
also, it's important to consider if you might have instances that don't contain - and what would be the output in this case
Use regexp_replace(text,text,text) function to do so giving the pattern to match and replacement string.
First argument is the value to be replaced, second is the POSIX regular expression and third is a replacement text.
Example
SELECT regexp_replace('1189-13627273', '.*-', '140');
Output: 14013627273
Sample data set query
SELECT regexp_replace(value_to_replace, '.*-', '140')
FROM (
VALUES ('1189-13627273'), ('89-13706681'), ('118-13708388')
) t(value_to_replace);
Caution! Pattern .*- will replace every character until it finds last occurence of - with text 140.
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)
In my table for the rows containing values like
sample>test Y10,
Sample> y21
I want to get a substring like y10,y21 from all rows. May I pls know how to get it. I tried with regexp_substr,Instr but not able to find the solution.
I am supposing that your string from column is devided by a single space .
It will give you last occurances which will be splited by ' ' a space
substr(your_string, 1, instr(yourString,' ') - 1)
OR you can achive this using regexp_substr
regexp_substr(your_String, '[^[:space:]]+', 1, -1 )
Assuming that yxx is always preceded by a space, it should be as easy as doing this:
TRIM(REGEXP_SUBSTR(mycolumn, ' y\d+', 1, 1, 'i'))
The above regular expression will grab y (note that it is case-insensitive, so it will grab Y as well) followed by an indefinite number (one or more) of digits. If you want to grab just two digits, replace \d+ with \d{2}.
Also, please note that it will get the first occurrence only. Getting multiple occurrences is a bit more complicated, but it can still be done.