I need to get a substring from a table column that is after a colon and before a whitespace. The length of the substring can vary, but the length of the data before the colon and after the whitespace is constant.
So the data in my table column named "Subject" consists of 5 words, immediately followed by a colon, immediately followed by the substring I need (which can vary in length), followed by a whitespace and a date. The substring I need is a course name. Examples:
Payment Due for Upcoming Course:FIN/370T 11/26/2019
Payment Due for Upcoming Course:BUS/475 11/26/2019
Payment Due for Upcoming Course:ADMIN9/475TG 11/26/2019
I have tried using REGEXP function with REGEXP_SUBSTR(COLUMN_NAME,'[^:]+$') to get everything after the colon, and REGEXP_SUBSTR(COLUMN_NAME, '[^ ]+' , 1 , 5 ) to get data before the last whitespace, but I need to combine them.
I have tried the following:
select
REGEXP_SUBSTR(SUBJECT,'[^:]+$') COURSE_ID
from TABLE
Result:
FIN/370T 11/26/2019
and this:
select
REGEXP_SUBSTR (SUBJECT, '[^ ]+' , 1 , 5 ) COURSE_ID2
from TABLE
Result:
Course:FIN/370T
I need the output to return FIN/370T
In short use:
select regexp_replace(str,'(.*:)(.*)( )(.*)$','\2') as short_course_id
from tab
I prefer regexp_replace, because there are more possibilities to extract part of strings.
If you don't want to mess with regex, you can use a combo of substr and instr.
select
substr(part1,1,instr(part1, ' ',-1,1) ) as course,
part1
from (
select
substr(<your column>,instr(<your column>,':',1,1) +1) as part1
from
<your table>
) t
Fiddle
One option would be
select replace(regexp_substr(str,'[^:]+$'),
regexp_substr(str,'[^:][^ ]+$'),'') as course_id
from tab
Demo
where first regexp_substr() extracts the substring starting from the colon to the end, and the second one from the last space to the end.
Related
I have a string like T_44B56T4 that I'd like to make T_B56T4. I can't use positional logic because the string could instead be TE_2BMT that I'd like to make TE_BMT.
What is the most concise Oracle SQL logic to remove the leftmost grouping on consecutive numbers from the string?
EDIT:
regex_replace is unavailable but I have LTRIM,REPLACE,SUBSTR, etc.
would this fit the bill? I am assuming there are alphanumeric characters, then underscore, and then the numbers you want to remove followed by anything.
select regexp_replace(s, '^([[:alnum:]]+)_\d*(.*)$', '\1_\2')
from (
select 'T_44B56T4' s from dual union all
select 'TXM_1JK7B' from dual
)
It uses regular expressions with matched groups.
Alphanumeric characters before underscore are matched and stored in first group, then underscore followed by 0-many digits (it will match as many digits as possible) followed by anything else that is stored in second group.
If we have a match, the string will be replaced by content of the first group followed by underscore and content of the second group.
if there is no match, the string will not be changed.
It seems that you must use standard string functions, as regular expression functions are not available to you. (Comment under Gordon Linoff's answer; it would help if you would add the same at the bottom of your original question, marked clearly as EDIT).
Also, it seems that the input will always have at least one underscore, and any digits that must be removed will always be immediately after the first underscore.
If so, here is one way you could solve it:
select s, substr(s, 1, instr(s, '_')) ||
ltrim(substr(s, instr(s, '_') + 1), '0123456789') as result
from (
select 'T_44B56T4' s from dual union all
select 'TXM_1JK7B' from dual union all
select '34_AB3_1D' from dual
)
S RESULT
--------- ------------------
T_44B56T4 T_B56T4
TXM_1JK7B TXM_JK7B
34_AB3_1D 34_AB3_1D
I added one more test string, to show that only digits immediately following the first underscore are removed; any other digits are left unchanged.
Note that this solution would very likely be faster than regexp solutions, too (assuming that matters; sometimes it does, but often it doesn't).
If I understand correctly, you can use regexp_replace():
select regexp_replace('T_44B56T4', '_[0-9]+', '_')
Here is a db<>fiddle with your two examples.
Note: Your questions says the left most grouping, but the examples all have the number following an underscore, so the underscore seems to be important.
EDIT:
If you really just want the first string of digits replaced without reference to the underscore:
select regexp_replace(code, '[0-9]+', '', 1, 1)
from (select 'T_44B56T4' as code from dual union all select 'TE_2BMT' from dual ) t
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 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 run the following query:
select * from my_temp_table
And get this output:
PNRP1-109/RT
PNRP1-200-16
PNRP1-209/PG
013555366-IT
How can I alter my query to strip the last two characters from each value?
Use the SUBSTR() function.
SELECT SUBSTR(my_column, 1, LENGTH(my_column) - 2) FROM my_table;
Another way using a regular expression:
select regexp_replace('PNRP1-109/RT', '^(.*).{2}$', '\1') from dual;
This replaces your string with group 1 from the regular expression, where group 1 (inside of the parens) includes the set of characters after the beginning of the line, not including the 2 characters just before the end of the line.
While not as simple for your example, arguably more powerful.
I am trying to fetch an id from an oracle table. It's something like TN0001234567890345. What I want is to sort the values according to the right most 10 positions (e.g. 4567890345). I am using Oracle 11g. Is there any function to cut the rightmost 10 places in Oracle SQL?
You can use SUBSTR function as:
select substr('TN0001234567890345',-10) from dual;
Output:
4567890345
codaddict's solution works if your string is known to be at least as long as the length it is to be trimmed to. However, if you could have shorter strings (e.g. trimming to last 10 characters and one of the strings to trim is 'abc') this returns null which is likely not what you want.
Thus, here's the slightly modified version that will take rightmost 10 characters regardless of length as long as they are present:
select substr(colName, -least(length(colName), 10)) from tableName;
Another way of doing it though more tedious. Use the REVERSE and SUBSTR functions as indicated below:
SELECT REVERSE(SUBSTR(REVERSE('TN0001234567890345'), 1, 10)) FROM DUAL;
The first REVERSE function will return the string 5430987654321000NT.
The SUBSTR function will read our new string 5430987654321000NT from the first character to the tenth character which will return 5430987654.
The last REVERSE function will return our original string minus the first 8 characters i.e. 4567890345
SQL> SELECT SUBSTR('00000000123456789', -10) FROM DUAL;
Result: 0123456789
Yeah this is an old post, but it popped up in the list due to someone editing it for some reason and I was appalled that a regular expression solution was not included! So here's a solution using regex_substr in the order by clause just for an exercise in futility. The regex looks at the last 10 characters in the string:
with tbl(str) as (
select 'TN0001239567890345' from dual union
select 'TN0001234567890345' from dual
)
select str
from tbl
order by to_number(regexp_substr(str, '.{10}$'));
An assumption is made that the ID part of the string is at least 10 digits.