I have a string value in ColumnA separated by '_'.
ColumnA
ColumnB
1_2_AB34-E1
8
2_3_CD56-F1
9
I need to modify ColumnA by replacing the value after 2nd '_' with the value in ColumnB
ColumnA
ColumnB
1_2_8
8
2_3_9
9
I tried using REGEXP_REPLACE(ColumnA, '[^|]+', ColumnB, 1, 3). But it is not working as expected. Can someone share their inputs?
You could just take everything up to the second _ and then concatenate the other column:
select regexp_replace(a, '^([^_]+_[^_]+_).*', '\1') || b
If you want to modify the column in place, the logic works in an update:
update t
set a = regexp_replace(a, '^([^_]+_[^_]+_).*', '\1') || b;
Here is a db<>fiddle.
Assuming that every input string has at least two underscores, and that everything after the second underscore must be replaced, you can do this much more efficiently than with regular expressions - use standard string functions instead.
select substr(columnA, 1, instr(columnA, '_', 1, 2)) || columnB
from ...
(or use similar in update). instr returns the position of the second underscore in the input string in columnA, and then substr returns the substring from the first position up to and including that second underscore. Then concatenate columnB to that substring. The code follows the logic exactly, in every detail.
If the input string may sometimes have fewer than two underscores, you need to explain the requirement. The query above, in those cases only, will replace the entire string from columnA with the string from columnB - perhaps not the desired outcome. The query can be modified in those cases, to implement the required handling - while still being much more efficient than a regular expressions solution.
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
I have a SQL query which returns some rows having the below format:
DB_host
DB_host_instance
How can i filter to get rows which only have the format of 'DB_host' (place a condition to return values with only one occurrence of '_')
i tried using [0-9a-zA-Z_0-9a-zA-Z], but seems like its not right. Please suggest.
One option would be using REGEXP_COUNT and at most one underscore is needed then use
WHERE REGEXP_COUNT( col, '_' ) <= 1
or strictly one underscore should exist then use
WHERE REGEXP_COUNT( col, '_' ) = 1
A simple method is a regular expression:
where regexp_like(col, '^[^_]+_[^_]+$')
This matches the full string when there is a string with no underscores followed by an underscore followed by another string with no underscores.
You could also do this with LIKE, but it is more complicated:
where col like '%\_%' and col not like '%\_%\_%'
That is, has one underscore but not two. The \ is needed because _ is a wildcard for LIKE patterns.
You can suppress underscores in the string, and ensure that the length of the result is just one character less than the original:
where len(replace(col, '_', '')) = len(col) - 1
I wonder how this method would compare to a regex or two likes in terms of efficiency on a large dataset. I would not be surprised it it was more efficient.
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 want to get only those rows that contain ONLY certain characters in a column.
Let's say the column name is DATA.
I want to get all rows where in DATA are ONLY (must have all three conditions!):
Numeric characters (1 2 3 4 5 6 7 8 9 0)
Dash (-)
Comma (,)
For instance:
Value "10,20,20-30,30" IS OK
Value "10,20A,20-30,30Z" IS NOT OK
Value "30" IS NOT OK
Value "AAAA" IS NOT OK
Value "30-" IS NOT OK
Value "30," IS NOT OK
Value "-," IS NOT OK
Try patindex:
select * from(
select '10,20,20-30,30' txt union
select '10,20,20-30,40' txt union
select '10,20A,20-30,30Z' txt
)x
where patindex('%[^0-9,-]%', txt)=0
For you table, try like:
select
DATA
from
YourTable
where
patindex('%[^0-9,-]%', DATA)=0
As per your new edited question, the query should be like:
select
DATA
from
YourTable
where
PATINDEX('%[^0-9,-]%', DATA)=0 and
PATINDEX('%[0-9]%', LEFT(DATA, 1))=1 and
PATINDEX('%[0-9]%', RIGHT(DATA, 1))=1 and
PATINDEX('%[,-][-,]%', DATA)=0
Edit: Your question was edited, so this answer is no longer correct. I won't bother updating it since someone else already has updated theirs. This answer does not fulfil the condition that all three character types must be found.
You can use a LIKE expression for this, although it's slightly convoluted:
where data not like '%[^0123456789,!-]%' escape '!'
Explanation:
[^...] matches any character that is not in the ... part. % matches any number (including zero) of any character. So [^0123456789-,] is the set of characters that you want to disallow.
However: - is a special character inside of [], so we must escape it, which we do by using an escape character, and I've chosen !.
So, you match rows that do not contain (not like) any character that is not in your disallowed set.
Use option with PATINDEX and LIKE logic operator
SELECT *
FROM dbo.test70
WHERE PATINDEX('%[A-Z]%', DATA) = 0
AND PATINDEX('%[0-9]%', DATA) > 0
AND DATA LIKE '%-%'
AND DATA LIKE '%,%'
Demo on SQLFiddle
As already mentioned u can use a LIKE expression but it will only work with some minor modifications, otherwise too many rows will be filtered out.
SELECT * FROM X WHERE T NOT LIKE '%[^0-9!-,]%' ESCAPE '!'
see working example here:
http://sqlfiddle.com/#!3/474f5/6
edit:
to meet all 3 conditions:
SELECT *
FROM X
WHERE T LIKE '%[0-9]%'
AND T LIKE '%-%'
AND T LIKE '%,%'
see: http://sqlfiddle.com/#!3/86328/1
Maybe not the most beautiful but a working solution.