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.
Related
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 have a column with assignment numbers like - 11827,27266,91717,09818-2,726252-3,8716151-0,827272,18181
Now i am selecting the records like
select assignment_number from table;
But now i want that the column detail is retreived in such a way that numbers are only retrieved without -2 -3 etc like
726252-3---> 726252 8716151-0-->8716151
I know i can use regex for this but i do not know how to use it
This will select everthing before the character -:
^([^-]+)
From 726252-3 will match 726252
You would use regexp() substr:
select regexp_substr(assignmentnumber, '[0-9]+')
This will return the first string of numbers encountered in the string.
I am using this query to replace one character in a cell
select replace(id,',','')id from table
But I want to replace two characters in a cell.
If the cell is having this data (1,3.1), and I want it to look like this (131).
How can I replace two different characters in one cell?
Use TRANSLATE instead of REPLACE(). It replaces each occurrence of a character in the first pattern with its matched character in the second. To remove characters, simply leave cut short the replacement string:
select translate(id, '1,.', '1') id from table
Note that the second string cannot be null. Hence the need to include 1 (or some other character) in both strings.
Find out more.
Obviously the more characters you need to convert/remove the more attractive TRANSLATE() becomes. The main use for REPLACE is changing patterns (such as words) rather than individual characters.
Can use
select replace(translate(id,',.',' '),' ','') from table;
or
select regexp_replace('1,3.1','[,.]','') from dual;
or
select replace(replace(id,',',''),'.','') from table;
Call the replace again.
select replace(replace(id,',',''), '.','') id from table
Do this:
select REPLACE(REPLACE(id,',',''),'.','')
Or use a regular expression:
select regexp_replace(id, '[.,]', '') id from table
Find out more
I know I'm close to figuring this out but need a little help. What I'm trying to do is all grab a column from a particular table, but chop off the first 4 characters. For example if in a column the value is "KPIT08L", the result I was is 08L. Here is what I have so far but not getting the desired results.
SELECT LEFT(FIELD_NAME, 4)
FROM TABLE_NAME
First up, left will give you the leftmost characters. If you want the characters starting at a specific location, you need to look into mid:
select mid (field_name,5) ...
Secondly, if you value performance,portability and scalability at all, this sort of "sub-column" manipulation should generally be avoided. It's usually far easier (and faster) to patch columns together than to split them apart.
In other words, keep the first four characters in their own column and the rest in a separate column, and do your selects on the relevant one. If you're using anything less than a full column, then it's technically not one attribute of the row.
Try with
SELECT MID(FIELD_NAME, 5) FROM TABLE_NAME
Mid is very powerfull, it let you select the starting point and all the remainder, or,
if specified, the length desidered as in
SELECT MID(FIELD_NAME, 5, 2) FROM TABLE_NAME ' gives 08 in your example text
SELECT RIGHT(FIELD_NAME,LEN(FIELD_NAME)-4)
FROM TABLE_NAME;
If it is for a generic string then the above one will work...
Don't have Access at my current location, but please try this.
SELECT RIGHT(FIELD_NAME, LEN(FIELD_NAME)-4)
FROM TABLE_NAME
The LEFT(FIELD_NAME, 4) will return the first 4 caracters of FIELD_NAME.
What you need to do is :
SELECT MID(FIELD_NAME, 5)
FROM TABLE_NAME
If you have a FIELD_NAME of 10 caracters, the function will return the 6 last caracters (chopping the first 4)!