Regular expression for gettin data after - in sql - sql

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.

Related

Substring of a specific occurence

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.

replace all occurrences of a sub string between 2 charcters using sql

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.

Regular expression to return number after matched string in oracle

I have a query:
select ITEM_ID from system_items where id=4020;
I want a regular expression that takes the above query as input and matches for pattern "id=" and returns 4020.
Please let me know if you have any suggestions, as I have been trying with REGEXP_SUBSTR in Oracle and couldn't get it.
REGEX_SUBSTR won't allow a look-behind like (?<=id=\s*)\d+ so I suspect you need to do this in two operations. First get id=4020, then strip the id=.
One possible way of doing that would be:
REGEXP_SUBSTR(REGEXP_SUBSTR(a, 'id=\s*\d+'), '\d+')
SQLFiddle
This should do it
/id=(\d+)/
id is literal match
() are used for making the capture groups
\d is more numbers
+ ensures 1 or more
demo here http://rubular.com/r/GBxfhID5hS

How to get part of the string that matched with regular expression in Oracle SQL

Lets say I have following string: 'product=1627;color=45;size=7' in some field of the table.
I want to query for the color and get 45.
With this query:
SELECT REGEXP_SUBSTR('product=1627;color=45;size=7', 'color\=([^;]+);?') "colorID"
FROM DUAL;
I get :
colorID
---------
color=45;
1 row selected
.
Is it possible to get part of the matched string - 45 for this example?
One way to do it is with REGEXP_REPLACE. You need to define the whole string as a regex pattern and then use just the element you want as the replace string. In this example the ColorID is the third pattern in the entire string
SELECT REGEXP_REPLACE('product=1627;color=45;size=7'
, '(.*)(color\=)([^;]+);?(.*)'
, '\3') "colorID"
FROM DUAL;
It is possible there may be less clunky regex solutions, but this one definitely works. Here's a SQL Fiddle.
Try something like this:
SELECT REGEXP_SUBSTR(REGEXP_SUBSTR('product=1627;color=45;size=7', 'color\=([^;]+);?'), '[[:digit:]]+') "colorID"
FROM DUAL;
From Oracle 11g onwards we can specify capture groups in REGEXP_SUBSTR.
SELECT REGEXP_SUBSTR('product=1627;color=45;size=7', 'color=(\d+);', 1, 1, 'i', 1) "colorID"
FROM DUAL;

How to extract group from regular expression in Oracle?

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;