Retrieve the characters before a matching pattern - sql

Question

135 ;1111776698 ;AB555678765
I have the above string and what I am looking for is to retrieve all the digits before the first occurrence of ;.
But the number of characters before the first occurrence of ; varies i.e. it may be a 4 digit number or 3 digit number.
I have played with regex_instr and instr, but I unable to figure this out.
The query should return all the digits before the first occurrence of ;

Answer 1

This answer assumes that you are using Oracle database. I don't know of way to do this using REGEX_INSTR alone, but we can do with REGEXP_REPLACE using capture groups:
SELECT REGEXP_REPLACE('135 ;1111776698 ;AB555678765', '^\s*(\d{3,4})\s*;.*', '\1')
FROM dual;
Demo
Here is the regex pattern being used:
^\s*(\d{3,4})\s*;.*
This allows, from the start of the string, any amount of leading whitespace, followed by a 3 or 4 digit number, followed again by any amount of whitespace, then a semicolon. The .* at the end of the pattern just consumes whatever remains in your string. Note (\d{3,4}), which captures the 3-4 digit number, which is then available in the replacement as \1.

Answer 2

Using INSTR,SUBTSR and TRIM should work ( based on your comment that there are "just white spaces and digits" )
select TRIM(SUBSTR(s,1, INSTR(s,';')-1)) FROM t;
Demo

Answer 3

The following using regexp_substr() should work:
SELECT s, REGEXP_SUBSTR(s, '^[^;]*')

Answer 4

Make sure you try all possible values in that first position, even those you don't expect and make sure they are handled as you want them to be. Always expect the unexpected! This regex matches the first subgroup of zero or more optional digits (allows a NULL to be returned) when followed by an optional space then a semi-colon, or the end of the line. You may need to tighten (or loosen) up the matching rules for your situation, just make sure to test even for incorrect values, especially if the input comes from user-entered data.
with tbl(id, str) as (
select 1, '135 ;1111776698 ;AB555678765' from dual union all
select 2, ' 135 ;1111776698 ;AB555678765' from dual union all
select 3, '135;1111776698 ;AB555678765' from dual union all
select 4, ';1111776698 ;AB555678765' from dual union all
select 5, ';135 ;1111776698 ;AB555678765' from dual union all
select 6, ';;1111776698 ;AB555678765' from dual union all
select 7, 'xx135 ;1111776698 ;AB555678765' from dual union all
select 8, '135;1111776698 ;AB555678765' from dual union all
select 9, '135xx;1111776698 ;AB555678765' from dual
)
select id, regexp_substr(str, '(\d*?)( ?;|$)', 1, 1, NULL, 1) element_1
from tbl
order by id;
ID ELEMENT_1
---------- ------------------------------
1 135
2 135
3 135
4
5
6
7 135
8 135
9
9 rows selected.

Answer 5

To get the desired result, you should use REGEX_SUBSTR as it will substring your desired data from the string you give. Here is the example of the Query.
Solution to your example data:
SELECT REGEXP_SUBSTR('135 ;1111776698 ;AB555678765','[^;]+',1,1) FROM DUAL;
So what it does, Regex splits the string on the basis of ; separator. You needed the first occurrence so I gave arguments as 1,1.
So if you need the second string 1111776698 as your output you can give an argument as 1,2.
The syntax for Regexp_substr is as following:
REGEXP_SUBSTR( string, pattern [, start_position [, nth_appearance [, match_parameter [, sub_expression ] ] ] ] )
Here is the link for more examples:
https://www.techonthenet.com/oracle/functions/regexp_substr.php
Let me know if this works for you. Best luck.