Update ID value to format XXXXXXXX-X using oracle SQL - sql

Question

Table name: TEST
Column name: ID [VARCHAR(200)]
The format of ID is ‘XXXXXXXX-X’, where ‘X’ is a number from 0 to 9.
Additional operations in case above format is not satisfied:
if the ID consists of 9 digits and there is a double dash between eighth and ninth digit , the extra dash is removed (e.g. 08452142--6 -> 08452142-6)
if the ID consists of 9 digits and there is/are space(s) between eighth and ninth digit and/or non-digits and/or non-letter symbol(s) then replace them to dash (e.g. 08452142 - . 3 -> 08452142-3)
if the ID consists 9 digits and starts/ends with non-digits and/or non-letter symbol(s) then delete that symbol(s) up to digit (e.g. 08452142-2.. -> 08452142-2)
if the ID contains only 9 digits then put a dash before the last digit (e.g. 123456789 -> 12345678-9)
I have achieved the necessary format by using the below snippet.
UPDATE TEST
SET ID = (SELECT REGEXP_REPLACE(ID,'^\d{8}-\d{1}$','') AS "ID"
from TEST
WHERE PK = 11;
)
What are the possible ways to add transformations as mentioned in points[1-4] above in a single query?
Using REGEXP_REPLACE, I can achieve ID in above format. But in case format is incorrect, and ID needs to be transformed[like removing extra dash, or adding dash in case 9 digits are received] to achieve satisfactory format, how can that be achieved in a single UPDATE query?

Answer 1

In any case, you need to extract 9 digits from your string in the first step. And then
add a hyphen before the last character. For both steps use regexp_replace() function
with test(id) as
(
select '08452142--6' from dual union all
select '08452142 - . 3' from dual union all
select '08452142-2..' from dual union all
select '123456789' from dual union all
select '1234567890' from dual
)
select case when length(regexp_replace(id,'(\D)'))=9 then
regexp_replace(regexp_replace(id,'(\D)'),
'(^[[:digit:]]{8})(.*)([[:digit:]]{1}$)','\1-\3')
end as id
from test;
ID
----------
08452142-6
08452142-3
08452142-2
12345678-9
<null>
Demo

Answer 2

You can use the following I think:
UPDATE TEST
SET ID = REGEXP_REPLACE(ID,'^\D*(\d{8})\D*(\d)\D*$','\1-\2')
WHERE REGEXP_LIKE(ID,'^\D*(\d{8})\D*(\d)\D*$')
This way you ignore all non-digit charcters and search for a 8-digit number and then an 1-digit number. Take these 2 numbers and put a single '-' in between.
This is a little more generous as you might need but should work with all your provided examples.

Answer 3

I think you want the first 8 digits, then a hyphen, then the 9th digit:
select ( substr(regexp_replace(id, '[^0-9]', ''), 1, 8) ||
'-' ||
substr(regexp_replace(id, '[^0-9]', ''), 9, 1)
)

Answer 4

I tried an approach based on the suggestion by #BarbarosÖzhan:
with source as (
select '02426467--6' id from dual union all
select '02426467-6' id from dual union all
select '02597718 -- . 3' id from dual union all
select '02597718 --dF5 . 3' id from dual union all
select '00120792-2..' id from dual union all
select '..00120792-2..' id from dual union all
select '123456789' id from dual union all
select '1234567890' id from dual
)
select
case
when regexp_like(id, '\d{8}-\d{1}')
then id
else
case
when regexp_like(id, '\d{8}-\d{1}')
then id
else
case
when regexp_count(id, '\d') = 9
then
case
when
regexp_like(
regexp_replace(
regexp_replace(
id, '(\d{8}-)(-)(\d{1})', '\1\3'
), '(\d{8})([^A-Za-z1-9])(\d{1})', '\1-\3'
)
, '\d{8}-\d{1}')
then
regexp_replace(
regexp_replace(
id, '(\d{8}-)(-)(\d{1})', '\1\3'
), '(\d{8})([^A-Za-z1-9])(\d{1})', '\1-\3'
)
else id
end
else id
end
end id_tr
from source
However, in cases 3 and 4, I cannot get rid of the space, dot and alphabets. I think something wrong with the logic in case length is more than 9. I end with "id" as it is so the result is the same without any modifications.
Any suggestions to impprove this?