License check with REGEXP in PL-SQL Oracle Apex - sql

Question

create or replace trigger "KENTEKEN_CHECK"
after insert or update of kenteken
on auto
for each row
declare
kenteken varchar2;
teller number := 0;
tellerletter number := 0;
tellercijfer number := 0;
begin
kenteken := lower(:NEW.kenteken);
loop
if substr(kenteken, teller, 1) = REGEXP ("[eoiau]") then
raise_application_error (-20502, 'Kenteken kan geen klinkers bevatten.');
elsif substr(kenteken, teller, 1) = REGEXP ("[0987654321]") then
tellercijfer := tellercijfer + 1;
elsif substr(kenteken, teller, 1) = REGEXP ("[qwrtypsdfghjklzxcvbnm]") then
tellerletter := tellerletter + 1;
else raise_application_error (-20502, 'Er is een ongeldig kenteken ingevoerd.');
end if;
teller := teller + 1;
exit when teller = 5;
end loop;
end;
I need to check a license plate (has six characters). It needs at least 2 letters and 2 numbers and the e, a, o, u and i are not allowed. How should I use the REGEXP the right way to check this?
Teller stands for counter
Tellernumber stands for a counter for the numbers
Tellerletter stands for a counter for the letters
Yes, I'm a starter with this so don't blame my style of coding... Never worked with REGEXP before so don't know how to use it.
Hope it's clear.

Answer 1

This could be an approach:
with test(s) as (
select 'axx11a' from dual union all
select 'xxxx1x' from dual union all
select '111111' from dual union all
select 'xx1111' from dual union all
select 'x1x1x1' from dual
)
select s
from test
where regexp_count(s, '[aeiou]') = 0
and regexp_count(s, '[qwrtypsdfghjklzxcvbnm]') >= 2
and regexp_count(s, '[0-9]') >= 2
The regexp_count counts the number of occurrences of the regular expression in the string; you may use it both to check that you have at least 2 characters of a given set and to check that you do not have unwanted characters.
You can rewrite this in different ways, I used the regexp_count to check all the cases to make it clearer.

Answer 2

You can do this with standard string functions (instead of regexp), which should result in improved performance - if that is a consideration.
To test whether a character is present, you can use TRANSLATE to remove that character from the string, and then compare the length of the resulting string to the length of the original. You need a small trick - TRANSLATE will remove (delete) the characters from the FROM list that do not have a correspondent in the TO list, but you can't have an empty TO list (if you do, the result will be the NULL string).
So, something like this:
with test( s ) as (
select 'axx11a' from dual union all
select 'xxxx1x' from dual union all
select '111111' from dual union all
select 'xx1111' from dual union all
select 'x1x1x1' from dual
)
-- end of test data; solution (SQL query) begins below this line
select s
from test
where length(translate(s, '~aeiou' , '~')) = length(s)
and length(translate(s, '~0123456789' , '~')) <= length(s) - 2
and length(translate(s, '~qwrtypsdfghjklzxcvbnm', '~')) <= length(s) - 2
;
S
------
xx1111
x1x1x1
In your procedure you can use these tests separately, in your IF... clauses.

Answer 3

It's important to realize that Oracle supports POSIX Extended Regular Expressions, which is more restricted than PCRE Regexes that you find most places.
Because POSIX ERE does not support so-called lookaheads, I would use these regexes:
.*[A-Z].*[A-Z]
To check if there are two letters (anywhere in the license plate).
.*[0-9].*[0-9]
To check if there are two numbers.
^[A-Z0-9]{6}$
To check if there are vowels I would use NOT REGEX_LIKE() with this regex:
[AEIOU]
To check whether the license plate consists of 6 numbers or letters from the start (^) to the end ($) of the string. This will fail if there are any other characters, or a different number of characters other than exactly 6.

Answer 4

([qwrtypsdfghjklzxcvbnm]{2}\d{2}) ==> dd02
(\d{2}[qwrtypsdfghjklzxcvbnm]{2}) ==> 02dd
([qwrtypsdfghjklzxcvbnm]{2}-\d{2}) ==> dd-02
I think that this will do your job.
Succes!!
Groet