oracle regexp_like , specific number format without negative lookahead - regex

Question

I'm using regexp_like function in Oracle in order to match the following number format : xxxyxxx
I'm trying this :
select 1 "val"
from dual
where regexp_like('5553555','^(\d){3}(?!\1)\d\1{3}$')
but as I realized, negative lookahead is not supported in Oracle.
how to do it without negative lookahead?

Answer 1

Indeed, no look around is possible. Please note that you also have another issue: (\d){3} will match also 3 different digits. You would need (\d)\1\1 to match only three of the same digits.
For your particular case you could still use a regular expression. What I could think of is using a particular property: numbers with all the same 7 digits (xxxxxxx) will be dividable by 1111111.
With regexp_like and an additional modulo test:
with tbl(val) as (
select '5555555' from dual union
select '5553555' from dual union
select 'nothing' from dual
)
select val
from tbl
where regexp_like(val,'^(\d)\1\1\d\1{3}$') and mod(val, 1111111) > 0;
Or you could use two regexes:
with tbl(val) as (
select '5555555' from dual union
select '5553555' from dual union
select 'nothing' from dual
)
select val
from tbl
where regexp_like(val,'^(\d)\1\1\d\1{3}$') and not regexp_like(val,'^(\d)..\1');
Admittedly, neither is really elegant, and also not the most efficient. For more efficiency you should not use regular expressions.

Answer 2

Maybe oldfashioned SUBSTR might help. Something like this: split input string (COL) into two equal pieces, and compare whether they match. LEN is used to distinguish odd from even lengths and what to do with the second part of the string (i.e. which is its starting point).
A few examples:
SQL> WITH test (col) AS (SELECT '5554555' FROM DUAL),
2 len AS (SELECT LENGTH (col) len FROM test)
3 SELECT CASE
4 WHEN SUBSTR (col, 1, TRUNC (LENGTH (col) / 2)) =
5 SUBSTR (
6 col,
7 TRUNC (LENGTH (col) / 2)
8 + CASE WHEN MOD (l.len, 2) = 0 THEN 1 ELSE 2 END)
9 THEN
10 'OK'
11 ELSE
12 'Not OK'
13 END
14 result
15 FROM test t, len l;
RESULT
------
OK
SQL> l1
1* WITH test (col) AS (SELECT '5554555' FROM DUAL),
SQL> c/5554/2234/
1* WITH test (col) AS (SELECT '2234555' FROM DUAL),
SQL> /
RESULT
------
Not OK
SQL> l1
1* WITH test (col) AS (SELECT '2234555' FROM DUAL),
SQL> c/2234555/1221/
1* WITH test (col) AS (SELECT '1221' FROM DUAL),
SQL> /
RESULT
------
Not OK
SQL> l1
1* WITH test (col) AS (SELECT '1221' FROM DUAL),
SQL> c/1221/8888/
1* WITH test (col) AS (SELECT '8888' FROM DUAL),
SQL> /
RESULT
------
OK
SQL>

Answer 3

Use of the Trim Function to Trim Off The 'X' Values from 'Y'
This is just another approach to solving this subset of numeric palindrome problems.
If this were just a numeric palindrome, the undocumented function, reverse, could be used. Since we have a Y for the midvalue and we are testing to make sure that Y is not equal to X, the reverse function does not help us a lot here.
Borrowing on the use of subexpressions (aka character grouping) approach that Trincot uses, I just create a second subexpression for the midvalue and then I trim off the midvalue. If the trimmed expression is equal to original value, then we can be assured that Y != X.
SCOTT#db>WITH tst ( val ) AS (
2 SELECT '5555555' FROM DUAL UNION ALL
3 SELECT '12121' FROM DUAL UNION ALL
4 SELECT '5553555' FROM DUAL UNION ALL
5 SELECT 'amanaplanpanama' FROM DUAL UNION ALL
6 SELECT '' FROM DUAL
7 ) SELECT
8 val,
9 REGEXP_SUBSTR(val,'^(\d)\1\1(\d)\1{3}$',1,1,NULL,2) midval,
10 TRIM(BOTH REGEXP_SUBSTR(val,'^(\d)\1\1(\d)\1{3}$',1,1,NULL,2) FROM val) trim_midval
11 FROM
12 tst
13 WHERE
14 1 = 1
15 AND val = TRIM(BOTH regexp_substr(val,'^(\d)\1\1(\d)\1{3}$',1,1,NULL,2) FROM val);
-----------------------------
VAL MIDVAL TRIM_MIDVAL
5553555 3 5553555
-----------------------------
Littlefoot's non-regular expression solution appears to be the most straightforward here.