Set flag column based on a regular expression - sql

Question

I have developed the following query but it does not work as expected:
WITH TABLE1 AS
(
SELECT 613414473 as ID, 1706014200964 as P_NUM, 119539 as d_id, 'F20.0' AS CODE FROM DUAL UNION ALL
SELECT 613414473 as ID, 1706014200964 as P_NUM, 119539 as d_id, 'F22.0' AS CODE FROM DUAL UNION ALL
SELECT 613415801 as ID, 1707045167741 as P_NUM, 115182 as d_id, 'A94.0' AS CODE FROM DUAL UNION ALL
SELECT 613415801 as ID, 1707045167741 as P_NUM, 115182 as d_id, NULL AS CODE FROM DUAL UNION ALL
SELECT 613417084 as ID, 1702038456441 as P_NUM, 6541 as d_id, 'E79' AS CODE FROM DUAL UNION ALL
SELECT 613417084 as ID, 1702038456421 as P_NUM, 6541 as d_id, 'I10' AS CODE FROM DUAL UNION ALL
SELECT 613418372 as ID, 1706226211517 as P_NUM, 25727 as d_id, 'F32.9' AS CODE FROM DUAL )
SELECT T1.*
, CASE when regexp_like( CODE, 'C0[5-9]|' ||
'A0[0-9]|A1[0-9]|A2[0-9]|A3[0-9]|A4[0-9]|A5[0-9]|A6[0-9]|A7[0-9]|A8[0-9]|A9[0-7]|' )
THEN 1
ELSE 0 END AS FOUND_CODE
FROM TABLE1 T1;
I want codes which are like C0[5-9]% or A0[0-97] to be flagged with value 1, and then for the same p_num, if at least one code was found to set all the flags for that p_num to 1.
Example output of the above:
| 613414473|1706014200964|119539|F20.0|0|
| 613414473|1706014200964|119539|F22.0|0|
| 613415801|1707045167741|115182|A94.0|1|
| 613415801|1707045167741|115182|NULL |1|
| 613417084|1702038456441|6541 |E79 |0|
| 613417084|1702038456421|6541 |I10 |0|
| 613418372|1706226211517|25727 |F32.9|0|
How can I modify my query to get that output? And is there a better regular expression I can use?

Answer 1

Based on your description, the regular expression pattern should be
'^(C0[5-9]|A[0-8][0-9]|A9[0-7])'
The ^ anchors to the start of the value, and the parentheses allow any of the pipe-separated patterns to match; and the patterns are simplified, as A00 to A89 can be handles in one go.
That flags the same single row as your original query. The next stage is to move that into a subquery, and then use an analytic function partitioned by the p_num which you want to be common:
max(found_code) over (partition by p_num)
So together that becomes (with additional rows to match a different rule):
with table1 (id, p_num, d_id, code) as
(
select 613414470, 1706014200960, 119530, 'D99' from dual union all
select 613414471, 1706014200960, 119531, 'C05' from dual union all
--
select 613414473, 1706014200964, 119539, 'F20.0' from dual union all
select 613414473, 1706014200964, 119539, 'F22.0' from dual union all
select 613415801, 1707045167741, 115182, 'A94.0' from dual union all
select 613415801, 1707045167741, 115182, null from dual union all
select 613417084, 1702038456441, 6541 , 'E79' from dual union all
select 613417084, 1702038456421, 6541 , 'I10' from dual union all
select 613418372, 1706226211517, 25727 , 'F32.9' from dual
)
select id, p_num, d_id, code, max(found_code) over (partition by p_num) as found_code
from (
select t1.*
, case when regexp_like( code, '^(C0[5-9]|A[0-8][0-9]|A9[0-7])' )
then 1
else 0
end as found_code
from table1 t1
);
ID P_NUM D_ID CODE FOUND_CODE
------------- ------------- ------------- ----- -------------
613414470 1706014200960 119530 D99 1
613414471 1706014200960 119531 C05 1
613414473 1706014200964 119539 F20.0 0
613414473 1706014200964 119539 F22.0 0
613415801 1707045167741 115182 A94.0 1
613415801 1707045167741 115182 1
613417084 1702038456441 6541 E79 0
613417084 1702038456421 6541 I10 0
613418372 1706226211517 25727 F32.9 0