check if the person have pervious record with certain date :db2 sql - sql

Question

i have the following table :
id name start end
1 Asla 2021-01-01 2021-12-31
1 Asla 2022-01-01 2022-04-15
2 Tina 2021-05-16 2021-09-23
3 Layla 2021-01-01 2021-09-27
3 Layla 2022-01-01 2022-07-18
2 Sim 2020-05-12 2020-08-13
3 Anderas 2021-07-01 2021-09-13
3 Anderas 2021-10-01 2021-11-18
3 Anderas 2022-01-01 2029-11-18
4 Klara 2022-01-01 null
what i want to do get persons that have work (date) under 2021 and create a new column that show status (if the person continue having work under 2022 -- ok else not ok and if the person is new like 'Klara' get new ) and show last record for every person . maybe too End = null ??????
i tried this .
select w.id ,w.name ,w.start ,w.end, max_date.end
from Work_date w
left join (select * from Work_date where start>='2022-01-01')max_date on max_date.id=id
where w.start>='2021-01-01'
``` but the problem i get the result as this
<pre>
id name start end
1 Asla 2021-01-01 null
1 Asla 2022-01-01 2022-04-15
2 Tina 2021-05-16 null
3 Layla 2021-01-01 null
3 Layla 2022-01-01 2022-07-18
3 Anderas 2021-07-01 null
3 Anderas 2021-10-01 2021-11-18
3 Anderas 2022-01-01 null
4 Klara 2022-01-01 null
</pre>
men i want to get result as <pre>
id name start end status
1 Asla 2022-01-01 2022-04-15 ok
2 Tina 2021-05-16 2021-09-23 not ok
3 Layla 2022-01-01 2022-07-18 ok
3 Anderas 2022-01-01 2029-11-18 ok
4 Klara 2022-01-01 null ok

Answer 1

Looks like you can simply aggregate.
Then use a CASE WHEN for the status.
select
w.id
, w.name
, max(w.start) as start
, max(w.end) as end
, case
when year(max(end)) < 2022 then 'not ok'
else 'ok'
end as status
from Work_date w
where w.start >= '2021-01-01'
group by w.id, w.name
order by w.id, max(w.start), max(w.end);
ID
NAME
START
END
STATUS
1
Asla
2022-01-01
2022-04-15
ok
2
Tina
2021-05-16
2021-09-23
not ok
3
Layla
2022-01-01
2022-07-18
ok
3
Anderas
2022-01-01
2029-11-18
ok
4
Klara
2022-01-01
null
ok
Demo on db<>fiddle here