Reputation:
I have a table which contains a lot of duplicated rows like this:
id_emp id date ch_in ch_out
1 34103 2019-09-01
1 34193 2019-09-01 17:00
1 34194 2019-09-02 07:03:21 16:59:26
1 34104 2019-09-02 07:03:21 16:59:26
1 33361 2019-09-02 NULL NULL
I want just one row for each date and others must delete with condition like I want the output must be:
id_emp id date ch_in ch_out
1 34193 2019-09-01 17:00
1 34104 2019-09-02 07:03:21 16:59:26
I tried to use distinct but nothing working:
select distinct id_emp, id, date_1, ch_in,ch_out
from ch_inout
where id_emp=1 order by date_1 asc
And I tried too using this query to delete:
select *
from (
select *, rn=row_number() over (partition by date_1 order by id)
from ch_inout
) x
where rn > 1;
But nothing is working the result is empty.
Upvotes: 0
Views: 83
Reputation: 1271171
You can use aggregation:
select id_emp, max(id) as id, date, min(ch_in), max(ch_out)
from ch_inout
group by id_emp, date;
This returns the maximum id
for each group of rows. That is not exactly what is returned in the question, but you don't specify the logic.
EDIT:
If you want to delete all but the largest id for each id_emp
/date
combination, you can use:
delete c from ch_inout c
where id < (select max(c2.id)
from ch_inout c2
where c2.id_emp = c.id_emp and c2.date = c.date
);
Upvotes: 1
Reputation: 222722
You can use ROW_NUMBER()
to identify the records you want to delete. Assuming that you want to keep the record with the lowest id
on each date
:
SELECT *
FROM (
SELECT
t.*,
ROW_NUMBER() OVER(PARTITION BY date ORDER BY id) rn
FROM ch_inout t
) x
WHERE rn > 1
You can easily turn this into a DELETE
statement:
WITH cte AS (
SELECT
t.*,
ROW_NUMBER() OVER(PARTITION BY date ORDER BY id) rn
FROM ch_inout t
)
DELETE FROM cte WHERE rn > 1
Upvotes: 0