EX :
ID Date(with time) Price
---- ------- ------
A 23-Aug-12 (09:25pm) 10(cosider this was the latest on this date)
A 25-May-10 20
A 23-Aug-12 (8:20pm) 30
A 23-Aug-12 (7:00pm) 35
B 03-Apr-09 45
B 05-Dec-10 60
I want to Retrieve ID,Date,Price i.e If for two same dates if der are multiple prices then I have to select the date that is latest Update on date based on Timestamp included.
Expected output :
A,23-Aug,12,10
A,25-May-10,20
B,03-Apr-09,45
B,05-Dec-10,60
Most versions of SQL support the row_number function. Extracting the date from a datetime varies between databases. Here is one way to do what you want:
select id, datetime, price
from (select t.*,
row_number() over (partition by id, cast(datetime as date) order by datetime desc
) as seqnum
from t
) t
where seqnum = 1;
This gives the general structure. The exact syntax varies by database.
Try something this:-
SELECT
ID,
MAX(date)
FROM
Some_Unnamed_Table
GROUP BY
ID
ORDER BY
ID
Related
I am looking for a way to fetch Data.
by
latest date In the same day
by UserId
UserId,Value1,Date
1, 2030,2020–09-07 10:58:58
1, 2020,2020–09-07 05:58:28
1, 2050,2020–09-08 19:58:28
2, 3000,2020–09-07 10:58:18
2, 2001,2020–09-06 10:58:55
3, 2400,2020–09-08 10:28:53
4, 2400,2020–09-07 13:28:53
e.g
where Date >= trunc(TO_DATE(’20200907’,’YYYYMMDD’)) and Date < trunc(TO_DATE(’20200908’,’YYYYMMDD’))
Ideal Result
UserId,Value
1,2050
2,3000
4,2400
select UserId, value
What should I use ?
max(Date) ? Distinct userId ? Group by userId?
If value is the only column you want, then you could use keep:
select userid, max(value1) keep(dense_rank last order by dt) value1
from mytable
where dt >= date '2020-09-07' and dt < date '2020-09-08'
group by userid
order by userid
Notes:
this uses the standard date syntax rather than to_date() to build literal dates
dateis a reserved word in Oracle, hence not a good choice for a column name; I renamed it to dt in the query.
If you want more columns in the resultset, then filtering with window functions is more appropriate:
select t.*
from (
select t.*, row_number() over(partition by userid order by dt desc) rn
from mytable t
where dt >= date '2020-09-07' and dt < date '2020-09-08'
) t
where rn = 1
This is a sample data file
Data Contains unique IDs with different latitudes and longitudes on multiple timestamps.I would like to select the rows of latest 30 days of coordinates for each unique ID.Please help me on how to run the query .This date is in Hive table
Regards,
Akshay
According to your example above (where no current year dates for id=2,3), you can numbering date for each id (order by date descending) using window function ROW_NUMBER(). Then just get latest 30 values:
--get all values for each id where num<=30 (get last 30 days for each day)
select * from
(
--numbering each date for each id order by descending
select *, row_number()over(partition by ID order by DATE desc)num from Table
)X
where num<=30
If you need to get only unique dates (without consider time) for each id, then can try this query:
select * from
(
--numbering date for each id
select *, row_number()over(partition by ID order by new_date desc)num
from
(
-- move duplicate using distinct
select distinct ID,cast(DATE as date)new_date from Table
)X
)Y
where num<=30
In Oracle this will be:
SELECT * FROM TEST_DATE1
WHERE DATEUPDT > SYSDATE - 30;
select * from MyTable
where
[Date]>=dateadd(d, -30, getdate());
To group by ID and perform aggregation, something like this
select ID,
count(*) row_count,
max(Latitude) max_lat,
max(Longitude) max_long
from MyTable
where
[Date]>=dateadd(d, -30, getdate())
group by ID;
I have the following db structure.
table
-----
id (uuids)
date (TIMESTAMP)
I want to write a query in postgres (actually cockroachdb which uses the postgres engine, so postgres query should be fine).
The query should return a count of records between 2 dates , id of the record with latest date and id of the record with latest earliest date within that range.
So the query should return the following:
count, id(of the earliest record in the range), id (of the latest record in the range)
thanks.
You can use row_number() twice, then conditional aggregation:
select
no_records,
min(id) filter(where rn_asc = 1) first_id
max(id) filter(where rn_desc = 1) last_id
from (
select
id,
count(*) over() no_records
row_number() over(order by date asc) rn_asc,
row_number() over(order by date desc) rn_desc
from mytable
where date >= ? and date < ?
) t
where 1 in (rn_asc, rn_desc)
The question marks represents the (inclusive) start and (exclusive) end of the date interval.
Of course, if ids are always increasing, simple aggregation is sufficient:
select count(*), min(id) first_id, max(id) last_id
from mytable
where date >= ? and date < ?
Unfortunately, Postgres doesn't support first_value() as an aggregation function. One method is to use arrays:
select count(*),
(array_agg(id order by date asc))[1] as first_id,
(array_agg(id order by date desc))[1] as last_id
from t
where date >= ? and date <= ?
I have a requirement to get values from a table based on an offset conditions on a date column.
Say for eg: for the below attached table, if there is any dates that comes close within 15 days based on effectivedate column I should return only the first one.
So my expected result would be as below:
Here for A1234 policy, it returns 6/18/16 entry and skipped 6/12/16 entry as the offset between these 2 dates is within 15 days and I took the latest one from the list.
If you want to group rows together that are within 15 days of each other, then you have a variant of the gaps-and-islands problem. I would recommend lag() and cumulative sum for this version:
select polno, min(effectivedate), max(expirationdate)
from (select t.*,
sum(case when prev_ed >= dateadd(day, -15, effectivedate)
then 1 else 0
end) over (partition by polno order by effectivedate) as grp
from (select t.*,
lag(expirationdate) over (partition by polno order by effectivedate) as prev_ed
from t
) t
) t
group by polno, grp;
I have gotten all rows for the latest date like this:
SELECT date, quarter, sales_region, revenue
FROM regions
WHERE date = (SELECT MAX(date) FROM regions)
ORDER BY 1
So how would I get the rows for the second latest date?
I have tried but no luck:
SELECT MAX(date), quarter, sales_region, revenue
FROM regions
WHERE date < (SELECT MAX(date) FROM regions)
ORDER BY 1
Here is one method:
SELECT date, quarter, sales_region, revenue
FROM regions
WHERE date = (SELECT DISTINCT date
FROM regions r2
ORDER BY date DESC
OFFSET 1 FETCH FIRST 1 ROW ONLY
)
ORDER BY 1;
Another method uses dense_rank():
select r.*
from (select r.*, dense_rank() over (order by date desc) as seqnum
from regions r
) r
where seqnum = 2;
Gordon answered your question precisely, but if you want to get the records for the last two dates in one query, you could use IN instead of =, and get the top two records with LIMIT 2:
SELECT date, quarter, sales_region, revenue
FROM regions
WHERE date IN (SELECT DISTINCT date
FROM regions r2
ORDER BY date DESC
LIMIT 2)
ORDER BY 1;
Starting with version 8.4, you can also use FETCH FIRST 2 ROW ONLY instead of LIMIT 2.