I'm querying a table and only want to select the end of quarter dates, I've done so like this:
select
yyyy_mm_dd,
id
from
t1
where
yyyy_mm_dd = cast(date_add(trunc(add_months(yyyy_mm_dd,3-pmod(month(yyyy_mm_dd)-1,3)),'MM'),-1) as date) --last day of q
With daily rows, from 2020-01-01 until 2020-12-31, the above works fine. However, 2021 rows end up being omitted as the quarter is incomplete. How could I modify the where clause so I select the last day of each quarter and the max date in the current quarter?
You can assign a row number for each quarter in descending order of date, and filter the rows with row number equals 1 (last date in each quarter):
select yyyy_mm_dd, id
from
(select
yyyy_mm_dd,
id,
row_number() over (partition by id, year(yyyy_mm_dd), quarter(yyyy_mm_dd) order by yyyy_mm_dd desc) as rn
from
t1
) t2
where rn = 1
It is not clear if you have multiple rows on the end-of-quarter dates. It might be safer to take the max and use that:
select t1.*
from (select t1.*,
max(yyyy_mm_dd) over (partition by id, year(yyyy_mm_dd), quarter(yyyy_mm_dd)) as max_yyyy_mm_dd
from t1
) t1
where yyyy_mm_dd = max_yyyy_mm_dd;
Note that this uses t1.* for the select. If you only wanted the maximum date, you can aggregate:
select id, max(yyyy_mm_dd)
from t1
group by id, year(yyyy_mm_dd), quarter(yyyy_mm_dd);
Related
How would I filter this SQL server database so only the green records are left aka the last recorded date every year for each Customer ID field.
If you want to get the rows, not only the date values, using ROW_NUMBER() is an option (you only need to use the appropriate PARTITON BY and ORDER BY clauses):
SELECT *
FROM (
SELECT
CustomerId,
[Date],
ROW_NUMBER() OVER (PARTITION BY CustomerId, YEAR[Date] ORDER BY [Date] DESC) AS Rn
FROM YourTable
) t
WHERE Rn = 1
To check the maximum date in the year, you can write a query to get for each year the date where not exists another (in the same year), as follow:
SELECT *
FROM yourtable t1
WHERE NOT EXISTS
(SELECT 1
FROM yourtable t2
WHERE t1.customerID = t2.customerID
AND t1.date > t2.date
AND DATEPART(YEAR, t1) = DATEPART(YEAR, t2))
If you have only two columns, then you can just use aggregation:
select customer_id, max(date)
from t
group by customer_id, year(date);
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 a table with a lot of cumulative columns, these columns reset to 0 at the end of each month. If I sum this data, I'll end up double counting. Instead, With Hive, I'm trying to select the max date of each month.
I've tried this:
SELECT
yyyy_mm_dd,
id,
name,
cumulative_metric1,
cumulative_metric2
FROM
mytable
WHERE
yyyy_mm_dd = last_day(yyyy_mm_dd)
mytable has daily data from the start of the year. In the output of the above, I only see the last date for January but not February. How can I select the last day of each month?
February is not over yet. Perhaps a window function does what you want:
SELECT yyyy_mm_dd, id, name, cumulative_metric1, cumulative_metric2
FROM (SELECT t.*,
MAX(yyyy_mm_dd) OVER (PARTITION BY last_day(yyyy_mm_dd)) as last_yyyy_mm_dd
FROM mytable t
) t
WHERE yyyy_mm_dd = last_yyyy_mm_dd;
This calculates the last day in the data.
use correlated subquery and date to month function in hive
SELECT
yyyy_mm_dd,
id,
name,
cumulative_metric1,
cumulative_metric2
FROM
mytable t1
WHERE
yyyy_mm_dd = select max(yyyy_mm_dd) from mytable t2 where
month(t1.yyyy_mm_dd)= month(t2.yyyy_mm_dd)
I have the sample data set below which list the water meters not working for specific reason for a certain range period (jan 2016 to december 2018).
I would like to have a query that retrieves the last maximum and minimum consecutive period where the meter was not working within that range of period.
any help will be greatly appreciated.
You have two options:
select code, to_char(min_period, 'yyyymm') min_period, to_char(max_period, 'yyyymm') max_period
from (
select code, min(period) min_period, max(period) max_period,
max(min(period)) over (partition by code) max_min_period
from (
select code, period, sum(flag) over (partition by code order by period) grp
from (
select code, period,
case when add_months(period, -1)
= lag(period) over (partition by code order by period)
then 0 else 1 end flag
from (select mrdg_acc_code code, to_date(mrdg_per_period, 'yyyymm') period from t)))
group by code, grp)
where min_period = max_min_period
Explanation:
flag rows where period is not equal previous period plus one month,
create column grp which sums flags consecutively,
group data using code and grp additionaly finding maximal start of period,
show only rows where min_period = max_min_period
Second option is recursive CTE available in Oracle 11g and above:
with
data(period, code) as (
select to_date(mrdg_per_period, 'yyyymm'), mrdg_acc_code from t
where mrdg_per_period between 201601 and 201812),
cte (period, code) as (
select to_char(period, 'yyyymm'), code from data
where (period, code) in (select max(period), code from data group by code)
union all
select to_char(data.period, 'yyyymm'), cte.code
from cte
join data on data.code = cte.code
and data.period = add_months(to_date(cte.period, 'yyyymm'), -1))
select code, min(period) min_period, max(period) max_period
from cte group by code
Explanation:
subquery data filters only rows from 2016 - 2018 additionaly converting period to date format. We need this for function add_months to work.
cte is recursive. Anchor finds starting rows, these with maximum period for each code. After union all is recursive member, which looks for the row one month older than current. If it finds it then net row, if not then stop.
final select groups data. Notice that period which were not consecutive were rejected by cte.
Though recursive queries are slower than traditional ones, there can be scenarios where second solution is better.
Here is the dbfiddle demo for both queries. Good luck.
use aggregate function with group by
select max(mdrg_per_period) mdrg_per_period, mrdg_acc_code,max(mrdg_date_read),rea_Desc,min(mdrg_per_period) not_working_as_from
from tablename
group by mrdg_acc_code,rea_Desc
This is a bit tricky. This is a gap-and-islands problem. To get all continuous periods, it will help if you have an enumeration of months. So, convert the period to a number of months and then subtract a sequence generated using row_number(). The difference is constant for a group of adjacent months.
This looks like:
select acc_code, min(period), max(period)
from (select t.*,
row_number() over (partition by acc_code order by period_num) as seqnum
from (select t.*, floor(period / 100) * 12 + mod(period, 100) as period_num
from t
) t
where rea_desc = 'METER NOT WORKING'
) t
group by (period_num - seqnum);
Then, if you want the last one for each account, you can use a subquery:
select t.*
from (select acc_code, min(period), max(period),
row_number() over (partition by acc_code order by max(period desc) as seqnum
from (select t.*,
row_number() over (partition by acc_code order by period_num) as seqnum
from (select t.*, floor(period / 100) * 12 + mod(period, 100) as period_num
from t
) t
where rea_desc = 'METER NOT WORKING'
) t
group by (period_num - seqnum)
) t
where seqnum = 1;
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.