Last day of the month with a twist in SQLPLUS - sql

I would appreciate a little expert help please.
in an SQL SELECT statement I am trying to get the last day with data per month for the last year.
Example, I am easily able to get the last day of each month and join that to my data table, but the problem is, if the last day of the month does not have data, then there is no returned data. What I need is for the SELECT to return the last day with data for the month.
This is probably easy to do, but to be honest, my brain fart is starting to hurt.
I've attached the select below that works for returning the data for only the last day of the month for the last 12 months.
Thanks in advance for your help!
SELECT fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,fd.column_name
FROM super_table fd,
(SELECT TRUNC(daterange,'MM')-1 first_of_month
FROM (
select TRUNC(sysdate-365,'MM') + level as DateRange
from dual
connect by level<=365)
GROUP BY TRUNC(daterange,'MM')) fom
WHERE fd.cust_id = :CUST_ID
AND fd.coll_date > SYSDATE-400
AND TRUNC(fd.coll_date) = fom.first_of_month
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)

You probably need to group your data so that each month's data is in the group, and then within the group select the maximum date present. The sub-query might be:
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY YEAR(coll_date) * 100 + MONTH(coll_date);
This presumes that the functions YEAR() and MONTH() exist to extract the year and month from a date as an integer value. Clearly, this doesn't constrain the range of dates - you can do that, too. If you don't have the functions in Oracle, then you do some sort of manipulation to get the equivalent result.
Using information from Rhose (thanks):
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY TO_CHAR(coll_date, 'YYYYMM');
This achieves the same net result, putting all dates from the same calendar month into a group and then determining the maximum value present within that group.

Here's another approach, if ANSI row_number() is supported:
with RevDayRanked(itemDate,rn) as (
select
cast(coll_date as date),
row_number() over (
partition by datediff(month,coll_date,'2000-01-01') -- rewrite datediff as needed for your platform
order by coll_date desc
)
from super_table
)
select itemDate
from RevDayRanked
where rn = 1;
Rows numbered 1 will be nondeterministically chosen among rows on the last active date of the month, so you don't need distinct. If you want information out of the table for all rows on these dates, use rank() over days instead of row_number() over coll_date values, so a value of 1 appears for any row on the last active date of the month, and select the additional columns you need:
with RevDayRanked(cust_id, server_name, coll_date, rk) as (
select
cust_id, server_name, coll_date,
rank() over (
partition by datediff(month,coll_date,'2000-01-01')
order by cast(coll_date as date) desc
)
from super_table
)
select cust_id, server_name, coll_date
from RevDayRanked
where rk = 1;
If row_number() and rank() aren't supported, another approach is this (for the second query above). Select all rows from your table for which there's no row in the table from a later day in the same month.
select
cust_id, server_name, coll_date
from super_table as ST1
where not exists (
select *
from super_table as ST2
where datediff(month,ST1.coll_date,ST2.coll_date) = 0
and cast(ST2.coll_date as date) > cast(ST1.coll_date as date)
)
If you have to do this kind of thing a lot, see if you can create an index over computed columns that hold cast(coll_date as date) and a month indicator like datediff(month,'2001-01-01',coll_date). That'll make more of the predicates SARGs.

Putting the above pieces together, would something like this work for you?
SELECT fd.cust_id,
fd.server_name,
fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,
fd.column_name
FROM super_table fd,
WHERE fd.cust_id = :CUST_ID
AND TRUNC(fd.coll_date) IN (
SELECT MAX(TRUNC(coll_date))
FROM super_table
WHERE coll_date > SYSDATE - 400
AND cust_id = :CUST_ID
GROUP BY TO_CHAR(coll_date,'YYYYMM')
)
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)

Related

Column neither grouped nor aggregated after introducing window query

I have trouble integrating a simple window function into my query. I work with this avocado dataset from Kaggle. I started off with a simple query:
SELECT
date,
SUM(Total_Bags) as weekly_bags,
FROM
`course.avocado`
WHERE
EXTRACT(year FROM date) = 2015
GROUP BY
date
ORDER BY
date
And it works just fine. Next, I want to add the rolling sum to the query to display along the weekly sum. I tried the following:
SELECT
date,
SUM(Total_Bags) as weekly_bags,
SUM(Total_Bags) OVER(
PARTITION BY date
ORDER BY date
ROWS BETWEEN 4 PRECEDING AND CURRENT ROW
)
FROM
`course.avocado`
WHERE
EXTRACT(year FROM date) = 2015
GROUP BY
date
ORDER BY
date
but im getting the common error:
SELECT list expression references column Total_Bags which is neither grouped nor aggregated at [4:7]
and im confused. Total_Bags in the first query was aggregated yet when it's introduced again in the second query, it's not aggregated anymore. How do I fix this query? Thanks.
In your query, which returns 2 columns: date and aggregate SUM(Total_Bags), the window function SUM() is evaluated after the aggregation when there is no column Total_Bags and this is why you can't use it inside the window function.
However, you can do want you want, without group by, by using only window functions and DISTINCT:
SELECT DISTINCT date,
SUM(Total_Bags) OVER(PARTITION BY date) AS weekly_bags,
SUM(Total_Bags) OVER(
PARTITION BY date
ORDER BY date
ROWS BETWEEN 4 PRECEDING AND CURRENT ROW
)
FROM course.avocado
WHERE EXTRACT(year FROM date) = 2015
ORDER BY date;
or, use window function on the the aggregated result:
SELECT date,
SUM(Total_Bags) AS weekly_bags,
SUM(SUM(Total_Bags)) OVER(
ORDER BY date
ROWS BETWEEN 4 PRECEDING AND CURRENT ROW
)
FROM course.avocado
WHERE EXTRACT(year FROM date) = 2015
GROUP BY date
ORDER BY date;
I tried to approach it from a different angle and seems I have figured it out, the results seem just right. Here's the code:
WITH daily_bags AS
(SELECT
Date,
CAST(SUM(Total_Bags) as int64) as all_bags
FROM
`course.avocado`
WHERE
EXTRACT(year from Date) = 2015
GROUP BY
Date
ORDER BY
Date)
SELECT
Date,
all_bags,
SUM(all_bags) OVER(
ROWS BETWEEN 4 PRECEDING AND CURRENT ROW
) as rolling_sum
FROM
daily_bags
Thanks everyone for your help.

Finding id's available in previous weeks but not in current week

How to find if an id which was present in previous weeks but not available in current week on a rolling basis. For e.g
Week1 has id 1,2,3,4,5
Week2 has id 3,4,5,7,8
Week3 has id 1,3,5,10,11
So I found out that id 1 and 2 are missing in week 2 and id 2,4,7,8 are missing in week 3 from previous 2 weeks But how to do this on a rolling window for a large amount of data distributed over a period of 20+ years
Please find the sample dataset and expected output. I am expecting the output to be partitioned based on the week_end Date
Dataset
ID|WEEK_START|WEEK_END|APPEARING_DATE
7152|2015-12-27|2016-01-02|2015-12-27
8350|2015-12-27|2016-01-02|2015-12-27
7152|2015-12-27|2016-01-02|2015-12-29
4697|2015-12-27|2016-01-02|2015-12-30
7187|2015-12-27|2016-01-02|2015-01-01
8005|2015-12-27|2016-01-02|2015-12-27
8005|2015-12-27|2016-01-02|2015-12-29
6254|2016-01-03|2016-01-09|2016-01-03
7962|2016-01-03|2016-01-09|2016-01-04
3339|2016-01-03|2016-01-09|2016-01-06
7834|2016-01-03|2016-01-09|2016-01-03
7962|2016-01-03|2016-01-09|2016-01-05
7152|2016-01-03|2016-01-09|2016-01-07
8350|2016-01-03|2016-01-09|2016-01-09
2403|2016-01-10|2016-01-16|2016-01-10
0157|2016-01-10|2016-01-16|2016-01-11
2228|2016-01-10|2016-01-16|2016-01-14
4697|2016-01-10|2016-01-16|2016-01-14
Excepted Output
Partition1: WEEK_END=2016-01-02
ID|MAX(LAST_APPEARING_DATE)
7152|2015-12-29
8350|2015-12-27
4697|2015-12-30
7187|2015-01-01
8005|2015-12-29
Partition1: WEEK_END=2016-01-09
ID|MAX(LAST_APPEARING_DATE)
7152|2016-01-07
8350|2016-01-09
4697|2015-12-30
7187|2015-01-01
8005|2015-12-29
6254|2016-01-03
7962|2016-01-05
3339|2016-01-06
7834|2016-01-03
Partition3: WEEK_END=2016-01-10
ID|MAX(LAST_APPEARING_DATE)
7152|2016-01-07
8350|2016-01-09
4697|2016-01-14
7187|2015-01-01
8005|2015-12-29
6254|2016-01-03
7962|2016-01-05
3339|2016-01-06
7834|2016-01-03
2403|2016-01-10
0157|2016-01-11
2228|2016-01-14
Please use below query,
select ID, MAX(APPEARING_DATE) from table_name
group by ID, WEEK_END;
Or, including WEEK)END,
select ID, WEEK_END, MAX(APPEARING_DATE) from table_name
group by ID, WEEK_END;
You can use aggregation:
select t.*, max(week_end)
from t
group by id
having max(week_end) < '2016-01-02';
Adjust the date in the having clause for the week end that you want.
Actually, your question is a bit unclear. I'm not sure if a later week end would keep the row or not. If you want "as of" data, then include a where clause:
select t.id, max(week_end)
from t
where week_end < '2016-01-02'
group by id
having max(week_end) < '2016-01-02';
If you want this for a range of dates, then you can use a derived table:
select we.the_week_end, t.id, max(week_end)
from (select '2016-01-02' as the_week_end union all
select '2016-01-09' as the_week_end
) we cross join
t
where t.week_end < we.the_week_end
group by id, we.the_week_end
having max(t.week_end) < we.the_week_end;

ORACLE SQL: Find last minimum and maximum consecutive period

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;

Redshift list 3 most recent values per year

I have a column of dates and I want to find the three maximum dates for each year I have tried the following.
select max(date, rank() over (partition by SPLIT_PART(date, '-', 1) order by date desc)
from table
;
My desired output would be
2013,2010-12-31
2013,2010-12-30
2013,2010-12-29
also there are repeats dates in the table so I would have to filter those out as well
Assuming there are no duplicate dates, you can partition by the year part of date and get the latest 3 dates per year. Use distinct (if needed) in the final query to remove the duplicates, if any.
select yr,date
from (select date_part(year,date) as yr,date
,dense_rank() over (partition by date_part(year,date) order by date desc) as rnk
from table
) t
where rnk<=3

SQL Statement Only latest entry of the day

seems it is too long ago that I needed create own SQL Statements. I have a table (GAS_COUNTER) with timestamps (TS) and values (VALUE).
There are hundreds of entries per day, but I only need the latest of the day. I tried different ways but never get what I need.
Edit
Thanks for the fast replies, but some do not meet my needs (I need the latest value of each day in the table) and some don't work. My best own statement was:
select distinct (COUNT),
from
(select
extract (DAY_OF_YEAR from TS) as COUNT,
extract (YEAR from TS) as YEAR,
extract (MONTH from TS) as MONTH,
extract (DAY from TS) as DAY,
VALUE as VALUE
from GAS_COUNTER
order by COUNT)
but the value is missing. If I put it in the first select all rows return. (logical correct as every line is distinct)
Here an example of the Table content:
TS VALUE
2015-07-25 08:47:12.663 0.0
2015-07-25 22:50:52.155 2.269999999552965
2015-08-10 11:18:07.667 52.81999999284744
2015-08-10 20:29:20.875 53.27999997138977
2015-08-11 10:27:21.49 54.439999997615814
2nd Edit and solution
select TS, VALUE from GAS_COUNTER
where TS in (
select max(TS) from GAS_COUNTER group by extract(DAY_OF_YEAR from TS)
)
This one would give you the very last record:
select top 1 * from GAS_COUNTER order by TS desc
Here is one that would give you last records for every day:
select VALUE from GAS_COUNTER
where TS in (
select max(TS) from GAS_COUNTER group by to_date(TS,'yyyy-mm-dd')
)
Depending on the database you are using you might need to replace/adjust to_date(TS,'yyyy-mm-dd') function. Basically it should extract date-only part from the timestamp.
Select the max value for the timestamp.
select MAX(TS), value -- or whatever other columns you want from the record
from GAS_COUNTER
group by value
Something like this would window the data and give you the last value on the day - but what happens if you get two TS the same? Which one do you want?
select *
from ( select distinct cast( TS as date ) as dt
from GAS_COUNTER ) as gc1 -- distinct days
cross apply (
select top 1 VALUE -- last value on the date.
from GAS_COUNTER as gc2
where gc2.TS < dateadd( day, 1, gc1.dt )
and gc2.TS >= gc1.dt
order by gc2.TS desc
) as x