Number field is the key but it can be modified several times a month (or not all all). Need to pull for last 6 complete month/year (Month_Year), only last reading before the 1st of next month.
So you want something like this?
select number, to_char(sysdate, 'Mon-YYYY') as month_year,
max(reading) keep (dense_rank first order by modified_date_time desc) as reading
from t
group by number;
The most recent reading for each number along with the current date formatted as Mon-YYYY.
Related
I have the following data:
it's a calendar that has date and year/week number there week is our internal calendar. what i want to do is sort by the most recent week number and have that be rel_week_index=1, and then the previous week be rel_week_index=2, etc. I got this to work by doing ROW_NUMBER() OVER(PARTITION BY YEAR ORDER BY YEAR DESC), however, once you get to the previous year, this rel_week_index column starts over at 1, which isn't what i want.
Ideally year_week=202111 would be rel_week_index=1, year_week=202110 would be rel_week_index=2 and so on.
If I understood you correctly this is the query to calculate row number based on year and week combination in descending order:
ROW_NUMBER() OVER( ORDER BY YEAR_week DESC)
WHERE to_date(smz_loanservicing.as_of_date) = last_day(add_months(current_date, -1))
The above will provide data only if the loanservicing.as_of_date occurs on the very last day of the month.
Last month (May 31 2020) the last day of the month fell on Sunday.
Is there a way to get the the first day of the month and say if this particular date occurs between the first and last day of the month, show the date? Essentially there were no activities on Sunday so the data was missed.
I tried
to_date(smz_loanservicing.as_of_date)
between first_day(add_month(current_date,-1))
and last_day(add_months(current_date, -1))`
However I get syntax error.
You seem to want to check if your date column belongs to the current month.
The syntax you use would work in Oracle, so let me assume this is the database that you are running. I also assume that column as_of_date is of datatype date (or timestamp).
What you ask for should be as simple as:
where
as_of_date >= trunc(current_date, 'mm')
and as_of_date < add_months(trunc(current_date, 'mm'), 1)
Actually, your syntax would also work in Snowflake - and so would the above code.
Note: if as_of_date actually is a string, then you need to_date() to convert it.
You could just truncate the date to the month, then you don’t need the know the first or last day.
where trunc(as_of_date, ‘MM’) = trunc(current_date, ‘MM’)
In SQL Server can you select the first set of values (i.e. week numbers 1 - 52) give them another value in a new column, then select the next lot of values.
The problem I am having is the data table I am working on has week numbers for each financial year, which starts the first Sunday after 1 October. So it simply iterates 1 - 52 for each financial year.
I am trying to make a column in a view that grabs the first 52 gives them the a financial year value of 1, then grabs the next 52 and gives them a financial year value of 2 etc (obviously with year 1 starting at the first record). I do have the Week Ending Date column to work with also.
Here is a snippet of the table:
Is this possible?
Leave the Sundays and Octobers. If I understand correctly, you only need to assign a rank to each occurrence of week number in order of the ending dates.
Please try this (but use copy of the table or transaction to check first; of course T is name of your table):
update T
set fiscal_year = YearNumbers.FiscalYear
from T
inner join
(
select WeekEndingDate, WeekNumber, DENSE_RANK() over (partition by WeekNumber order by WeekEndingDate) as FiscalYear
from T
) as YearNumbers
on T.WeekEndingDate = YearNumbers.WeekEndingDate and T.WeekNumber = YearNumbers.WeekNumber
My report gets data for the 1st of the current month. Let's say the 1st has still not come then how would I make the report show the data for the 1st of the previous month.
Thanks.
Simply use a select top 1 from your table, filtering by extract(day from yourDateColumn) = 1 to get only the rows with the data for the 1st day of any month, and order them in descending order by your date column (order by yourDateColumn desc), so that you always get the 1st day of the last available month in your table.
Docs for Oracle EXTRACT function
I need to show distinct users per week. I have a date-visit column, and a user id, it is a big table with 1 billion rows.
I can change the date column from the CSVs to year,month, day columns. but how do I deduce the week from that in the query.
I can calculate the week from the CSV, but this is a big process step.
I also need to show how many distinct users visit day after day, looking for workaround as there is no date type.
any ideas?
To get the week of year number:
SELECT STRFTIME_UTC_USEC(TIMESTAMP('2015-5-19'), '%W')
20
If you have your date as a timestamp (i.e microseconds since the epoch) you can use the UTC_USEC_TO_DAY/UTC_USEC_TO_WEEK functions. Alternately, if you have an iso-formatted date string (e.g. "2012/03/13 19:00:06 -0700") you can call PARSE_UTC_USEC to turn the string into a timestamp and then use that to get the week or day.
To see an example, try:
SELECT LEFT((format_utc_usec(day)),10) as day, cnt
FROM (
SELECT day, count(*) as cnt
FROM (
SELECT UTC_USEC_TO_DAY(PARSE_UTC_USEC(created_at)) as day
FROM [publicdata:samples.github_timeline])
GROUP BY day
ORDER BY cnt DESC)
To show week, just change UTC_USEC_TO_DAY(...) to UTC_USEC_TO_WEEK(..., 0) (the 0 at the end is to indicate the week starts on Sunday). See the documentation for the above functions at https://developers.google.com/bigquery/docs/query-reference for more information.