Why am I getting different results when I compare weekly results into using a code individually per week. Does it have something to do with the timestamp?
This is the code for all the weeks:
select date_trunc('week',date_joined) as week, COUNT(*) as count from auth_user
where date_joined>='01-01-2019' and date_joined<='31-03-2019'
group by week
order by week
This is the resulting table:
first result
This is the code for getting an individual week:
select COUNT(*) from auth_user where date_joined>='31-12-2018' and date_joined<='06-01-2019'
This is the result for the first week: second result
I'd say that date_joined is a timestamp, and your second query misses the entries from January 6th.
Try with
AND date_joined < '2019-01-07'
Also, you should use ISO notation: YYYY-MM-DD
Related
The following query selects rows where the date_field is a particular month and day:
Table.where("DATE_FORMAT(date_field, '%m/%d') = ?", "03/05")
I was wondering if there is a way to do something similar but instead, I supply a list of month/day strings ex: ["03/05","02,12", "12/25"] and get all rows where date_field matches that particular day and month.
You can use a SQL IN statement:
Table.where("DATE_FORMAT(date_field, '%m/%d') IN ?", ["03/05", "02/12", "12/25"])
So I want to make a query to show me if a certain calendar week has all 7 Day.
It would be okay if it just returns the numbers 1-7.
The table that I have contains articles of the 3 month of 2020 but even so the first week just contains Wednesday to Sunday it still counts it as a calendar week.
With that select I would make pl/sql Script to check it and if yes something happens.
This is an example of the Table:
Date Articel_Id
14.10.2020 78
15.10.2020 80
16.10.2020 96
17.10.2020 100
18.10.2020 99
Can I Use to_char() to check if Calendar Week has all 7 Days ?
If yes, how ?
The challenging is actually defining the weeks. If you want to define them using the ISO standard, then aggregate:
select to_char(date, 'IYYYY-IW') as yyyyww,
count(distinct trunc(date)) as num_days
from t
group by to_char(date, 'IYYYY-IW')
order by yyyyww;
This counts the number of days per week. I'm not sure if you want to filter, have a flag, or what the result set should look like. For filtering, using a having clause, such as having count(distinct trunc(date)) = 7.
I've been working on this query and I'm about 90% where I need to be however there is one piece of this that I'm unable to figure out. Basically I'm looking for the Sum of Net flows by month, starting 12/31/2014 through the current date. I'm able to extract data by day and the sum of net flows for that day, however now I really need to be able to group all the dates in to their respective months. Ex. If I have 01/01/2015, 01/02/2015 and 01/03/2015 I just want both of them to be grouped together and show up as 01/2015. Bellow is the query that I have written. Please help with this last step.
SELECT
DATE,
SUM(NET_FLOWS/1000000.00) AS YTD_NET_FLOWS
FROM
HISTORY_TBL
WHERE
DATE >= TO_DATE ('12312014','MMDDYYYY')
GROUP BY
DATE
ORDER BY
DATE
You can truncate a date to a given format (day, year, month, etc.), as shown here: http://docs.oracle.com/cd/B19306_01/server.102/b14200/functions230.htm#i1002084
SELECT TRUNC(DATE,'MM'), SUM(NET_FLOWS/1000000.00) AS YTD_NET_FLOWS
FROM HISTORY_TBL
WHERE DATE >= TO_DATE ('12312014','MMDDYYYY')
GROUP BY TRUNC(DATE,'MM')
ORDER BY DATE
I am looking for a query which gives me the daily playing time. The start (first_date) and end date(last_update) are given as shown in the Table. The following query gives me the sum of playing time on given date. How can I extend it to get a table from first day to last day and plot the query data in it and show 0 on dates when no game is played.
SELECT startTime, SUM(duration) as sum
FROM myTable
WHERE startTime = endTime
GROUP BY startTime
To show date when no one play you will need create a table days with a date field day so you could do a left join. (100 years is only 36500 rows).
Using select Generate days from date range
This use store procedure in MSQL
I will assume if a play pass the midnight a new record begin. So I could simplify my code and remove the time from datetime field
SELECT d.day, SUM(duration) as sum
FROM
days d
left join myTable m
on CONVERT(date, m.starttime) = d.day
GROUP BY d.day
If I understand correctly, you could try:
SELECT SUM(duration) AS duration, date
FROM myTable
WHERE date <= 20140430
AND date => 20140401
GROUP BY date
This would get the total time played for each date between april 1 and april 30
As far as showing 0 for dates not in the table, I don't know.
Also, the table you posted doesn't show a duration column, but the query you posted does, so I went ahead and used it.
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.