I have a below query which gives me data for past 12 hour. Instead of that I wanted to get data for past X days which should be configurable if possible? Is there any way to do it?
SELECT processId,
houroftheday,
minuteofhour,
listagg(clientId, ',') within group (order by minuteofhour) as clientIds,
count(*) as psg
FROM data.process
where kite = 'BULLS'
and code is null
and timestampinepochsecond > date_part(epoch, sysdate) - 3600 * 12
group by 1, 2, 3
How can I use timestampinepochsecond column to get data for past X days?
The passed N days would be:
timestampinepochsecond > date_part(epoch, sysdate) - N*24*60*60
Related
I am new to Big Query. I am trying to do a where condition to only select yesterday's data and that of same day last year (in this case, 10/25/2021 data and 10/25/2020 data). I know how to select a range of data, but I couldn't figure out a way to only select those 2 days of data. Any help is appreciated.
I recommend using BigQuery functions to define dates. You can read about them here.
WHERE DATE(your_date_field) IN ((DATE_SUB(CURRENT_DATE(), INTERVAL 1 DAY),
DATE_SUB(DATE_SUB(CURRENT_DATE(), INTERVAL 1 DAY), INTERVAL 1 YEAR))
This is dynamic to any day that you run the query. It will take the current date, then subtract 1 day. For the other date, it will take the current date and subtract 1 day and then 1 year, making it yesterday's date 1 year prior.
WHERE date_my_field IN (DATE('2021-10-25'), DATE('2020-10-25'))
Use IN which is a short cut for OR operator
Consider below (less verbose approach - especially if you remove time zone)
select
current_date('America/Los_Angeles') - 1 as yesterday,
date(current_date('America/Los_Angeles') - 1 - interval 1 year) same_day_last_year
with output
So, now you can use it in your WHERE clause as in below example (with dummy data via CTE)
with data as (
select your_date_field
from unnest(generate_date_array(current_date() - 1000, current_date())) your_date_field
)
select *
from data
where your_date_field in (
current_date('America/Los_Angeles') - 1,
date(current_date('America/Los_Angeles') - 1 - interval 1 year)
)
with output
I'm trying to write a query where i want to get every 10 days of data between 2 years.
i'm getting data between particular date to current date. but i don't know how to get each 10 days of data
in between the selected date.
Here's my current query. I'd appreciate any help. Thanks!
SELECT "Name","id",cast(sum("amount")/100 as money) AS "AMOUNT", sum(1) AS "COUNT" from table1
WHERE "DATETIME" BETWEEN '01-01-2019' AND now() - interval '10 days'
GROUP BY "Name","id"
HAVING sum("amount") > 3000
Expecting it to get data of each 10 days from 01-01-2019 to current date
If you mean you have a load of data that is day by day and you want to group it up into blocks of 10 days starting from 2019-01-01, then you can generate a time series of every 10 days and join the real data to it:
select
tendays.day,
"Name",
"id",
cast(sum("amount")/100 as money) as "AMOUNT",
sum(1) as "COUNT"
from
generate_series ('2019-01-01'::timestamp, now()::timestamp, '10 day'::interval) AS tendays(day)
LEFT JOIN
(select '2019-01-02'::timestamp as "DATETIME", 'a' as "Name", 'b' as "id", 1 as "amount" ) t1
ON
t1."DATETIME" >= tendays.day AND
t1."DATETIME" < (tendays.day + interval '10 days')
group by "Name","id", tendays.day
having sum("amount") > 3000
I wasn't really clear on whether you wanted to keep the existing grouping of Name and ID or not. Also, be careful with your HAVING - you'll only get results if a name/id pair has total amount over 3000 in a ten day block
I've made it a left join so you can remove the HAVING and see if any ten day blocks have no data meeting the criteria. It can be swapped for INNER JOIN if this is not important to you
Essentially to get items, separated by a specific interval of time, you can GROUP BY them by using a specific date difference:
SELECT ...
FROM ...
GROUP BY DATEADD(DAY, 10, DATEDIFF(day, 0, '2019-01-01'), 0)
...
Please try experimenting with the above to match your exact needs.
Update:
For Postgres the DATEDIFF alternative could be using the EXTRACT or DATEPART.
Thanks, #a_horse_with_no_name, for noticing that.
I will try to be simple as possible to make my question crystal-clear. I have a table that's called 'fb_ads' (it's about different facebook compaigns for different stores in USA) on BigQuery, it contains the following columns:
STORE : name of store
CLICKS: number of clicks.
IMPRESSIONS: number of impressions of the ad
COST: the ad cost
DATE: AAAA-MM-DD
Frequency: number of visitors of a store
So, I'm trying to calculate the variance between two years 2017 and 2018.
Here is the variance I'm trying to calculate:
Variance_Of_Frequency = ((Frequency in 2018 at date X) - ((Frequency in 2017 at date X))/((Frequency in 2017 at date X)
The problem is, that I'll have to compare the same day of the week close to Date X;
For example, if I have a compaign run on a Monday 2017-08-13, I'll need to compare to another monday in 2018 close to 2018-08-13 (it might be a monday on 2018-08-15 for example).
This is a daily variance!
I tried to make a weekly variance calculating and I don't know if it's correct, here is how I did it:
I first started with aggregating my daily table to a weekly tables using the following query:
creating my weekly_table
SELECT
year_week,
STORE,
min(DATE ) as DATE ,
SUM(IMPRESSIONS ) AS FB_IMPRESSIONS ,
SUM(CLICKS ) AS FB_CLICKS ,
SUM(COST) AS FB_COST ,
SUM(Frequency) AS FREQUENCY,
FROM (
SELECT
*,
CONCAT(cast(ANNEE as string), LPAD(cast((extract(WEEK from date)) as string), 2, '0') ) AS year_week
FROM `fb_ads`)
GROUP BY
year_week,
STORE,
ORDER BY year_week
Then I tried to calculate the variance using this:
SELECT
base.*, (base.frequency-lw.frequency) / lw.frequency as VAR_FF
FROM
`weekly_table` base
JOIN (
SELECT
* EXCEPT (date),
DATE_ADD(DATE(TIMESTAMP(date)) , INTERVAL 1 Week)AS date
FROM
`weekly_table` ) lw
ON
base.date = lw.date
AND base.store= lw.store
Anyone has any idea how to do the daily thing or if my weekly queries are correct ?
Thanks!
For a given date, you want to know the date of the nearest Monday to the same date in the following year...
SET #dt = '2017-08-17';
SELECT CASE WHEN WEEKDAY(#dt + INTERVAL 1 YEAR) > 3
THEN ADDDATE(ADDDATE(#dt + INTERVAL 1 YEAR,INTERVAL 1 WEEK),INTERVAL - WEEKDAY(#dt + INTERVAL 1 YEAR) DAY)
ELSE ADDDATE(#dt + INTERVAL 1 YEAR,INTERVAL - WEEKDAY(#dt + INTERVAL 1 YEAR) DAY)
END x;
Obviously, I could remove all those + INTERVAL 1 YEAR bits by defining #dt that way to begin with.
I'm trying to fix some problems in my database and i want to re-calculate column in my db based on other 2 date columns. This col is float and i want to get the difference between 2 dates in months with decimal point for days.
For example if i have 2 dates '2016-01-15', '2015-02-01' the difference should be 12.5 best of 12 months differences and 0.5 for the remaining 15 days
Here is what i tried so far based on my searches but i think there is something i'm missing as it tells me there is an error with my date col as it doesn't exist
Select EXTRACT(year FROM vehicle_delivery(date, vehicle_received_date))*12 + EXTRACT(month FROM vehicle_delivery(date, vehicle_received_date));
Where vehicle_delivery is my table name & date is my end date and vehicle_received_date is my start date
same thing happes with this sql :
select extract('years' from vehicle_delivery) * 12 + extract('months' from vehicle_delivery) + extract('days' from vehicle_delivery) / 30
from (select age(date::timestamp, vehicle_received_date::timestamp)) a;
The SQL should look like this:
select extract(year from diff) * 12 + extract(month from diff) + extract(day from diff) / 30
from (select age(date::timestamp, vehicle_received_date::timestamp) as diff
from vehicle_delivery
) vd;
I don't know what the purpose of the / 30 is, but you appear to want it.
Notes:
The FROM clause references the table.
The first argument in extract() is a keyword, not a string.
You want to reference the age() value in the extract().
extract() returns an interval, so it is rather redundant to take out the parts (only needed if you want them in separate columns).
Hi Ive been having an issue with getting the correct difference in a date from the current month not including the day.
ie if the month when the query is run is march 2013
then the following should be the result
EXECUTION_DATE, EXEC_DIFF
01-FEB-13, 1
31-JAN-13, 2
30-JAN-13, 2
however using the below sql statement im getting
EXECUTION_DATE, EXEC_DIFF
01-FEB-13, 1
31-JAN-13, 2
30-JAN-13, 1
select EXECUTION_DATE,
floor(MONTHS_BETWEEN (trunc(sysdate,'MM')-1, EXECUTION_DATE))+1 "EXEC_DIFF"
from V_CERT_LIST
WHERE EXECUTION_DATE < TO_DATE('02/02/2013','DD/MM/YYYY')
ORDER BY EXECUTION_DATE DESC
Please can someone put me right ive been bashing my head with this for some time now
thanks
select EXECUTION_DATE,
MONTHS_BETWEEN (trunc(sysdate,'MM'), trunc(EXECUTION_DATE,'MM')) "EXEC_DIFF"
from V_CERT_LIST
WHERE EXECUTION_DATE < TO_DATE('02/02/2013','DD/MM/YYYY')
ORDER BY EXECUTION_DATE DESC
Not looking for scores but cannot understand what is the problem with months_between? In my understanding it does not matter when in month execution takes place - Jan-31 or Jan 30... The difference is still 2 months between Jan and Mar as in your example. I can add more days in month in the query but mo_betw. will still be the same...:
SELECT to_char(exec_date, 'DD-MON-YYYY') exec_date, MONTHS_BETWEEN(run_date, exec_date) months_btwn
FROM
(
SELECT to_date('01/03/2013', 'DD/MM/YYYY') run_date
, Add_Months(Trunc(sysdate,'YEAR'),Level-1) exec_date -- first day of each month
FROM dual
CONNECT BY LEVEL <= 3
)
/
EXEC_DATE MONTHS_BTWN
------------------------
01-JAN-2013 2
01-FEB-2013 1
01-MAR-2013 0
Months_Between has complex logic that takes the day of the month into account.
Perhaps what you want is this:
select EXECUTION_DATE,
((year(sysdate)*12+month(sysdate)) - (year(execution_date)*12 + month(execution_date))
) as Exec_Diff
from V_CERT_LIST
WHERE EXECUTION_DATE < TO_DATE('02/02/2013','DD/MM/YYYY')
ORDER BY EXECUTION_DATE DESC
This converts the year/month combination into the number of months since 0 time and then subtracts the results.