i am looking to only get records from exactly 3 days ago, not the records from now and as far back as 3 days ago... i only want records for that 24hour period and when i use the > its selecting everything from now until then
this is my query
SELECT user_id, nickname, user_email, date_of_register, verification_code, user_photo_url FROM `tbl_users` WHERE verified='N' AND date_of_register > DATE_SUB(CURDATE(), INTERVAL 3 DAY) ORDER BY date_of_register ASC
column date_of_register is date time and this is the format
2016-08-26 08:57:52
thank you for your help
Compare dates instead of date and time, CAST your date_of_register as DATE only.
WHERE verified='N' AND
CAST( date_of_register AS DATE) = CAST(DATE_SUB(CURDATE(), INTERVAL 3 DAY) AS DATE)
Try:
$date = "2016-08-26";
SELECT * FROM `tbl_users` WHERE `verified`='N' AND `date_of_register` = '{$date}' ORDER BY `date_of_register` ASC
Setting a variable of $date will always be the better way as it can be quicker to search rather than routing through numerous lines of code to find your query.
Rather than selecting a set few, why not just SELECT * (all) ?
You have to match date and ignore time to get result
use DATE(date_of_register) to get record from 3 days
SELECT user_id, nickname, user_email, date_of_register, verification_code, user_photo_url FROM `tbl_users` WHERE verified='N' AND DATE(date_of_register) > DATE_SUB(CURDATE(), INTERVAL 3 DAY) ORDER BY date_of_register ASC
Adding this as the current accepted solution would prevent indexes from working on the date_of_register field.
set #date=DATE_SUB(CURDATE(), INTERVAL 3 DAY)
SELECT user_id, nickname, user_email, date_of_register, verification_code, user_photo_url
FROM `tbl_users`
WHERE verified='N'
AND date_of_register > #date
AND date_of_register < DATE_ADD(#date,INTERVAL 1 DAY)
ORDER BY date_of_register ASC
Related
Can someone help me with a query in Teradata/SQL that I can use to extract all users that have more than 3 transactions in a timestamp of 10 minutes. Below is an extract of the table in question.
Kind regards,
You can use lag()/lead() and time comparisons. To get the rows where there are 2 such transactions before:
select t.*
from t
qualify transaction_timestamp < lag(transaction_timestamp, -2) over (partition by userid order by transaction_timestamp) + interval '10' minute
If you only want the users:
select distinct userid
from (select t.*
from t
qualify transaction_timestamp < lag(transaction_timestamp, 2) over (partition by userid order by transaction_timestamp) + interval '10' minute
) t
I am doing a query where the query will append the data from previous date as the outcome in BigQuery.
So, the result data for today will be higher than yesterdays as the data is appending by days.
So far, what I only managed to get the outcome is the data by days (where you can see the number of ID declining and is not appending from previous day) as this result:
What should I do to add appending function in the query so each day will get the result of data from the previous day in bigquery?
code:
WITH
table1 AS (
SELECT
ID,
...
FROM t
WHERE DATE_SUB('2020-01-31', INTERVAL 31 DAY) and '2020-01-31'
),
table2 AS (
SELECT
ID,
COUNTIF((rating < 7) as bad,
COUNTIF((rating >= 7 AND SAFE_CAST(NPS_Rating as INT64) < 9) as intermediate,
COUNTIF((rating as good
FROM
t
WHERE DATE_SUB('2020-01-31', INTERVAL 31 DAY) and '2020-01-31'
)
SELECT
DATE_SUB('2020-01-31', INTERVAL 31 DAY) as date,
*
FROM table1
FULL OUTER JOIN table2 USING (ID)
If you have counts that you want to accumulate, then you want a cumulative sum. The query would look something like this:
select datecol, count(*), sum(count(*)) over (order by datecol)
from t
group by datecol
order by datecol;
I have a Bigquery table with a date partition key.
I get daily records in that table and I try to find if there's any missing day for like 3 years of historical data.
So I tried to use the following query:
SELECT KeyPartitionDate
FROM (
SELECT KeyPartitionDate, DATE(KeyPartitionDate) as day, DATE_ADD(date(KeyPartitionDate), INTERVAL 1 DAY) AS dayplusone
FROM `project.dataset.table`
)
WHERE DATE_DIFF(day, dayplusone , DAY) > 1
GROUP BY KeyPartitionDate
ORDER BY KeyPartitionDate
The query is valid but returns no results while I know there are some...
My guess is that I'm messing with the DATE_ADD function but cant tell how
Below is for BigQuery Standard SQL and just gives you the list of missing days
#standardSQL
SELECT day AS missing_days
FROM (
SELECT MIN(KeyPartitionDate) min_day, MAX(KeyPartitionDate) max_day
FROM `project.dataset.table`
), UNNEST(GENERATE_DATE_ARRAY(min_day, max_day)) day
LEFT JOIN (
SELECT DISTINCT KeyPartitionDate AS day
FROM `project.dataset.table`
) t
USING(day)
WHERE t.day IS NULL
You went about this the wrong way:
day = DATE(KeyPartitionDate)
then you did
dayplusone = DATE_ADD(date(KeyPartitionDate), INTERVAL 1 DAY)
which is basically saying dayplusone = day +(1 day)
Then you do :
WHERE DATE_DIFF(day, dayplusone , DAY) > 1
which is like saying : dayplusone - day > (1 day) which would mean
day + (1 day) - day > (1 day)
You can clearly see why that is wrong.
What you needed to do instead is compare the current row date with the preivous row date. That is achieved using window functions:
SELECT KeyPartitionDate FROM (
SELECT DISTINCT KeyPartitionDate,
LAG(KeyPartitionDate)
OVER (ORDER BY KeyPartitionDate ASC) AS PreviousKeyPartitionDate
FROM `project.dataset.table`)
WHERE DATE_DIFF(DATE(PreviousKeyPartitionDate),DATE(KeyPartitionDate), DAY ) > 1
ORDER BY KeyPartitionDate
Dear Stack Overflow community,
I am looking for the patient id where the two consecutive dates after the very first one are less than 7 days.
So differences between 2nd and 1st date <= 7 days
and differences between 3rd and 2nd date <= 7 days
Example:
ID Date
1 9/8/2014
1 9/9/2014
1 9/10/2014
2 5/31/2014
2 7/20/2014
2 9/8/2014
For patient 1, the two dates following it are less than 7 days apart.
For patient 2 however, the following date are more than 7 days apart (50 days).
I am trying to write an SQL query that just output the patient id "1".
Thanks for your help :)
You want to use lead(), but this is complicated because you want this only for the first three rows. I think I would go for:
select t.*
from (select t.*,
lead(date, 1) over (partition by id order by date) as next_date,
lead(date, 2) over (partition by id order by date) as next_date_2,
row_number() over (partition by id order by date) as seqnum
from t
) t
where seqnum = 1 and
next_date <= date + interval '7' day and
next_date2 <= next_date + interval '7' day;
You can try using window function lag()
select * from
(
select id,date,lag(date) over(order by date) as prevdate
from tablename
)A where datediff(day,date,prevdate)<=7
I need to schedule some items in a postgres query based on a requested delivery date for an order. So for example, the order has a requested delivery on a Monday (20120319 for example), and the order needs to be prepared on the prior working day (20120316).
Thoughts on the most direct method? I'm open to adding a dates table. I'm thinking there's got to be a better way than a long set of case statements using:
SELECT EXTRACT(DOW FROM TIMESTAMP '2001-02-16 20:38:40');
This gets you previous business day.
SELECT
CASE (EXTRACT(ISODOW FROM current_date)::integer) % 7
WHEN 1 THEN current_date-3
WHEN 0 THEN current_date-2
ELSE current_date-1
END AS previous_business_day
To have the previous work day:
select max(s.a) as work_day
from (
select s.a::date
from generate_series('2012-01-02'::date, '2050-12-31', '1 day') s(a)
where extract(dow from s.a) between 1 and 5
except
select holiday_date
from holiday_table
) s
where s.a < '2012-03-19'
;
If you want the next work day just invert the query.
SELECT y.d AS prep_day
FROM (
SELECT generate_series(dday - 8, dday - 1, interval '1d')::date AS d
FROM (SELECT '2012-03-19'::date AS dday) x
) y
LEFT JOIN holiday h USING (d)
WHERE h.d IS NULL
AND extract(isodow from y.d) < 6
ORDER BY y.d DESC
LIMIT 1;
It should be faster to generate only as many days as necessary. I generate one week prior to the delivery. That should cover all possibilities.
isodow as extract parameter is more convenient than dow to test for workdays.
min() / max(), ORDER BY / LIMIT 1, that's a matter of taste with the few rows in my query.
To get several candidate days in descending order, not just the top pick, change the LIMIT 1.
I put the dday (delivery day) in a subquery so you only have to input it once. You can enter any date or timestamp literal. It is cast to date either way.
CREATE TABLE Holidays (Holiday, PrecedingBusinessDay) AS VALUES
('2012-12-25'::DATE, '2012-12-24'::DATE),
('2012-12-26'::DATE, '2012-12-24'::DATE);
SELECT Day, COALESCE(PrecedingBusinessDay, PrecedingMondayToFriday)
FROM
(SELECT Day, Day - CASE DATE_PART('DOW', Day)
WHEN 0 THEN 2
WHEN 1 THEN 3
ELSE 1
END AS PrecedingMondayToFriday
FROM TestDays) AS PrecedingMondaysToFridays
LEFT JOIN Holidays ON PrecedingMondayToFriday = Holiday;
You might want to rename some of the identifiers :-).