I have to find distinct IDs throughout the whole history of each ID whose due dates are always on the last day of each month.
Suppose I have the following dataset:
ID DUE_DT
1 1/31/2014
1 2/28/2014
1 3/31/2014
1 6/30/2014
2 1/30/2014
2 2/28/2014
3 1/29/2016
3 2/29/2016
I want to write a code in SQL so that it gives me ID = 1 as for this specific ID the due date is always on the last day of each given month.
What would be the easiest way to approach it?
You can do:
select id
from t
group by id
having sum(case when extract(day from due_dt + interval '1 day') = 1 then 1 else 0 end) = count(*);
This uses ANSI/ISO standard functions for date arithmetic. These tend to vary by database, but the idea is the same in all databases -- add one day and see if the day of the month is 1 for all the rows.
If your using SQL Server 2012+ you can use the EOMONTH() function to achieve this:
SELECT DISTINCT ID FROM [table]
WHERE DUE_DT = EOMONTH(DUE_DT)
http://rextester.com/VSPQR78701
The idea is quite simple:
you are on the last day of the month if (the month of due date) is not the same as (the month of due date + 1 day). This covers all cases across year, leap year and so on.
from there on, if (the count of rows for one id) is the same as (the count of rows for this id which are the last day of the month) you have a winner.
I tried to write an example (not tested). You do not specify which DB so I will assume that cte (common table expression) are available. If not just put the cte as subquery.
In the same way, I am not sure that dateadd and interval work the same in all dialect.
with addlastdayofmonth as (
select
id
-- adding a 'virtualcolumn', 1 if last day of month 0 otherwise
, if(month(dateadd(due_date, interval '1' day)) != month(due_date), 1 ,0) as onlastday
from
table
)
select
id
, count(*) - sum(onlastday) as alwayslastday
from
addlastdayofmonth
group by
id
having
-- if count(rows) == count(rows with last day) we have a winner
halwayslastday = 0
MySQL-Version (credits to #Gordon Linoff)
SELECT
ID
FROM
<table>
GROUP BY
ID
HAVING
SUM(IF(day(DUE_DT + interval 1 Day) = 1, 1, 0)) = COUNT(ID);
Original Answer:
SELECT MAX(DUE_DT) FROM <table> WHERE ID = <the desired ID>
or if you want all MAX(DUE_DT) for each unique ID
SELECT ID, MAX(DATE) FROM <table> GROUP BY ID
Related
Have a requirement where I would need to rope the calculated value of the previous row for calculation in the current row.
The following is a sample of how the data currently looks :-
ID
Date
Days
1
2022-01-15
30
2
2022-02-18
30
3
2022-03-15
90
4
2022-05-15
30
The following is the output What I am expecting :-
ID
Date
Days
CalVal
1
2022-01-15
30
2022-02-14
2
2022-02-18
30
2022-03-16
3
2022-03-15
90
2022-06-14
4
2022-05-15
30
2022-07-14
The value of CalVal for the first row is Date + Days
From the second row onwards it should take the CalVal value of the previous row and add it with the current row Days
Essentially, what I am looking for is means to access the previous rows calculated value for use in the current row.
Is there anyway we can achieve the above via Postgres SQL? I have been tinkering with window functions and even recursive CTEs but have had no luck :(
Would appreciate any direction!
Thanks in advance!
select
id,
date,
coalesce(
days - (lag(days, 1) over (order by date, days))
, days) as days,
first_date + cast(days as integer) as newdate
from
(
select
-- get a running sum of days
id,
first_date,
date,
sum(days) over (order by date, days) as days
from
(
select
-- get the first date
id,
(select min(date) from table1) as first_date,
date,
days
from
table1
) A
) B
This query get the exact output you described. I'm not at all ready to say it is the best solution but the strategy employed is to essential create a running total of the "days" ... this means that we can just add this running total to the first date and that will always be the next date in the desired sequence. One finesse: to put the "days" back into the result, we calculated the current running total less the previous running total to arrive at the original amount.
assuming that table name is table1
select
id,
date,
days,
first_value(date) over (order by id) +
(sum(days) over (order by id rows between unbounded preceding and current row))
*interval '1 day' calval
from table1;
We just add cumulative sum of days to first date in table. It's not really what you want to do (we don't need date from previous row, just cumulative days sum)
Solution with recursion
with recursive prev_row as (
select id, date, days, date+ days*interval '1 day' calval
from table1
where id = 1
union all
select t.id, t.date, t.days, p.calval + t.days*interval '1 day' calval
from prev_row p
join table1 t on t.id = p.id+ 1
)
select *
from prev_row
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;
I have a table with just two columns: User_ID and fail_date. Each time somebody's card is rejected they are logged in the table, their card is automatically tried again 3 days later, and if they fail again, another entry is added to the table. I am trying to write a query that counts unique failures by month so I only want to count the first entry, not the 3 day retries, if they exist. My data set looks like this
user_id fail_date
222 01/01
222 01/04
555 02/15
777 03/31
777 04/02
222 10/11
so my desired output would be something like this:
month unique_fails
jan 1
feb 1
march 1
april 0
oct 1
I'll be running this in Vertica, but I'm not so much looking for perfect syntax in replies. Just help around how to approach this problem as I can't really think of a way to make it work. Thanks!
You could use lag() to get the previous timestamp per user. If the current and the previous timestamp are less than or exactly three days apart, it's a follow up. Mark the row as such. Then you can filter to exclude the follow ups.
It might look something like:
SELECT month,
count(*) unique_fails
FROM (SELECT month(fail_date) month,
CASE
WHEN datediff(day,
lag(fail_date) OVER (PARTITION BY user_id,
ORDER BY fail_date),
fail_date) <= 3 THEN
1
ELSE
0
END follow_up
FROM elbat) x
WHERE follow_up = 0
GROUP BY month;
I'm not so sure about the exact syntax in Vertica, so it might need some adaptions. I also don't know, if fail_date actually is some date/time type variant or just a string. If it's just a string the date/time specific functions may not work on it and have to be replaced or the string has to be converted prior passing it to the functions.
If the data spans several years you might also want to include the year additionally to the month to keep months from different years apart. In the inner SELECT add a column year(fail_date) year and add year to the list of columns and the GROUP BY of the outer SELECT.
You can add a flag about whether this is a "unique_fail" by doing:
select t.*,
(case when lag(fail_date) over (partition by user_id order by fail_date) > fail_date - 3
then 0 else 1
end) as first_failure_flag
from t;
Then, you want to count this flag by month:
select to_char(fail_date, 'Mon'), -- should aways include the year
sum(first_failure_flag)
from (select t.*,
(case when lag(fail_date) over (partition by user_id order by fail_date) > fail_date - 3
then 0 else 1
end) as first_failure_flag
from t
) t
group by to_char(fail_date, 'Mon')
order by min(fail_date)
In a Derived Table, determine the previous fail_date (prev_fail_date), for a specific user_id and fail_date, using a Correlated subquery.
Using the derived table dt, Count the failure, if the difference of number of days between current fail_date and prev_fail_date is greater than 3.
DateDiff() function alongside with If() function is used to determine the cases, which are not repeated tries.
To Group By this result on Month, you can use MONTH function.
But then, the data can be from multiple years, so you need to separate them out yearwise as well, so you can do a multi-level group by, using YEAR function as well.
Try the following (in MySQL) - you can get idea for other RDBMS as well:
SELECT YEAR(dt.fail_date) AS year_fail_date,
MONTH(dt.fail_date) AS month_fail_date,
COUNT( IF(DATEDIFF(dt.fail_date, dt.prev_fail_date) > 3, user_id, NULL) ) AS unique_fails
FROM (
SELECT
t1.user_id,
t1.fail_date,
(
SELECT t2.fail_date
FROM your_table AS t2
WHERE t2.user_id = t1.user_id
AND t2.fail_date < t1.fail_date
ORDER BY t2.fail_date DESC
LIMIT 1
) AS prev_fail_date
FROM your_table AS t1
) AS dt
GROUP BY
year_fail_date,
month_fail_date
ORDER BY
year_fail_date ASC,
month_fail_date ASC
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 :-).
I have a table that contains multiple records for each day of the month, over a number of years. Can someone help me out in writing a query that will only return the last day of each month.
SQL Server (other DBMS will work the same or very similarly):
SELECT
*
FROM
YourTable
WHERE
DateField IN (
SELECT MAX(DateField)
FROM YourTable
GROUP BY MONTH(DateField), YEAR(DateField)
)
An index on DateField is helpful here.
PS: If your DateField contains time values, the above will give you the very last record of every month, not the last day's worth of records. In this case use a method to reduce a datetime to its date value before doing the comparison, for example this one.
The easiest way I could find to identify if a date field in the table is the end of the month, is simply adding one day and checking if that day is 1.
where DAY(DATEADD(day, 1, AsOfDate)) = 1
If you use that as your condition (assuming AsOfDate is the date field you are looking for), then it will only returns records where AsOfDate is the last day of the month.
Use the EOMONTH() function if it's available to you (E.g. SQL Server). It returns the last date in a month given a date.
select distinct
Date
from DateTable
Where Date = EOMONTH(Date)
Or, you can use some date math.
select distinct
Date
from DateTable
where Date = DATEADD(MONTH, DATEDIFF(MONTH, -1, Date)-1, -1)
In SQL Server, this is how I usually get to the last day of the month relative to an arbitrary point in time:
select dateadd(day,-day(dateadd(month,1,current_timestamp)) , dateadd(month,1,current_timestamp) )
In a nutshell:
From your reference point-in-time,
Add 1 month,
Then, from the resulting value, subtract its day-of-the-month in days.
Voila! You've the the last day of the month containing your reference point in time.
Getting the 1st day of the month is simpler:
select dateadd(day,-(day(current_timestamp)-1),current_timestamp)
From your reference point-in-time,
subtract (in days), 1 less than the current day-of-the-month component.
Stripping off/normalizing the extraneous time component is left as an exercise for the reader.
A simple way to get the last day of month is to get the first day of the next month and subtract 1.
This should work on Oracle DB
select distinct last_day(trunc(sysdate - rownum)) dt
from dual
connect by rownum < 430
order by 1
I did the following and it worked out great. I also wanted the Maximum Date for the Current Month. Here is what I my output is. Notice the last date for July which is 24th. I pulled it on 7/24/2017, hence the result
Year Month KPI_Date
2017 4 2017-04-28
2017 5 2017-05-31
2017 6 2017-06-30
2017 7 2017-07-24
SELECT B.Year ,
B.Month ,
MAX(DateField) KPI_Date
FROM Table A
INNER JOIN ( SELECT DISTINCT
YEAR(EOMONTH(DateField)) year ,
MONTH(EOMONTH(DateField)) month
FROM Table
) B ON YEAR(A.DateField) = B.year
AND MONTH(A.DateField) = B.Month
GROUP BY B.Year ,
B.Month
SELECT * FROM YourTableName WHERE anyfilter
AND "DATE" IN (SELECT MAX(NameofDATE_Column) FROM YourTableName WHERE
anyfilter GROUP BY
TO_CHAR(NameofDATE_Column,'MONTH'),TO_CHAR(NameofDATE_Column,'YYYY'));
Note: this answer does apply for Oracle DB
Here's how I just solved this. day_date is the date field, calendar is the table that holds the dates.
SELECT cast(datepart(year, day_date) AS VARCHAR)
+ '-'
+ cast(datepart(month, day_date) AS VARCHAR)
+ '-'
+ cast(max(DATEPART(day, day_date)) AS VARCHAR) 'DATE'
FROM calendar
GROUP BY datepart(year, day_date)
,datepart(month, day_date)
ORDER BY 1