In my table I have:
Activity : Date
---------------
doSomething1 : June 1, 2020
doSomething2 : June 14, 2020
I want to be able to make a query so that I can get the following result (assuming today is June 1, 2020):
Today : ThisMonth
1 : 2
I looked at group by but I wasn't sure how to do that without a lot of additional code and I think there's very likely a much simpler solution that I'm missing. Something that will just return a single row with two results. Is this possible and if so how?
you can write subqueries to get data in single row,
Select today , month
from
(
( query to get today's count ) as today,
( query to get month's count ) as month
) t;
yes, u can do group by on dates to get todays nd months count.
Hope this will give u some perception to go on.
Is this what you want?
select array_agg(activity) filter (where date = current_date) as today,
array_agg(activity) filter (where date <> current_date) as rest_of_month
from t
where date_trunc('month', date) = current_date;
This uses arrays so it can handle more than one activity in either category.
Assume you want to query based on a particular date -
select count(case when d.date = :p_query_date then 0 end) day_count
,count(0) month_count
from d -- your table name
where d.date between date_trunc('month', :p_query_date)
and date_trunc('month', :p_query_date + interval '1 month') - interval '1 day'
The above query assumes you have index defined on d.date column. If you have index defined on date_trunc('month', date), the query condition can be simplified to:
date_trunc('month', d.date) = date_trunc('month', :p_query_date)
Related
I am trying to get all the paid contracts from my contracts table and group them by month. I can get the data but for months where there is no new paid contract I want to get a zero instead of missing month. I have tried coalesce and generate_series but I cannot seem to get the missing row.
Here is my query:
with months as (
select generate_series(
'2019-01-01', current_date, interval '1 month'
) as series )
select date(months.series) as day, SUM(contracts.price) from months
left JOIN contracts on date(date_trunc('month', contracts.to)) = months.series
where contracts.tier='paid' and contracts.trial=false and (contracts.to is not NULL) group by day;
I want the results to look like:
|Contract Value| Month|
| 20 | 01-2020|
| 10 | 02-2020|
| 0 | 03-2020|
I can get the rows where there is a contract but cannot get the zero row.
Postgres Version 10.9
I think that you want:
with months as (
select generate_series('2019-01-01', current_date, interval '1 month' ) as series
)
select m.series as day, coalesce(sum(c.price), 0) sum_price
from months m
left join contracts c
on c.to >= m.series
and c.to < m.series + interval '1' month
and co.tier = 'paid'
and not c.trial
group by m.series;
That is:
you want the condition on the left joined table in the on clause of the join rather than in the where clause, otherwise they become mandatory, and evict rows where the left join came back empty
the filter on the date can be optimized to avoid using date functions; this makes the query SARGeable, ie the database may take advantage of an index on the date column
table aliases make the query easier to read and write
You need to move conditions to the on clause:
with months as (
select generate_series( '2019-01-01'::date, current_date, interval '1 month') as series
)
select dm.series as day, coalesce(sum(c.price), 0)
from months m left join
contracts c
on c.to >= m.series and
c.to < m.series + interval '1 month' and
c.tier = 'paid' and
c.trial = false
group by day;
Note some changes to the query:
The conditions on c that were in the where clause are in the on clause.
The date comparison uses simple data comparisons, rather than truncating to the month. This helps the optimizer and makes it easier to use an index.
Table aliases make the query easier to write and to read.
There is no need to convert day to a date. It already is.
to is a bad choice for a column name because it is reserved. However, I did not change it.
In postgresql, how do I perform a query that returns the sum amounts of rows created of a particular table by month? I would like the result to be something like:
month: January
count: 67
month: February
count: 85
....
....
Let's suppose a I have a table, users. This table has a primary key, id, and a created_at column with time stored in ISO8601 formatting. Last year n number of users were created, and now I want to know how many were created by month, and I want the data returned to me in the above format -- grouped by month and an associated count reflecting how many users were created that month.
Does anyone know how to perform the above SQL query in postgresql?
The query would look something like this:
select date_trunc('month', created_at) as mm, count(*)
from users u
where subscribed = true and
created_at >= '2016-01-01' and
created_at < '2017-01-01'
group by date_trunc('month', created_at);
I don't know where the constant '2017-03-20 13:38:46.688-04' is coming from.
Of course you can make the year comparison dynamic:
select date_trunc('month', created_at) as mm, count(*)
from users u
where subscribed = true and
created_at >= date_trunc('year', now()) - interval '1 year' and
created_at < date_trunc('year', now())
group by date_trunc('month', created_at);
I have a Postgres table that I'm trying to analyze based on some date columns.
I'm basically trying to count the number of rows in my table that fulfill this requirement, and then group them by month and year. Instead of my query looking like this:
SELECT * FROM $TABLE WHERE date1::date <= '2012-05-31'
and date2::date > '2012-05-31';
it should be able to display this for the months available in my data so that I don't have to change the months manually every time I add new data, and so I can get everything with one query.
In the case above I'd like it to group the sum of rows which fit the criteria into the year 2012 and month 05. Similarly, if my WHERE clause looked like this:
date1::date <= '2012-06-31' and date2::date > '2012-06-31'
I'd like it to group this sum into the year 2012 and month 06.
This isn't entirely clear to me:
I'd like it to group the sum of rows
I'll interpret it this way: you want to list all rows "per month" matching the criteria:
WITH x AS (
SELECT date_trunc('month', min(date1)) AS start
,date_trunc('month', max(date2)) + interval '1 month' AS stop
FROM tbl
)
SELECT to_char(y.mon, 'YYYY-MM') AS mon, t.*
FROM (
SELECT generate_series(x.start, x.stop, '1 month') AS mon
FROM x
) y
LEFT JOIN tbl t ON t.date1::date <= y.mon
AND t.date2::date > y.mon -- why the explicit cast to date?
ORDER BY y.mon, t.date1, t.date2;
Assuming date2 >= date1.
Compute lower and upper border of time period and truncate to month (adding 1 to upper border to include the last row, too.
Use generate_series() to create the set of months in question
LEFT JOIN rows from your table with the declared criteria and sort by month.
You could also GROUP BY at this stage to calculate aggregates ..
Here is the reasoning. First, create a list of all possible dates. Then get the cumulative number of date1 up to a given date. Then get the cumulative number of date2 after the date and subtract the results. The following query does this using correlated subqueries (not my favorite construct, but handy in this case):
select thedate,
(select count(*) from t where date1::date <= d.thedate) -
(select count(*) from t where date2::date > d.thedate)
from (select distinct thedate
from ((select date1::date as thedate from t) union all
(select date2::date as thedate from t)
) d
) d
This is assuming that date2 occurs after date1. My model is start and stop dates of customers. If this isn't the case, the query might not work.
It sounds like you could benefit from the DATEPART T-SQL method. If I understand you correctly, you could do something like this:
SELECT DATEPART(year, date1) Year, DATEPART(month, date1) Month, SUM(value_col)
FROM $Table
-- WHERE CLAUSE ?
GROUP BY DATEPART(year, date1),
DATEPART(month, date1)
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