Given this scenario , I have the following employment records
id | user_id | Month | Active
1 | 1 | June 2014 | true
2 | 1 | September 2014 | false
3 | 2 | June 2014 | true
How can i make a query to return the total active users for each month, the return data should be:
active_count | month
2 | June 2014
2 | July 2014
2 | August 2014
1 | September 2014
Any help is highly appreciated
You are looking for a conditional aggregate:
SELECT count(case when active then 1 end) as active_count,
month
FROM employment
GROUP BY month;
With Postgres 9.4 this can be written a bit more concise using the filter() operator:
SELECT count(*) filter (where active) as active_count,
month
FROM employment
GROUP BY month;
It is sql query try to this
SELECT
count(id) active_count,
month
FROM
employment
GROUP BY
month;
Related
This is TSQL and I'm trying to calculate repeat purchase rate for last 12 months. This is achieved by looking at sum of customers who have bought more than 1 time last 12 months and the total number of customers last 12 months.
The SQL code below will give me just that; but i would like to dynamically do this for the last 12 months. This is the part where i'm stuck and not should how to best achieve this.
Each month should include data going back 12 months. I.e. June should hold data between June 2018 and June 2018, May should hold data from May 2018 till May 2019.
[Order Date] is a normal datefield (yyyy-mm-dd hh:mm:ss)
DECLARE #startdate1 DATETIME
DECLARE #enddate1 DATETIME
SET #enddate1 = DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE())-1, 0) -- Starting June 2018
SET #startdate1 = DATEADD(mm,DATEDIFF(mm,0,GETDATE())-13,0) -- Ending June 2019
;
with dataset as (
select [Phone No_] as who_identifier,
count(distinct([Order No_])) as mycount
from [MyCompany$Sales Invoice Header]
where [Order Date] between #startdate1 and #enddate1
group by [Phone No_]
),
frequentbuyers as (
select who_identifier, sum(mycount) as frequentbuyerscount
from dataset
where mycount > 1
group by who_identifier),
allpurchases as (
select who_identifier, sum(mycount) as allpurchasescount
from dataset
group by who_identifier
)
select sum(frequentbuyerscount) as frequentbuyercount, (select sum(allpurchasescount) from allpurchases) as allpurchasecount
from frequentbuyers
I'm hoping to achieve end result looking something like this:
...Dec, Jan, Feb, March, April, May, June each month holding both values for frequentbuyercount and allpurchasescount.
Here is the code. I made a little modification for the frequentbuyerscount and allpurchasescount. If you use a sumif like expression you don't need a second cte.
if object_id('tempdb.dbo.#tmpMonths') is not null drop table #tmpMonths
create table #tmpMonths ( MonthID datetime, StartDate datetime, EndDate datetime)
declare #MonthCount int = 12
declare #Month datetime = DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0)
while #MonthCount > 0 begin
insert into #tmpMonths( MonthID, StartDate, EndDate )
select #Month, dateadd(month, -12, #Month), #Month
set #Month = dateadd(month, -1, #Month)
set #MonthCount = #MonthCount - 1
end
;with dataset as (
select m.MonthID as MonthID, [Phone No_] as who_identifier,
count(distinct([Order No_])) as mycount
from [MyCompany$Sales Invoice Header]
inner join #tmpMonths m on [Order Date] between m.StartDate and m.EndDate
group by m.MonthID, [Phone No_]
),
buyers as (
select MonthID, who_identifier
, sum(iif(mycount > 1, mycount, 0)) as frequentbuyerscount --sum only if count > 1
, sum(mycount) as allpurchasescount
from dataset
group by MonthID, who_identifier
)
select
b.MonthID
, max(tm.StartDate) StartDate, max(tm.EndDate) EndDate
, sum(b.frequentbuyerscount) as frequentbuyercount
, sum(b.allpurchasescount) as allpurchasecount
from buyers b inner join #tmpMonths tm on tm.MonthID = b.MonthID
group by b.MonthID
Be aware, that the code was tested only syntax-wise.
After the test data, this is the result:
MonthID | StartDate | EndDate | frequentbuyercount | allpurchasecount
-----------------------------------------------------------------------------
2018-08-01 | 2017-08-01 | 2018-08-01 | 340 | 3702
2018-09-01 | 2017-09-01 | 2018-09-01 | 340 | 3702
2018-10-01 | 2017-10-01 | 2018-10-01 | 340 | 3702
2018-11-01 | 2017-11-01 | 2018-11-01 | 340 | 3702
2018-12-01 | 2017-12-01 | 2018-12-01 | 340 | 3703
2019-01-01 | 2018-01-01 | 2019-01-01 | 340 | 3703
2019-02-01 | 2018-02-01 | 2019-02-01 | 2 | 8
2019-03-01 | 2018-03-01 | 2019-03-01 | 2 | 3
2019-04-01 | 2018-04-01 | 2019-04-01 | 2 | 3
2019-05-01 | 2018-05-01 | 2019-05-01 | 2 | 3
2019-06-01 | 2018-06-01 | 2019-06-01 | 2 | 3
2019-07-01 | 2018-07-01 | 2019-07-01 | 2 | 3
I have the following table "sales"
date revenue
2018-06-01 300
2018-06-02 400
2019-06-01 500
2019-06-02 700
and I want to compare sales current year vs past year with current date and get the following result:
date revenue 2019 revenue 2018
2019-06-01 500 300
2019-06-02 700 400
What query I should use?
The problem is that I should group 2018 revenue somehow to 2019 revenue.
You can do it with a self join:
select
s1.date,
s1.revenue revenue2019,
s2.revenue revenue2018
from sales s1 inner join sales s2
on s1.date = s2.date + interval '1' year
where date_part('year', s1.date) = 2019
See the demo.
Results:
> date | revenue2019 | revenue2018
> :--------- | ----------: | ----------:
> 2019-06-01 | 500 | 300
> 2019-06-02 | 700 | 400
I need help writing this SQL query (PostgresSQL) to display results in the form below:
--------------------------------------------------------------------------------
State | Jan '17 | Feb '17 | Mar '17 | Apr '17 | May '17 ... Dec '18
--------------------------------------------------------------------------------
Principal Outs. |700,839 |923,000 |953,000 |6532,293 | 789,000 ... 913,212
Disbursal Amount |23,000 |25,000 |23,992 | 23,627 | 25,374 ... 23,209
Interest |113,000 |235,000 |293,992 |322,627 |323,374 ... 267,209
There are multiple tables but I would be okay joining them.
Is there a function that calculates the total count of the complete month like below? I am not sure if postgres. I am looking for the grand total value.
2012-08=# select date_trunc('day', time), count(distinct column) from table_name group by 1 order by 1;
date_trunc | count
---------------------+-------
2012-08-01 00:00:00 | 22
2012-08-02 00:00:00 | 34
2012-08-03 00:00:00 | 25
2012-08-04 00:00:00 | 30
2012-08-05 00:00:00 | 27
2012-08-06 00:00:00 | 31
2012-08-07 00:00:00 | 23
2012-08-08 00:00:00 | 28
2012-08-09 00:00:00 | 28
2012-08-10 00:00:00 | 28
2012-08-11 00:00:00 | 24
2012-08-12 00:00:00 | 36
2012-08-13 00:00:00 | 28
2012-08-14 00:00:00 | 23
2012-08-15 00:00:00 | 23
2012-08-16 00:00:00 | 30
2012-08-17 00:00:00 | 20
2012-08-18 00:00:00 | 30
2012-08-19 00:00:00 | 20
2012-08-20 00:00:00 | 24
2012-08-21 00:00:00 | 20
2012-08-22 00:00:00 | 17
2012-08-23 00:00:00 | 23
2012-08-24 00:00:00 | 25
2012-08-25 00:00:00 | 35
2012-08-26 00:00:00 | 18
2012-08-27 00:00:00 | 16
2012-08-28 00:00:00 | 11
2012-08-29 00:00:00 | 22
2012-08-30 00:00:00 | 26
2012-08-31 00:00:00 | 17
(31 rows)
--------------------------------
Total | 12345
As best I can guess from your question and comments you want sub-totals of the distinct counts by month. You can't do this with group by date_trunc('month',time) because that'll do a count(distinct column) that's distinct across all days.
For this you need a subquery or CTE:
WITH day_counts(day,day_col_count) AS (
select date_trunc('day', time), count(distinct column)
from table_name group by 1
)
SELECT 'Day', day, day_col_count
FROM day_counts
UNION ALL
SELECT 'Month', date_trunc('month', day), sum(day_col_count)
FROM day_counts
GROUP BY 2
ORDER BY 2;
My earlier guess before comments was: Group by month?
select date_trunc('month', time), count(distinct column)
from table_name
group by date_trunc('month', time)
order by time
Or are you trying to include running totals or subtotal lines? For running totals you need to use sum as a window function. Subtotals are just a pain, as SQL doesn't really lend its self to them; you need to UNION two queries then wrap them in an outer ORDER BY.
select
date_trunc('day', time)::text as "date",
count(distinct column) as count
from table_name
group by 1
union
select
'Total',
count(distinct column)
from table_name
group by 1, date_trunc('month', time)
order by "date" = 'Total', 1
I need to use a where clause within an over clause. How?
SELECT SUM(amount) OVER(WHERE dateval > 12)
Or something like that.
--EDIT--
More details
My table is formatted with a year, month, and amount column.
I want to select all the year, month, and amount rows AND create a fourth 'virtual column' that has the sum of the past 12 months of amount column.
For example:
YEAR | MONTH | AMOUNT
2001 | 03 | 10
2001 | 05 | 25
2001 | 07 | 10
Should create:
YEAR | MONTH | AMOUNT | ROLLING 12 MONTHS
2001 | 03 | 10 | 10
2001 | 05 | 25 | 35
2001 | 07 | 10 | 45
Given a query against your three-column resultset, does the below work for you?
SELECT
SUM(amount) OVER(ORDER BY YEAR ASC, MONTH ASC
ROWS BETWEEN 11 PRECEDING AND CURRENT ROW)
...
select a,(select sum(a) from foo fa where fa.a > fb.a) from foo fb;
Doesn't use over, is pretty inefficient since it is running new sub-query for each query, but it works.