I have many users and Date time column. I would like to know the min and max values of each users for each year and each month?
name date Income Expense
Vijay 12-10-2017 10 8
Vijay 16-04-2017 25 12
year(date) as Y_year,
month(date) as M_Month,
I tried the code below but no use either
min(Income)over( PARTITION by (name, Y_year,M_Month)) as min_income_of_month,
Max(Expense)over( PARTITION by (name, Y_year,M_Month)) as Max_Expense_of__month
Is this what you want?
select name, year(date) as Y_year, month(date) as M_Month,
min(income), max(income), min(expense), max(expense)
from t
group by name, year(date), month(date)
order by name, Y_year, M_Month;
Just use min and max function on the columns you want to aggregate and group by user and the month of the date
No need to use window functions
Related
i am trying to work with this query to produce a list of all 11 years and 12 months within the years with the sales data for each month. Any suggestions? this is my query so far.
SELECT
distinct(extract(year from date)) as year
, sum(sale_dollars) as year_sales
from `project-1-349215.Dataset.sales`
group by date
it just creates a long list of over 2000 results when i am expecting 132 max one for each month in the years.
You should change your group by statement if you have more results than you expected.
You can try:
group by YEAR(date), MONTH(date)
or
group by EXTRACT(YEAR_MONTH FROM date)
A Grouping function is for takes a subsection of the date in your case year and moth and collect all rows that fit, and sum it up,
So a s´GROUp BY date makes no sense, what so ever as you don't want the sum of every day
So make this
SELECT
extract(year from date) as year
,extract(MONTH from date) as month
, sum(sale_dollars) as year_sales
from `project-1-349215.Dataset.sales`
group by 1,2
Or you can combine both year and month
SELECT
extract(YEAR_MONTH from date) as year
, sum(sale_dollars) as year_sales
from `project-1-349215.Dataset.sales`
group by 1
This is what my table looks like:
NOTE: Don't worry about the BMI field being empty in some rows. We assume that each row is a reading. I have omitted some columns for privacy reasons.
I want to get a count of the number of active customers per month. A customer is active if they have at least 18 readings in total (1 reading per day for 18 days in a given month). How do I write this SQL query? Assume the table name is 'cust'. I'm using SQL Server. Any help is appreciated.
Presumably a patient is a customer in your world. If so, you can use two levels of aggregation:
select yyyy, mm, count(*)
from (select year(createdat) as yyyy, month(createdat) as mm,
patient_id,
count(distinct convert(date, createdat)) as num_days
from t
group by year(createdat), month(createdat), patient_id
) ymp
where num_days >= 18
group by yyyy, mm;
You need to group by patient and the month, then group again by just the month
SELECT
mth,
COUNT(*) NumPatients
FROM (
SELECT
EOMONTH(c.createdat) mth
FROM cust c
GROUP BY EOMONTH(c.createdat), c.patient_id
HAVING COUNT(*) >= 18
-- for distinct days you could change it to:
-- HAVING COUNT(DISTINCT CAST(c.createdat AS date)) >= 18
) c
GROUP BY mth;
I just have to omit those records whose sum of sales in all 53 weeks is 0 and would need the output without group by
You cannnot really get that in one query.
To get all years without any sum of sales, you have to sum the sales.
That is:
Firstly:
select YEAR(date) from YourTable group by YEAR(date) having sum(sales) > 0
Then:
select * from YourTable where Year in (<firstquery>) as aliasname
order by <anydatecolumn>
If you are using mssql you can do that in one query using the OVER clause and partitioning
I am trying to write an SQL statement based on the following code.
CREATE TABLE mytable (
year INTEGER,
month INTEGER,
day INTEGER,
hoursWorked INTEGER )
Assuming that each employee works multiple days over each month in a 3 year period.
I need to write an sql statement that returns the total hours worked in each month, grouped by earliest year/month first.
I tried doing this, but I don't think it is correct:
SELECT Sum(hoursWorked) FROM mytable
ORDER BY(year,month)
GROUP BY(month);
I am a little confused about how to operate the sum function in conjunction with thee GROUP BY or ORDER BY function. How does one go about doing this?
Try this:
SELECT year, month, SUM(hoursWorked)
FROM mytable
GROUP BY year, month
ORDER BY year, month
This way you will have for example:
2014 December 30
2015 January 12
2015 February 40
Fields you want to group by always have be present in SELECT part of query. And vice-versa - what you put in SELECT part, need be also in GROUP BY.
SELECT year, month, Sum(hoursWorked)as workedhours
FROM mytable
GROUP BY year,month
ORDER BY year,month;
You have to group by year and month.
Is this what you are trying to do. This will sum by Year/Month and Order by Year/Month.
Select [Year], [Month], Sum(HoursWorked) as WorkedHours
From mytable
Group By [Year], [Month]
Order by [Year], [Month]
You have to group by year and month, otherwise you will have the hours you worked on March 2014 and 2015 in one record :)
SELECT Sum(hoursWorked) as hoursWorked, year, month
FROM mytable
GROUP BY(year, month)
ORDER BY(year,month)
;
I have a table called values that contains 3 columns - Location, value, date
I want to work out the average value per month
so far I have
SELECT Location, Avg(value), date
FROM Value
GROUP BY Location, date
This returns the average values but a value is entered on a daily basis so I have an average per day rather than per month, how can I achieve a monthly average?
try this:
SELECT Location,
Avg(value),
month(date),
year(date)
FROM Value
GROUP BY Location,
month(date),
year(date)
You can use the following, if you want month only grouping:
SELECT Location, Avg(value) AvgVal, Month(date) Mnth
FROM Value
GROUP BY Location, Month(date)
You can even use GROUPING SETS, which will GROUP BY Month, year, location and then give you a total for all:
SELECT Location,
Avg(value) AvgVal,
Month(dt) Mnth,
Year(dt) Yr
FROM yourtable
GROUP BY
GROUPING SETS((Month(dt), Year(dt), Location), (Location));
See SQL Fiddle with Demo
SELECT
Location,
year(date),
month(date),
Avg(value)
FROM
Value
GROUP BY
Location,
year(date),
month(date)
Can also do it as the following if you want to combine the dates into one column with the first of the month as the day.
SELECT Location,
Avg(value),
DateFromParts(Year(date), Month(date) , 1) AS FirstOfMonthDate
FROM Value
GROUP BY Location,
DateFromParts(Year(date), Month(date) , 1)