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
Related
I'm curious as to find the daily average sales for the month of December 1998 not greater than 100 as a where clause. So what I imagine is that since the table consists of the date of sales (sth like 1 december 1998, consisting of different date, months and year), amount due....First I'm going to define a particular month.
DEFINE a = TO_DATE('1-Dec-1998', 'DD-Month-YYYY')
SELECT SUBSTR(Sales_Date, 4,6), (SUM(Amount_Due)/EXTRACT(DAY FROM LAST_DAY(Sales_Date))
FROM ......
WHERE SUM(AMOUNT_DUE)/EXTRACT(DAY FROM LAST_DAY(&a)) < 100
I'm stuck as to extract the sum of amount due in the month of december 1998 for the where clause....
How can I achieve the objective?
To me, it looks like this:
select to_char(sales_date, 'mm.yyyy') month,
avg(amount_due) avg_value
from your_table
where sales_date >= trunc(date '1998-12-01', 'mm')
and sales_date < add_months(trunc(date '1998-12-01', 'mm'), 1)
group by to_char(sales_date, 'mm.yyyy')
having avg(amount_due) < 100;
WHERE clause can be simplified; it shows how to fetch certain period:
trunc to mm returns first day in that month
add_months to the above value (first day in that month) will return first day of the next month
the bottom line: give me all rows whose sales_date is >= first day of this month and < first day of the next month; basically, the whole this month
Finally, the where clause you used should actually be the having clause.
As long as the amount_due column only contains numbers, you can use the sum function.
Below SQL query should be able to satisfy your requirement.
Select SUM(Amount_Due) from table Sales where Sales_Date between '1-12-1998' and '31-12-1998'
OR
Select SUM(Amount_Due) from table Sales where Sales_Date like '%-12-1998'
I want to know the trick to find the list of customers who are transacting for consecutive 3 months ,that could be any 3 consecutive months with any number of occurrence.
example: suppose there is customer who transact in January then keep transacting till march then he stopped transacting.I want the list of these customer from my database .
I am working on AWS Athena.
One method uses aggregation and window functions:
select customer_id, yyyymm_2
from (select date_trunc(month, transactdate) as yyyymm, customer_id,
lag(date_trunc(month, transactdate), 2) over (partition by customer_id order by date_trunc(month, transactdate)) as prev_yyyymm_2
from t
where transactdate >= '2017-01-01' and
transactadte < '2019-01-01'
)
where prev_dt_2 = yyyymm - interval '2' month;
This aggregates transactions by month and looks at the transaction date two rows earlier. The outer filter checks that that date is exactly 2 months earlier.
I need to calculate an average based on historical data for a graph in SSRS:
Current Month
Previous Month
2 Months ago
6 Months ago
This query returns the average for each month:
SELECT
avg_val1, month, year
FROM
(SELECT
(sum_val1 / count) as avg_val1, month, year
FROM
(SELECT
SUM(val1) AS sum_val1, SUM(count) AS count, month, year
FROM
(SELECT
COUNT(val1) AS count, SUM(val1) AS val1,
MONTH([SnapshotDate]) AS month,
YEAR([SnapshotDate]) AS year
FROM
[DC].[dbo].[KPI_Values]
WHERE
[SnapshotKey] = 'Some text here'
AND No = '001'
AND Channel = '999'
GROUP BY
[SnapshotDate]) AS sub3
GROUP BY
month, year, count) AS sub2
GROUP BY sum_val1, count, month, year) AS sub1
ORDER BY
year, month ASC
When I add the following WHERE clause I get the average for March (2 months ago):
WHERE month = MONTH(GETDATE())-2
AND year = YEAR(GETDATE())
Now the problem is when I want to retrieve data from 6 months ago; MONTH(GETDATE()) - 6 will output -1 instead of 12. I also have an issue with the fact that the year changes to 2016 and I am a bit unsure of how to implement the logic in my query.
I think I might be going about this wrong... Any suggestions?
Subtract the months from the date using the DATEADD function before you do your comparison. Ex:
WHERE SnapshotDate BETWEEN DATEADD(month, -6, GETDATE()) AND GETDATE()
MONTH(GETDATE()) returns an int so you can go to 0 or negative values. you need a user scalar function managing this, adding 12 when <= 0
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)
;
My table structure is this:
ID,
country,
month,
year,
total amt in previous period,
total amt during period,
incr/decr in total amt in previous period,
incr/decr in total amt during (month, year)
The ID, month, year and total amt fields are available in table abc.
The incr/decr in total amt in previous period is the difference between total amt in previous period and total amt during period columns.
I wrote this query:
select m.id, m.month, m.year, m.total_amt
from abc m
order by year, month desc;
For the total amt in previous period I could not use Between Date( ) And DateAdd("M", -1, Date( )); as I have no date but just year and month.
How to compare the two columns with the columns year and month and how to have the last two derived columns using subqueries?
For comparing Month there is an specific function (that works for MS SQL and ORACLE). There's one function for days and years too. See links below:
YEAR
MONTH
DAY
Examples of this and getting derived columns from subqueries can be found in this topic already discussed in the forum:
Stackoverflow topic