I'm not sure if this has been asked before, as I don't know how to best phrase this question.
Given a query like:
SELECT DATE_FORMAT(i.created, '%Y-%m') as 'period',
COUNT(id) as 'total',
i.company_id
FROM invoice i
GROUP BY period, i.company_id
ORDER BY period DESC, total DESC
How can I return the average and/or mean count per month, grouped by company_id? It is important to only count those periods where there actually are any invoices.
If you want to exclude zero months then add HAVING condition and then select AVG() for each company using your query as a base:
SELECT company_id, AVG(total)
FROM
(SELECT DATE_FORMAT(i.created, '%Y-%m') as 'period',
COUNT(id) as 'total',
i.company_id
FROM invoice i
GROUP BY period, i.company_id
HAVING COUNT(id)>0
) as T1
GROUP BY company_id
Related
If i have a database with 2 columns, date and account and i want to first count account per day and then group by week. How wrong is my code and how to do it?
I edited my code a little bit, i was not thinking right from the beginning. I want the sum to be 9 for week 48.
SELECT date, account,
(SELECT date, COUNT(DISTINCT account)
FROM t1
GROUP BY date
) AS sum
FROM t1
GROUP BY YEARWEEK(date)
You seem to be looking for a simple aggregate query with count(distinct ...):
select yearweek(date) year_week, count(distinct account) cnt_account
from t1
group by yearweek(date)
order by year_week
Note: yearweek() gives you the year and week; this is better than week(), if your data spreads over several years.
EDIT
From the comments, you need two levels of aggregation:
select yearweek(dy) year_week, sum(cnt) cnt_account
from (
select date(t1.date) dy, count(distinct t1.account) cnt
from t1
group by date(t1.date)
) t
group by yearweek(dy)
order by year_week
I want to get a statistic for every month of the years i have in DB
SELECT monthname(created_at) AS month, YEAR(created_at) AS year, count(*) AS number
FROM tableName
WHERE type_of_user = "someType"
GROUP BY year, month(created_at)
ORDER BY created_at DESC
Now it gives me only month that I have, but I need to get statistics for every month, even if I don't have any stored data for that month
Create a calendar table. This will need one entry per month, for every year that you intend to use.
Then select from the calendar table, and join in the values that you get from your current query. Use COALESCE() to put a zero-value where the entry is NULL (e.g. when there are no records in the tableName for that month and year).
SELECT MONTHNAME(date) as month,
YEAR(date) as year,
COALESCE(number, 0) as number
FROM calendar AS C
LEFT JOIN (
SELECT created_at, COUNT(*) as number
FROM tableName AS T
WHERE T.type_of_user = 'someType'
GROUP BY YEAR(created_at), MONTH(created_at)
) AS T
ON MONTH(T.created_at) = MONTH(C.date) AND YEAR(T.created_at) = YEAR(C.date)
GROUP BY month, YEAR(created_at)
ORDER BY MONTH(date), YEAR(date)
SQL fiddle at http://sqlfiddle.com/#!9/e0a4dc/
SELECT month(created_at) as month
FROM tableName
RIGHT JOIN (select row_number() over (order by 1) as i
from someTableWithMoreThan12Records limit 12) x
ON x.i=month(created_at)
ORDER BY I;
JOINing with a table that has all the records will give you every month.
Currently i am able to get the sum for the highest amount for each month of the year. But what i want to do, is to be able to get the SUM of the month that has the highest value in amount for each year.
SELECT year(paymentDate), month(paymentDate) , SUM(amount)
FROM classicmodels.payments
GROUP BY year(paymentDate), month(paymentDate)
ORDER BY SUM(amount) DESC;
This orders the highest SUM(amount) in descending order but i only want to get the highest month for each year. there are only 3 years in my database.
Here's what happening on mysql workbench
One method uses a having clause:
SELECT year(p.paymentDate), month(p.paymentDate), SUM(p.amount)
FROM classicmodels.payments p
GROUP BY year(p.paymentDate), month(p.paymentDate)
HAVING SUM(p.amount) = (SELECT SUM(p2.amount)
FROM classicmodels.payments p2
WHERE year(p2.paymentDate) = year(p.paymentDate)
GROUP BY month(p2.paymentDate)
ORDER BY SUM(p2.amount) DESC
LIMIT 1
)
ORDER BY SUM(amount) DESC;
I need a mysql query to find Which week of month is the most expensive week (week_of_month) and how much in this week
is spent (WOM_AMT) than other weeks of month each year against each supplier
This query will give you max spent per week:
SELECT supplier, year, week_of_month, MAX(wom_amt)
FROM table t
GROUP BY supplier, year, week_of_month;
Now, in order to compare this amount against the SUM of the rest, you can wrap this query into outer query and calculate the SUM, e.g.:
SELECT a.supplier, a.year, a.week_of_month, a.wom_amt as 'max_amount',
(SELECT SUM(wom_amt) FROM table WHERE supplier = a.supplier AND year = a.year
AND week_of_month = a.week_of_month AND wom_amt != a.wom_amt) as 'other_amounts'
FROM (SELECT supplier, year, week_of_month, MAX(wom_amt)
FROM table t
GROUP BY supplier, year, week_of_month) a;
I think that what you are looking for is MAX and GROUP BY.
Something like this should work (not tested):
SELECT
`Supplier`,
`Year`,
`Week_Of_Month`,
MAX(`WOM_AMT`) AS WOM_AMT
FROM `table`
GROUP BY `Supplier`, `Year`
I have a sales table from which I select the total sales per month , highest sale , number of sale for all the months in the current year, using
select monthname(date),sum(amt_c),MAX(amt_c)
from sales where year(date)= year(now())
group by monthname(date) ;
I want to also select the customer who has done the highest purchase , i.e the customer correponding to the MAX(amt_c).
amt_c is the purchase done by the customer,
One way is a filtering join:
select filter.mn
, filter.sum_sales
, filter.max_sales
, sales.cust
from (
select monthname(date) as mn
, sum(amt_c) as sum_sales
, max(amt_c) as max_sales
from sales
where year(date) = year(now())
group by
mn
) filter
join sales
on monthname(sales.date) = filter.mn
and sales.amt_c = filter.max_sales
For more approaches, browse the greatest-n-per-group tag.
select v.monthname,
v.sum_amt_c,
v.max_amt_c,
count(s.amt_c) as num_of_amounts,
group_concat(s.cust) as customers
from (select monthname(date) as monthname,
sum(amt_c) as sum_amt_c,
max(amt_c) as max_amt_c
from sales
where date between concat(year(now()), '-01-01') and concat(year(now()), '-12-31')
group by monthname(date)) v
join sales s
on v.max_amt_c = s.amt_c
and v.monthname = monthname(s.date)
and s.date between concat(year(now()), '-01-01') and concat(year(now()), '-12-31')
group by v.monthname, v.sum_amt_c, v.max_amt_c
order by month(s.date)
This is similar to Andomar's answer however it provides the following benefits:
If your DATE field is indexed (it should be) the above query will use that index. You should not have criteria on a date field with a function applied to it. MySQL does not support function based indexes, so it is a given that year(date) is not indexed. date may be indexed, however.
This sorts the results by month # (1-12) but shows the month name.
In the event that the same 2+ customers are tied, this will list all of them, and show only one row for that month. You would otherwise potentially have 2, 3, 4+ rows for a single month in the event of a tie. This is done via MySQL's GROUP_CONCAT function.