How to get the maximum from averages values in MySQL? The following query returns average values of amounts from table orders grouped by customers.
SELECT AVG(amount)
FROM orders
GROUP BY cust;
I want to receive a maximum value from average values using a single query with aggregate functions. Using ORDER BY ... DESC LIMIT 1 surely works, but what I am interested in is getting the maximum average value using aggregate functions solely. Is it possible at all? Thanks
select max(avg_value)
from
(
SELECT AVG(amount) avg_value FROM orders GROUP BY cust
) tmp
I would do this with order by and limit:
SELECT AVG(o.amount) as avg_value
FROM orders o
GROUP BY cust
ORDER BY avg_value DESC
LIMIT 1;
This allows you to get the cust for the maximum as well.
Related
Hi I have a query like this
select avg(TotalSale) as value from Sales group by SalesPerson;
which gives me the output as
value
50.0000
250.0000
62.5000
I want to just get the max out of the values. What should i do?
I tried
select max(avg(TotalSale)) as value from Sales group by SalesPerson; which is wrong.
select avg(TotalSale) as value from Sales group by SalesPerson order by value desc limit 1; which works partially.
What would be better approach to do this?
Wrap your query....
select max(avgvalue)
from
(
select avg(totalsale) as avgvalue
from sales
group by salesperson
)z
Use order by and fetch one row:
select avg(TotalSale)
from Sales
group by SalesPerson
order by avg(TotalSale) desc
limit 1;
You can also get the SalePerson involved with this as well.
I guess it can be done in a single query .Try this :
SELECT
Max(Totalsale) as Value
FROM
Sales
ORDER BY avg(Totalsale)
LIMIT 1;
select orderNumber
, priceEach
, sum(priceEach) over(partition by orderNumber order by priceEach) as `running total`
, sum(priceEach) over(partition by orderNumber) as `total`
from orderdetails;
From above query, why don't I get cumulative sum in total and cumulative some only when I add order by?
from https://www.mysqltutorial.org/mysql-window-functions/
order by clause is "The ORDER BY clause specifies how the rows are ordered within a partition" Even though rows are not ordered why does that stop us from getting cumulative sum?
I want to find out the company with the maximum capital[price*quantity] (Including all entries for that company in the whole table). Any suggestions?
I've done this:
SELECT symbol, SUM(amount*price) AS total
FROM orders
GROUP BY symbol
If FETCH FIRST WITH TIES is supported:
SELECT symbol, SUM(amount*price) AS total
FROM orders
GROUP BY symbol
ORDER BY total DESC
FETCH FIRST 1 ROW WITH TIES
If not:
SELECT symbol, SUM(amount*price) AS total
FROM orders
GROUP BY symbol
HAVING SUM(amount*price) = (SELECT SUM(amount*price) AS total
FROM orders
GROUP BY symbol
ORDER BY total DESC
LIMIT 1)
Your query should look like this.
SELECT symbol,SUM(amount*price) AS total FROM orders GROUP BY symbol order by total desc limit 1;
I am adding limit 1 as you want to find out the company with maximum capital.
I'm trying to make a query with max and count like this one: (taken from http://www.w3resource.com/sql/aggregate-functions/max-count.php)
SELECT MAX (mycount)
FROM (SELECT agent_code,COUNT(agent_code) mycount
FROM orders
GROUP BY agent_code);
this query returns a column with the name 'MAX(MYCOUNT)' with the max value: '7',the simple change I want is that I would like to get the agent code of the one who got the maximum, instead of the max records of agent code.
tried to do this in some ways but no luck so far,
hope you can help me to do this right.
If you don't have to worry about a tie for two agents having the max number of orders, then you can try the following:
SELECT agent_code, COUNT(*) AS mycount
FROM orders
GROUP BY agent_code
ORDER BY COUNT(*) DESC
LIMIT 1
If you do have to worry about a tie for the max number of orders, and you want all ties, then you can use a subquery:
SELECT agent_code, COUNT(*) AS mycount
FROM orders
GROUP BY agent_code
HAVING COUNT(*) = (SELECT MAX(t.mycount) FROM
(SELECT COUNT(*) AS mycount FROM orders GROUP BY agent_code) t)
You can use order by and limit:
SELECT agent_code, COUNT(agent_code) as mycount
FROM orders
GROUP BY agent_code
ORDER BY mycount DESC
LIMIT 1;
I have table_schedule(id,instructorNumber,time,date,cpNumber,user_id);
Is it possible to output the values of instructorNumber,time, etc. with the same highest occurence with same values?
sorry I am new in sql, so I am hoping that someone can help me to the query
Just group by all the fields you need for "same value comparison", order by count desc (so the result with most occurences will be first), and take first.
select
instructorNumber, time, date, cpNumber
from table_schedule
group by instructorNumber, time, date, cpNumber
order by count(*) desc
LIMIT 1
you may use this as a join on a main query if you want more than one result.
First group by the values you want to compare by and count. Get the maximum count (in MySQL you can use LIMIT for that). Then do the same group-by query again and take only the results HAVING the maximum count. Use GROUP_CONCAT to get a list of IDs in a string.
select instructorNumber, time, date, cpNumber, user_id, group_concat(id) as ids
from table_schedule
group by instructorNumber, time, date, cpNumber, user_id
having count(*) =
(
select count(*)
from table_schedule
group by instructorNumber, time, date, cpNumber, user_id
order by count(*) desc limit 1
);