Average after group by in SQL query - mysql

select avg(select count(aid)
from athlete
group by codepays)
I get a "more than one row error".
How with I go about getting the average of the result from my fist select ?

You need to use a table expression (subquery).
For example:
select avg(cnt)
from (
select count(aid) as cnt
from athlete
group by codepays
) x

You can do this using division:
select count(aid) / count(distinct codepay)
from athlete;
No subqueries are necessary.
(Although the arithmetic needs to be tweaked if codepay can actually be NULL.)

Related

Two select statements on same table and get Count(*)

Im trying to do two queries on the same table to get the Count(*) value.
I have this
SELECT `a`.`name`, `a`.`points` FROM `rank` AS a WHERE `id` = 1
And in the same query I want to do this
SELECT `b`.`Count(*)` FROM `rank` as b WHERE `b`.`points` >= `a`.`points`
I tried searching but did not find how to do a Count(*) in the same query.
Typically you would not intermingle a non aggregate and aggregate query together in MySQL. You might do this in databases which support analytic functions, such as SQL Server, but not in (the current version of) MySQL. That being said, your second query can be handled using a correlated subquery in the select clause the first query. So you may try the following:
SELECT
a.name,
a.points,
(SELECT COUNT(*) FROM rank b WHERE b.points >= a.points) AS cnt
FROM rank a
WHERE a.id = 1;
As I understand from the question, you want to find out in a table for a given id how many rows have the points greater than this row. This can be achieved using full join.
select count(*) from rank a join rank b on(a.id != b.id) where a.id=1 and b.points >= a.points;

MySQL 5.7 How to do GROUP BY with sorting?

Similar to this issue: MySQL 5.7 group by latest record
I'm not sure how to do this properly in 5.7. Also with possibility of 2nd sort column. Working query in 5.6 that I'm trying to replicate in 5.7:
SELECT id FROM test
GROUP BY category
ORDER BY sort1 DESC, sort2 DESC
id is not always the highest, so MAX(id) does not work.
Looking into the link above, the solution for single sort should be:
SELECT t1.*
FROM test t1
INNER JOIN (
SELECT category, max(sort) AS sort FROM test GROUP BY category
) t2 ON t2.category = t1.category AND t2.sort = t1.sort
But how will it work with 2 sorting?
You are using GROUP BY the wrong way.
Think of group by as a way to separate data row into different groups. Each group has multiple rows, based on the value of group by column.
Once you get those groups, selecting table columns (as in: select *) is like picking any row from that group randomly. This is not helpful nor useful.
Usually once we group records (or rows), we need to find meta information about those records. For example: get us the count of records in that group (as in: select count(*)), or the sum of values of a specific column in that group (as in: select sum(price)), or get the min, max or avg values.
So in a nutshell, when you use group by you should use on of the aggregation functions with it, otherwise it's not going to do you any good.
Why don't you have the ORDER BY at your outer query, instead?
SELECT *
FROM (
SELECT 100 AS id, 1 AS category, NULL AS sort
UNION
SELECT 200 AS id, 1 AS category, 2 AS sort
) dt
GROUP BY category
ORDER BY sort DESC;
It seems that what happened to the data when it was grouped, it took the first data while neglecting the ORDER BY DESC. On your first query, it ordered descending first then group by took the first record which is 200. And yes, this shouldn't be the way you should use GROUP BY. It is used in conjunction with aggregate functions.
when you select a column in a group by query that is not one of the columns you are grouping by, (ie, your id) you have no control over the value unless you use another aggregate function. If you want to sort, use MIN or MAX:
SELECT MAX(id), category, FROM `test2`
GROUP BY category; -- always returns 200
SELECT MIN(id), category, FROM `test2`
GROUP BY category; -- always returns 100

Nested SQL Query

I have problems with the following SQL Query:
SELECT job
FROM (SELECT job, COUNT(*) AS cnt
FROM Employee
GROUP BY job)
WHERE cnt=1
As Result it should only shows all jobs where cnt (count of jobs) equals 1.
When I test the select query above on Fiddle, I get following error :
Incorrect syntax near the keyword 'WHERE'.
SQLFiddle: http://sqlfiddle.com/#!6/d812a/7
No need to increase complexity by using sub-query when it is not require
SELECT job, count(job)
FROM Employee
GROUP BY job
having count(job)=1;
You need to provide alias name to the nested query
SELECT A.job
FROM (SELECT job, COUNT(*) AS cnt
FROM Employee
GROUP BY job)A
WHERE A.cnt=1
You forget to add the alias name.
Please change the query like this
SELECT job
FROM (SELECT job, COUNT(*) AS cnt
FROM Employee
GROUP BY job) As MyTable
WHERE cnt=1
You should have to give the alias name for the inner query when you are using the select and where clauses outside.
You should use the HAVING syntax :
SELECT job, COUNT(*) AS cnt
FROM Employee
GROUP BY job
HAVING cnt = 1;
You should use the HAVING clause which is done for that kind of thing. Your request will be simply :
SELECT job FROM Employee GROUP BY job
HAVING COUNT(id)=1
The documentation states that
The SQL standard requires that HAVING must reference only columns in
the GROUP BY clause or columns used in aggregate functions. However,
MySQL supports an extension to this behavior, and permits HAVING to
refer to columns in the SELECT list and columns in outer subqueries as
well.
The important thing to note is that contrary to the WHERE clause, you can use aggregate funcitons (like count, max, min ...) in the HAVING clause.

COUNT(*) returning multiple rows instead of just one

Why does COUNT() return multiple rows when I just need the total count of how many rows my query generates?
Should return 1078.
The COUNT() is working as expected. When you put a group by clause, the count() gives you the result for GROUP BY. If you wish to get the count of rows in a query that includes group by, use it as a subquery instead.
Something like:
SELECT COUNT(*) FROM (SELECT * FROM `table`
GROUP BY `column1`) AS `a`
Well, simple answer. Don't GROUP BY if you don't need groups.
Either use COUNT(DISTINCT articles.company) without GROUP BY or keep the the GROUP BY and wrap the whole query in a SELECT COUNT(*) FROM (...) AS data, if you want to count the groups.
Do not use group by, it will count the number of each group.
Write the above query as subquery then it will give proper result and dont use group by
select count(*) from (select articles.id from 'contract_prices' left join 'articles' on
(arcticles.id = contract_prices.article)
where 'contract_to' >= curdate()
)

Is efficient the following SQL query?

Exist a better way to do what the following SQL query does? I have the feeling that table1 will be searched twice and may be that can be avoided with some trick and increase the efficient of the query, but I just can't figure out how ;( Here is the query (in MySQL):
SELECT a, SUM(count)
FROM table1
GROUP BY a
HAVING SUM(count) = (SELECT SUM(count) as total FROM table1 GROUP BY a ORDER BY total DESC LIMIT 1)
The goal is return the number(s) with the major accumulate, with its accumulate.
being table1 a two field table like:
a,count
1,10
1,30
1,0
2,1
2,100
2,4
3,10
4,50
4,55
The result with that data sample is:
2,105
4,105
Thanks in advance.
SELECT a, total FROM
(SELECT a AS a, SUM(COUNT) AS total
FROM table1
GROUP BY a) AS xyz
HAVING total = MAX(total)
Hope this will work for you
This sub-query is executed only once, and you don't have to bother with creating any pre-query as other answers may suggest (although doing so this is still correct, just not needed). Database engine will realise, that the sub-query is not using any variable dependent on the other part of the query. You can use EXPLAIN to see how the query is executed.
More on the topic in this answer:
https://stackoverflow.com/a/658954/1821029
I think you could probably do it by moving your HAVING sub-select query into its on prequery. Since it will always include a single row, you won't require any "JOIN", and it does not have to keep recomputing the COUNT(*) every time the HAVING is applied. Do it once, then the rest
SELECT
a,
SUM(count)
FROM
table1,
( SELECT SUM(count) as total
FROM table1
GROUP BY a
ORDER BY total DESC
LIMIT 1 ) PreQuery
GROUP BY
a
HAVING
SUM(count) = PreQuery.Total
This query return one row with two columns:
1- a list of comma separated values of "a" column, which have the biggest "Total"
2- and the biggest Total value
select group_concat(a), Total
from
(select a, sum(count) as Total
from table1
group by a) OnTableQuery
group by Total
order by Total desc
limit 1
Note that it queries table1 just one time. The query was already tested.