I have a table which gives the no of rides by a rider at each stand point. I need to find the stand for each rider for which he has the maximum rides.
My first result is in this format: 1
I require my final result like this: 2
I'm currently using this query, but I know it can be done in a better manner. Any suggestions would be helpful.
select c.rider_id, c.end_stand, b.max_rides
from
(select rider_id, max(rides) as max_rides
from
(select rider_id, end_stand, count(id) as rides
from ride where end_stand is not null
group by 1,2) a
group by 1
order by 2 desc, 1) b
join
(select rider_id, end_stand, count(id) as rides
from ride where end_stand is not null
group by 1,2) c
on c.rider_id = b.rider_id and c.rides = b.max_rides
order by 3 desc, 2,1
Before window functions, one method is a correlated subquery in the having clause:
select rider_id, end_stand, count(*) as rides
from ride r
where end_stand is not null
group by rider_id, end_stand
having count(*) = (select count(*)
from ride r2
where r2.end_stand is not null and
r2.rider_id = r.rider_id
group by r2.rider_id, r2.end_stand
order by count(*) desc
limit 1
);
With window functions, this is, of course, much simpler:
select *
from (select rider_id, end_stand, count(*) as rides
rank() over (partition by rider_id order by count(*) desc) as seqnum
from ride r
where end_stand is not null
group by rider_id, end_stand
) r
where seqnum = 1;
Both these will return duplicates, if there are ties for the max. The second version is easy to fix, if you want only one row: use row_number() instead of rank().
Related
I'm attempting to create an SQL query that retrieves the total_cost for every row in a table. Alongside that, I also need to collect the most dominant value for both columnA and columnB, with their respective values.
For example, with the following table contents:
cost
columnA
columnB
target
250
Foo
Bar
XYZ
200
Foo
Bar
XYZ
150
Bar
Bar
ABC
250
Foo
Bar
ABC
The result would need to be:
total_cost
columnA_dominant
columnB_dominant
columnA_value
columnB_value
850
Foo
Bar
250
400
Now I can get as far as calculating the total cost - that's no issue. I can also get the most dominant value for columnA using this answer. But after this, I'm not sure how to also get the dominant value for columnB and the values too.
This is my current SQL:
SELECT
SUM(`cost`) AS `total_cost`,
COUNT(`columnA`) AS `columnA_dominant`
FROM `table`
GROUP BY `columnA_dominant`
ORDER BY `columnA_dominant` DESC
WHERE `target` = "ABC"
UPDATE: Thanks to #Barmar for the idea of using a subquery, I managed to get the dominant values for columnA and columnB:
SELECT
-- Retrieve total cost.
SUM(`cost`) AS `total_cost`,
-- Get dominant values.
(
SELECT `columnA`
FROM `table`
GROUP BY `columnA`
ORDER BY COUNT(*) DESC
LIMIT 1
) AS `columnA_dominant`,
(
SELECT `columnB`
FROM `table`
GROUP BY `columnB`
ORDER BY COUNT(*) DESC
LIMIT 1
) AS `columnB_dominant`
FROM `table`
WHERE `target` = "XYZ"
However, I'm still having issues figuring out how to calculate the respective values.
You might get close, if we want to get percentage values we can try to add COUNT(*) at subquery to get max count by columnA and columnB then do division by total count
SELECT
SUM(cost),
(
SELECT tt.columnA
FROM T tt
GROUP BY tt.columnA
ORDER BY COUNT(*) DESC
LIMIT 1
) AS columnA_dominant,
(
SELECT tt.columnB
FROM T tt
GROUP BY tt.columnB
ORDER BY COUNT(*) DESC
LIMIT 1
) AS columnB_dominant,
(
SELECT COUNT(*)
FROM T tt
GROUP BY tt.columnA
ORDER BY COUNT(*) DESC
LIMIT 1
) / COUNT(*) AS columnA_percentage,
(
SELECT COUNT(*)
FROM T tt
GROUP BY tt.columnB
ORDER BY COUNT(*) DESC
LIMIT 1
) / COUNT(*) AS columnB_percentage
FROM T t1
If your MySQL version supports the window function, there is another way which reduce table scan might get better performance than a correlated subquery
SELECT SUM(cost) OVER(),
FIRST_VALUE(columnA) OVER (ORDER BY counter1 DESC) columnA_dominant,
FIRST_VALUE(columnB) OVER (ORDER BY counter2 DESC) columnB_dominant,
FIRST_VALUE(counter1) OVER (ORDER BY counter1 DESC) / COUNT(*) OVER() columnA_percentage,
FIRST_VALUE(counter2) OVER (ORDER BY counter2 DESC) / COUNT(*) OVER() columnB_percentage
FROM (
SELECT *,
COUNT(*) OVER (PARTITION BY columnA) counter1,
COUNT(*) OVER (PARTITION BY columnB) counter2
FROM T
) t1
LIMIT 1
sqlfiddle
try this query
select sum(cost) as total_cost,p.columnA,q.columnB,p.columnA_percentage,q.columnB_percentage
from get_common,(
select top 1 columnA,columnA_percentage
from(
select columnA,count(columnA) as count_columnA,cast(count(columnA) as float)/(select count(columnA) from get_common) as columnA_percentage
from get_common
group by columnA)s
order by count_columnA desc
)p,
(select top 1 columnB,columnB_percentage
from (
select columnB,count(columnB) as count_columnB, cast(count(columnB) as float)/(select count(columnB) from get_common) as columnB_percentage
from get_common
group by columnB) t
order by count_columnB desc)q
group by p.columnA,q.columnB,p.columnA_percentage,q.columnB_percentage
so if you want to get the percent and dominant value you must make their own query like this
select top 1 columnA,columnA_percentage
from(
select columnA,count(columnA) as count_columnA,cast(count(columnA) as float)/(select count(columnA) from get_common) as columnA_percentage
from get_common
group by columnA)s
order by count_columnA desc
then you can join with the sum query to get all value you want
hope this can help you
I need to get my data set as this table
I am trying to get eligible set like this, need to group_concat pinged set also
x.id IN (SELECT MAX(x.id) FROM x WHERE ping rider id IS NULL GROUP BY orderId)
You can assign a group based on the cumulative number of non-null values in eligible_riders. Then aggregate and take the last value:
select og.*
from (select order_id, grp, max(eligible_riders) as eligible_riders,
group_concat(rider_id) as riders,
row_number() over (partition by order_id order by min(id) desc) as seqnum
from (select t.*,
sum(eligible_riders <> '') over (partition by order_id order by id) as grp
from t
) t
group by order_id, grp
) og
where seqnum = 1;
Hmmm . . . You could also do this with a correlated subquery, which might look a bit simpler:
select order_id, max(eligible_riders) as eligible_riders,
group_concat(rider_id) as riders
from t
where t.id >= (select max(t2.id)
from t t2
where t2.order_id = t.order_id and
t2.eligible_riders <> ''
)
group by order_id;
For performance, you want an index on (order_id, eligible_riders).
I have a table with name rating as this.
I am writing the query like this
SELECT user_id, sum(score) as score
FROM quiz_rashad.rating
group by user_id
order by score desc
then how I get rating index of the 12th user?
Thank you for helping.
Assuming, that "the 12th user" means the user with the ID 12:
In MySQL 8.0+ you can use dense_rank().
SELECT x.rating_index
FROM (SELECT r.user_id,
dense_rank() OVER (ORDER BY sum(r.score) DESC) rating_index
FROM quiz_rashad.rating r
GROUP BY r.user_id) x
WHERE x.user_id = 12;
Edit:
For MySQL 5.7 you have to use subqueries getting the distinct count of total scores greater than or equal the total score for the user with ID 12.
SELECT count(DISTINCT x.score) rating_index
FROM (SELECT r.user_id,
sum(r.score) score
FROM quiz_rashad.rating r
GROUP BY r.user_id) x
WHERE x.score >= (SELECT sum(r.score)
FROM quiz_rashad.rating r
WHERE r.user_id = 12)
We can try using LIMIT with OFFSET here:
SELECT user_id, SUM(score) AS score
FROM quiz_rashad.rating
GROUP BY user_id
ORDER BY score DESC
LIMIT 1 OFFSET 11;
This answer assumes that what you really want here is the record with the twelfth rank. It also assumes that no two users would be tied for the same score.
I'm working with score ranking on my app for all user score. My problem is I don't know how to return one row for each stud_num.
My query:
SELECT * FROM score WHERE assess_type = 'professional' ORDER BY total_score DESC.
Result:
As you can see I have 3 stud_num and I only want one row per stud_num and the highest score of it.
You can use correlated query like this:
SELECT * FROM score t
WHERE t.assess_type = 'professional'
AND t.total_score = (select max(s.total_score)
from score s
where t.stud_num = s.stud_num)
group by stud_num
The option given by #sagi is good:
SELECT * FROM score t
WHERE t.assess_type = 'professional'
AND t.total_score = (select max(s.total_score)
from score s
where t.stud_num = s.stud_num)
group by stud_num
Another option would be to use an inner join and group by together.
The resulting query would become:
select * from score a
inner join (
SELECT stud_num, max(total_score) tscore FROM `score` group by stud_num) b
on a.stud_num = b.stud_num and total_score= tscore
group by a.stud_num
try it out at sqlfiddle
Use the MAX and GROUP BY functions like this:
SELECT score_id, stud_num, assess_type, total_item, MAX(total_score), average, date_taken
FROM score
WHERE assess_type = 'professional'
GROUP BY stud_num
ORDER BY 5 DESC
Here's my the ans:
SELECT score_id, stud_num, assess_type, total_item, MAX( total_score )
FROM score
WHERE assess_type = 'professional'
GROUP BY stud_num, total_item
ORDER BY MAX( total_score ) DESC
I need to show a ranking lists for a sport we manage.
It needs to sum up the 4 best results for each player (from a table that could have hundreds of results per player) and then sort the entire list from the player with the most points to least points.
The query below returns
ERROR 1054 (42S22): Unknown column 'r1.user_id' in 'where clause'
so I've gone off track somewhere.
SELECT r1.user_id, (
SELECT SUM(points)
FROM (
SELECT *
FROM ranking_matrix_points r2
WHERE user_id=r1.user_id
ORDER BY points DESC
LIMIT 4
) r3
) AS total_points
FROM ranking_matrix_points r1
WHERE
user.status IN('active')
GROUP BY r1.user_id
ORDER BY total_points DESC
One possible solution might be to number the rows for each user in order of points descending and then sum up the points with a rank <= 4. This might not perform very well though, and also you'll get a problem with ties (but you would have using limit too).
select user_id, sum(points) total_points
from (
select user_id, points,
(
case user_id
when #cur_user
then #row := #row + 1
else #row := 1 and #cur_user := user_id end
) as rank
from ranking_matrix_points,
(select #row := 0, #cur_user := '') r
order by user_id, points desc
) t
where rank <= 4
group by user_id;
I'm pretty sure there are better ways to do this but I can't think of any at the moment. This would have been very easy in just about any database with support for window functions, but sadly MySQL doesn't support any yet.
You don't need a double query, just
SELECT user_id, SUM(points)
FROM ranking_matrix_points
WHERE user.status in('active')
GROUP BY user_id
ORDER BY total_points DESC
LIMIT 4
or
SELECT TOP 4 user_id, SUM(points)
FROM ranking_matrix_points
WHERE user.status in('active')
GROUP BY user_id
ORDER BY total_points DESC