I am a newbie in MYSQL and had a question regarding the use of MAX and COUNT functions together in MYSQL. I have 2 tables worker and assignment and the primary key of worker is a foreign key in assignment table.
I need to show the employees name and id and the total assignment assigned to him, and only show the person with the most assignment that is the employee with the most assignment.
my code is
SELECT worker.Wrk_ID, worker.Wrk_LastName, MAX(a.count_id)
FROM worker,
(SELECT COUNT(assignment.Wrk_ID) as count_ID
FROM worker, assignment
WHERE worker.Wrk_ID = assignment.Wrk_ID
GROUP BY worker.Wrk_ID)as a
GROUP BY worker.Wrk_ID;
The code is giving an error no. #1054.
Please can anyone help me.
Thanking you in anticipation.
Try something like this:
SELECT worker.Wrk_ID, worker.Wrk_LastName, S.Count
FROM worker
JOIN
(SELECT Wrk_ID, COUNT(*) AS Count FROM Assignments
GROUP BY Wrk_Id ORDER BY COUNT(*) DESC LIMIT 1) S
ON worker.Wrk_ID = S.Wrk_ID
If you want a list of employees sorted by their total assignments:
SELECT w.WrkID, w.Wrk_LastName, COUNT(*) AS Assignments
FROM work w left join Assignments a
ON w.WrkID=a.WrkID
GROUP BY w.WrkID
ORDER BY COUNT(*) DESC;
To allow multiple winners:
SELECT s.*, w.Wrk_Lastname FROM
(
SELECT wrk_id , COUNT(*) AS tot_assignments
FROM Assignments
GROUP BY wrk_id
HAVING COUNT(*) =
(
SELECT MAX(tot) FROM
(
SELECT COUNT(*) AS TOT FROM Assignments GROUP BY wrk_id
) counts
)
) winners
INNER JOIN worker w ON s.wrk_id = w.wrk_id;
It can be slow since it does multiple GROUP BY. Doing it in separated steps in a procedure can be better.
Related
I have the following SQL Database structure:
Users are the registered users. Maps are like circuits or race tracks. When a user is driving a time a new time record will be created including the userId, mapId and the time needed to finish the racetrack.
I wish to create a view where all the users personal bests on all maps are listed.
I tried creating the view like this:
CREATE VIEW map_pb AS
SELECT MID, UID, TID
FROM times
WHERE score IN (SELECT MIN(score) FROM times)
ORDER BY registered
This does not lead to the wished result.
Thank you for your help!
I hope that you have 'times' table created as the above diagram and 'score' column in the table that you use to measure the best record.
(MIN(score) is the best record).
You can simply create a view to have the personal best records using sub-queries like this.
CREATE VIEW map_pb AS
SELECT a.MID, a.UID, a.TID
FROM times a
INNER JOIN (
SELECT TID, UID, MIN(score) score
FROM times
GROUP BY UID
) b ON a.UID = b.UID AND a.score= b.score
-- if you have 'registered' column in the 'times' table to order the result
ORDER BY registered
I hope this may work.
You probably need to use a query that will first return the minimum score for each user on each map. Something like this:
SELECT UID,
MID,
MIN(score) AS best_time
FROM times
GROUP BY UID, MID
Note: I used MIN(score) as this is what is shown in your example query, but perhaps it should be MIN(time) instead?
Then just use the subquery JOINed to your other tables to get the output:
SELECT *
FROM (
SELECT UID,
MID,
MIN(score) AS best_time
FROM times
GROUP BY UID, MID
) a
INNER JOIN users u ON u.UID = a.UID
INNER JOIN maps m ON m.MID = a.MID
Of course, replace SELECT * with the columns you actually want.
Note: code untested but does give an idea as to a solution.
Start with a subquery to determine each user's minimum score on each map
SELECT UID, TID, MIN(time) time
FROM times
GROUP BY UID, TID
Then join that subquery into a main query.
SELECT times.UID, times.TID,
mintimes.time
FROM times
JOIN (
) mintimes ON times.TID = mintimes.TID
AND times.UID = mintimes.UID
AND times.time = mintimes.time
JOIN maps ON times.MID = maps.MID
JOIN users ON times.UID = users.UID
This query pattern uses a GROUP BY function to find the outlying (MIN in this case) value for each combination. It then uses that subquery to find the detail record for each outlying value.
This is a slight variant of the question I asked here
SQL Query for getting maximum value from a column
I have a Person Table and an Activity Table with the following data
-- PERSON-----
------ACTIVITY------------
I have got this data in the database about users spending time on a particular activity.
I intend to get the data when every user has spent the maximum number of hours.
My Query is
Select p.Id as 'PersonId',
p.Name as 'Name',
act.HoursSpent as 'Hours Spent',
act.Date as 'Date'
From Person p
Left JOIN (Select MAX(HoursSpent), Date from Activity
Group By HoursSpent, Date) act
on act.personId = p.Id
but it is giving me all the rows for Person and not with the Maximum Numbers of Hours Spent.
This should be my result.
You have several issues with your query:
The subquery to get hours is aggregated by date, not person.
You don't have a way to bring in other columns from activity.
You can take this approach -- joins and group by, but it requires two joins:
select p.*, a.* -- the columns you want
from Person p left join
activity a
on a.personId = p.id left join
(select personid, max(HoursSpent) as max_hoursspent
from activity a
group by personid
) ma
on ma.personId = a.personId and
ma.max_hoursspent = a.hoursspent;
Note that this can return duplicates for a given person -- if there are ties for the maximum.
This is written more colloquially using row_number():
select p.*, a.* -- the columns you want
from Person p left join
(select a.*,
row_number() over (partition by a.personid order by a.hoursspent desc) as seqnum
from activity a
) a
on a.personId = p.id and a.seqnum = 1
ma.max_hoursspent = a.hoursspent;
Each student has some scores (and each score has student_id column).
I want to calculate student average, compare his average with other student, and find his position in his class.
Is it possible to find his position based on his average with 1 query? (may contains subqueries or joins)
I can sort all students by their average by this query:
SELECT s.*
FROM
scores s LEFT JOIN lessons lesson
ON lesson.id = s.lesson_id
WHERE lesson.display = 1
GROUP BY s.student_id
ORDER BY AVG(s.score) DESC
but it needs processing with PHP array_search function. (I think using MySQL functions is better, in this situation)
select student_id, AVG(scores) as 'average' from lessons as l, scores as s
where lessons.id = s.lesson_id and lesson.display = 1
GROUP BY s.student_id order by average desc`
Try this query
sample http://sqlfiddle.com/#!2/4fb8d/1
Hope this helps
select s.id, AVG(l.scores) as average from lessons as l, student as s
where l.id = s.lessonid and l.display = 1
GROUP BY s.id order by average desc
can solve your problem
Check this link
Thanks to Meherzad's sql query
I think you are looking for something like this:
SELECT COUNT(DISTINCT average)
FROM (
SELECT AVG(score) as average
FROM scores INNER JOIN lessons
ON scores.lesson_id = lessons.id
WHERE display=1
GROUP BY student_id
) sub_scores
WHERE average >= (SELECT AVG(score)
FROM scores INNER JOIN lessons
ON scores.lesson_id = lessons.id
WHERE display=1
AND student_id=1)
In the subquery sub_scores I'm calculating all averages of all students, in the outer query I'm calculating the number of distinct averages bigger than the average of student 1.
This will return the position of student 1.
Depending on what you are after, you might want to remove DISTINCT clause.
See this fiddle.
You need to use a user defined variable over the result set.
Try this:
SELECT *, (#rank := if(#rank is null, 1, #rank + 1)) as rank
FROM (SELECT s.*
FROM scores s
LEFT JOIN lessons lesson ON lesson.id = s.lesson_id
WHERE lesson.display = 1
GROUP BY s.student_id
ORDER BY AVG(s.score) DESC
) x
I have these tables and queries as defined in sqlfiddle.
First my problem was to group people showing LEFT JOINed visits rows with the newest year. That I solved using subquery.
Now my problem is that that subquery is not using INDEX defined on visits table. That is causing my query to run nearly indefinitely on tables with approx 15000 rows each.
Here's the query. The goal is to list every person once with his newest (by year) record in visits table.
Unfortunately on large tables it gets real sloooow because it's not using INDEX in subquery.
SELECT *
FROM people
LEFT JOIN (
SELECT *
FROM visits
ORDER BY visits.year DESC
) AS visits
ON people.id = visits.id_people
GROUP BY people.id
Does anyone know how to force MySQL to use INDEX already defined on visits table?
Your query:
SELECT *
FROM people
LEFT JOIN (
SELECT *
FROM visits
ORDER BY visits.year DESC
) AS visits
ON people.id = visits.id_people
GROUP BY people.id;
First, is using non-standard SQL syntax (items appear in the SELECT list that are not part of the GROUP BY clause, are not aggregate functions and do not sepend on the grouping items). This can give indeterminate (semi-random) results.
Second, ( to avoid the indeterminate results) you have added an ORDER BY inside a subquery which (non-standard or not) is not documented anywhere in MySQL documentation that it should work as expected. So, it may be working now but it may not work in the not so distant future, when you upgrade to MySQL version X (where the optimizer will be clever enough to understand that ORDER BY inside a derived table is redundant and can be eliminated).
Try using this query:
SELECT
p.*, v.*
FROM
people AS p
LEFT JOIN
( SELECT
id_people
, MAX(year) AS year
FROM
visits
GROUP BY
id_people
) AS vm
JOIN
visits AS v
ON v.id_people = vm.id_people
AND v.year = vm.year
ON v.id_people = p.id;
The: SQL-fiddle
A compound index on (id_people, year) would help efficiency.
A different approach. It works fine if you limit the persons to a sensible limit (say 30) first and then join to the visits table:
SELECT
p.*, v.*
FROM
( SELECT *
FROM people
ORDER BY name
LIMIT 30
) AS p
LEFT JOIN
visits AS v
ON v.id_people = p.id
AND v.year =
( SELECT
year
FROM
visits
WHERE
id_people = p.id
ORDER BY
year DESC
LIMIT 1
)
ORDER BY name ;
Why do you have a subquery when all you need is a table name for joining?
It is also not obvious to me why your query has a GROUP BY clause in it. GROUP BY is ordinarily used with aggregate functions like MAX or COUNT, but you don't have those.
How about this? It may solve your problem.
SELECT people.id, people.name, MAX(visits.year) year
FROM people
JOIN visits ON people.id = visits.id_people
GROUP BY people.id, people.name
If you need to show the person, the most recent visit, and the note from the most recent visit, you're going to have to explicitly join the visits table again to the summary query (virtual table) like so.
SELECT a.id, a.name, a.year, v.note
FROM (
SELECT people.id, people.name, MAX(visits.year) year
FROM people
JOIN visits ON people.id = visits.id_people
GROUP BY people.id, people.name
)a
JOIN visits v ON (a.id = v.id_people and a.year = v.year)
Go fiddle: http://www.sqlfiddle.com/#!2/d67fc/20/0
If you need to show something for people that have never had a visit, you should try switching the JOIN items in my statement with LEFT JOIN.
As someone else wrote, an ORDER BY clause in a subquery is not standard, and generates unpredictable results. In your case it baffled the optimizer.
Edit: GROUP BY is a big hammer. Don't use it unless you need it. And, don't use it unless you use an aggregate function in the query.
Notice that if you have more than one row in visits for a person and the most recent year, this query will generate multiple rows for that person, one for each visit in that year. If you want just one row per person, and you DON'T need the note for the visit, then the first query will do the trick. If you have more than one visit for a person in a year, and you only need the latest one, you have to identify which row IS the latest one. Usually it will be the one with the highest ID number, but only you know that for sure. I added another person to your fiddle with that situation. http://www.sqlfiddle.com/#!2/4f644/2/0
This is complicated. But: if your visits.id numbers are automatically assigned and they are always in time order, you can simply report the highest visit id, and be guaranteed that you'll have the latest year. This will be a very efficient query.
SELECT p.id, p.name, v.year, v.note
FROM (
SELECT id_people, max(id) id
FROM visits
GROUP BY id_people
)m
JOIN people p ON (p.id = m.id_people)
JOIN visits v ON (m.id = v.id)
http://www.sqlfiddle.com/#!2/4f644/1/0 But this is not the way your example is set up. So you need another way to disambiguate your latest visit, so you just get one row per person. The only trick we have at our disposal is to use the largest id number.
So, we need to get a list of the visit.id numbers that are the latest ones, by this definition, from your tables. This query does that, with a MAX(year)...GROUP BY(id_people) nested inside a MAX(id)...GROUP BY(id_people) query.
SELECT v.id_people,
MAX(v.id) id
FROM (
SELECT id_people,
MAX(year) year
FROM visits
GROUP BY id_people
)p
JOIN visits v ON (p.id_people = v.id_people AND p.year = v.year)
GROUP BY v.id_people
The overall query (http://www.sqlfiddle.com/#!2/c2da2/1/0) is this.
SELECT p.id, p.name, v.year, v.note
FROM (
SELECT v.id_people,
MAX(v.id) id
FROM (
SELECT id_people,
MAX(year) year
FROM visits
GROUP BY id_people
)p
JOIN visits v ON ( p.id_people = v.id_people
AND p.year = v.year)
GROUP BY v.id_people
)m
JOIN people p ON (m.id_people = p.id)
JOIN visits v ON (m.id = v.id)
Disambiguation in SQL is a tricky business to learn, because it takes some time to wrap your head around the idea that there's no inherent order to rows in a DBMS.
I am currently running the following query that returns a set of jobs along with their categories and the user name of the person who posted it.
SELECT job_id, user_id, title, profiles.user_name
FROM (jobs)
JOIN profiles ON jobs.user_id = profiles.user_id
JOIN job_categories ON jobs.cat_id = job_categories.cat_id
JOIN job_sub_categories ON jobs.sub_cat_id = job_sub_categories.sub_cat_id
WHERE `status` = 'open'
ORDER BY post_date desc
LIMIT 5
I have a table called feedback that holds rows of feedback for a particular employer based on their previous transactions (much like ebay).
feedback_id|employer_id|job_id|performance_score|quality_score|availability_score|communication_score
What I want to be able to do is to sort and filter my results based on an employers current feedback rating and I'm not sure how to add this into my query. It seems like I have to do some math within my query or run a sub-query perhaps?. Or should I modify my feedback table to include another field such as total feedback given for a particular rating?
Any help would be greatly appreciated.
Rating is calculated at all of the feedback scores added together divided by the number of rows then divided by the number 4 because there are 4 scored fields (performance, quality, availability and communication) so feedback_avg = (feedback_total/num_rows)/4
Let me give it a shot. I will assume you have only two tables, employers: [id, name] and feedback: [id, employer_id, score].
First off, the score subquery:
SELECT employer_id, SUM(score) AS total_score, COUNT(*) AS num_rows
FROM feedback GROUP BY employer_id;
Now the main query:
SELECT name, total_score/num_rows AS avg_score
FROM employers JOIN ([subquery]) AS sq ON(employers.id = sq.employer_id)
WHERE avg_score > 0.5;
Paste the entire subquery into the indicated position.
Tip: Views
If you like, you can make the sub-query a permanent view, and use that in the main query:
CREATE VIEW score_tally AS
SELECT employer_id, SUM(score) AS total_score, COUNT(*) AS num_rows
FROM feedback
GROUP BY employer_id;
SELECT name, total_score/num_rows AS avg_score
FROM employers JOIN score_tally ON(employers.id = score_tally.employer_id)
WHERE avg_score > 0.5;
Tip (again): The above tip was stupid, we should use the built-in AVG:
CREATE VIEW score_tally AS
SELECT employer_id, AVG(score) AS avg_score
FROM feedback
GROUP BY employer_id;
SELECT name, avg_score
FROM employers JOIN score_tally ON(employers.id = score_tally.employer_id)
WHERE avg_score > 0.5;
Let's guess what your complete query might look like:
SELECT job_id,
user_id,
title,
profiles.user_name AS user_name,
avg_score
FROM jobs
JOIN profiles ON(jobs.user_id = profiles.user_id)
JOIN job_categories ON(jobs.cat_id = job_categories.cat_id)
JOIN job_sub_categories ON(jobs.sub_cat_id = job_sub_categories.sub_cat_id)
JOIN (SELECT employer_id, AVG(score) AS avg_score FROM feedback GROUP BY employer_id) AS sq
ON(employers.id = sq.employer_id)
WHERE status = 'open' AND avg_score > 0.5
ORDER BY post_date desc
LIMIT 5