MySQL Left Join Count Not working - mysql

I have 2 tables (members, duty_sched)
I want to be able to count how many time each member appears in the duty_sched table.
This is what I have after several weeks searching
SELECT
members.fname,
members.lname,
(SELECT COUNT(duty_sched.id) FROM duty_sched
WHERE 'duty_sched.mbr_id' = 'members.mbr_id') AS shifts
FROM members
ORDER BY members.lname, members.fname
The output I'm getting is all zeros, what I would like it to show me is:
Name shifts
Bob Smith 4
Jane Johnson 2
Any help I can get will be greatly appreciated.

Remove the single quotes: WHERE duty_sched.mbr_id = members.mbr_id)
You can also write your query with a LEFT JOIN:
SELECT
m.fname,
m.lname,
COUNT(d.id) AS shifts
FROM members AS m
LEFT JOIN duty_sched AS d
ON d.mbr_id = m.mbr_id
GROUP BY
m.mbr_id -- the Primary Key of `members`
ORDER BY
m.lname, m.fname ;

You can try something like that
SELECT
members.fname,
members.lname,
count(*) FROM duty_sched JOIN members
WHERE duty_sched.mbr_id = members.mbr_id
Group by duty_sched.mbr_id
ORDER BY members.lname, members.fname

Try this:
SELECT m.fname, m.lname, IFNULL(A.shifts, 0) shifts
FROM members m
LEFT JOIN (SELECT mbr_id, COUNT(id) shifts
FROM duty_sched GROUP BY mbr_id) A ON m.mbr_id = A.mbr_id
ORDER BY m.lname, m.fname;

Related

Where is the error in my nested MySQL query?

I am guessing there is a mistake in the nested query however I cannot seem to find it myself.
Here is the query:
Select student.sid, name
from student join exam on exam.sid = student.sid
where student.sid in (select *
from course join exam on cid=courseid
group by exam.sid
having sum(credits) >= 20)
Thank you in advance!
You can use the group by as follows:
select s.sid, s.name
from student s
Join exam e on s.sid = e.sid
Join course c on c.cid = e.courseid
group by s.sid, s.name
having sum(c.credits) >= 20
Yes, there is a mistake. The nested query is returning multiple columns but the comparison is to only a single column.
This will obviously generate an error. Presumably, you don't want the sum() but without sample data, desired results, and an explanation of what the query is supposed to be doing, it is hard to make a concrete suggestion.
The query may have other errors, but presumably, you intend something like this:
select s.sid, s.name
from student s
where s.sid in (select e.sid,
from course c join
exam e
on c.cid = e.courseid
group by e.sid
having sum(c.credits) >= 20
)

Subquery left join refer to parent ID

I am trying to make a query to fetch the newest car for each user:
select * from users
left join
(select cars.* from cars
where cars.userid=users.userid
order by cars.year desc limit 1) as cars
on cars.userid=users.userid
It looks like it says Unknown column "users.userid" in where clause
I tried to remove cars.userid=users.userid part, but then it only fetches 1 newest car, and sticks it on to each user.
Is there any way to accomplish what I'm after? thanks!!
For this purpose, I usually use row_number():
select *
from users u left join
(select c.* , row_number() over (partition by c.userid order by c.year desc) as seqnum
from cars c
) c
on c.userid = u.userid and c.seqnum = 1;
One option is to filter the left join with a subquery:
select * -- better enumerate the columns here
from users u
left join cars c
on c.userid = u.userid
and c.year = (select max(c1.year) from cars c1 where c1.userid = c.userid)
For performance, consider an index on car(userid, year).
Note that this might return multiple cars per user if you have duplicate (userid, year) in cars. It would be better to have a real date rather than just the year.
Maybe there are better and more efficient way to query this. Here is my solution;
select users.userid, cars.*
from users
left join cars on cars.userid = users.userid
join (SELECT userid, MAX(year) AS maxDate
FROM cars
GROUP BY userid) as sub on cars.year = sub.maxDate;

mySQL Sub Select needed

I have three tables, libraryitems, copies and loans.
A libraryitem hasMany copies, and a copy hasMany loans.
I'm trying to get the latest loan entry for a copy only; The query below returns all loans for a given copy.
SELECT
libraryitems.title,
copies.id,
copies.qruuid,
loans.id AS loanid,
loans.status,
loans.byname,
loans.byemail,
loans.createdAt
FROM copies
INNER JOIN libraryitems ON copies.libraryitemid = libraryitems.id AND libraryitems.deletedAt IS NULL
LEFT OUTER JOIN loans ON copies.id = loans.copyid
WHERE copies.libraryitemid = 1
ORDER BY copies.id ASC, loans.createdAt DESC
I know there needs to be a sub select of some description in here, but struggling to get the correct syntax. How do I only return the latest, i.e MAX(loans.createdAt) row for each distinct copy? Just using group by copies.id returns the earliest, rather than latest entry.
Image example below:
in the subquery , getting maximum created time for a loan i.e. latest entry and joining back with loans to get other details.
SELECT
T.title,
T.id,
T.qruuid,
loans.id AS loanid,
loans.status,
loans.byname,
loans.byemail,
loans.createdAt
FROM
(
SELECT C.id, C.qruuid, L.title, MAX(LN.createdAt) as maxCreatedTime
FROM Copies C
INNER JOIN libraryitems L ON C.libraryitemid = L.id
AND L.deletedAt IS NULL
LEFT OUTER JOIN loans LN ON C.id = LN.copyid
GROUP BY C.id, C.qruuid, L.title) T
JOIN loans ON T.id = loans.copyid
AND T.maxCreatedTime = loans.createdAt
A self left join on loans table will give you latest loan of a copy, you may join the query to the other tables to fetch the desired output.
select * from loans A
left outer join loans B
on A.copyid = B.copyid and A.createdAt < B.createdAt
where B.createdAt is null;
This is your query with one simple modification -- table aliases to make it clearer.
SELECT li.title, c.id, c.qruuid,
l.id AS loanid, l.status, l.byname, l.byemail, l.createdAt
FROM copies c INNER JOIN
libraryitems li
ON c.libraryitemid = li.id AND
li.deletedAt IS NULL LEFT JOIN
loans l
ON c.id = l.copyid
WHERE c.libraryitemid = 1
ORDER BY c.id ASC, l.createdAt DESC ;
With this as a beginning let's think about what you need. You want the load with the latest createdAt date for each c.id. You can get this information with a subquery:
select l.copyid, max(createdAt)
from loans
group by l.copyId
Now, you just need to join this information back in:
SELECT li.title, c.id, c.qruuid,
l.id AS loanid, l.status, l.byname, l.byemail, l.createdAt
FROM copies c INNER JOIN
libraryitems li
ON c.libraryitemid = li.id AND
li.deletedAt IS NULL LEFT JOIN
loans l
ON c.id = l.copyid LEFT JOIN
(SELECT l.copyid, max(l.createdAt) as maxca
FROM loans
GROUP BY l.copyid
) lmax
ON l.copyId = lmax.copyId and l.createdAt = lmax.maxca
WHERE c.libraryitemid = 1
ORDER BY c.id ASC, l.createdAt DESC ;
This should give you the most recent record. And, the use of left join should keep all copies, even those that have never been leant.

Mysql SUM Float give wrong value [duplicate]

I'm looking for help using sum() in my SQL query:
SELECT links.id,
count(DISTINCT stats.id) as clicks,
count(DISTINCT conversions.id) as conversions,
sum(conversions.value) as conversion_value
FROM links
LEFT OUTER JOIN stats ON links.id = stats.parent_id
LEFT OUTER JOIN conversions ON links.id = conversions.link_id
GROUP BY links.id
ORDER BY links.created desc;
I use DISTINCT because I'm doing "group by" and this ensures the same row is not counted more than once.
The problem is that SUM(conversions.value) counts the "value" for each row more than once (due to the group by)
I basically want to do SUM(conversions.value) for each DISTINCT conversions.id.
Is that possible?
I may be wrong but from what I understand
conversions.id is the primary key of your table conversions
stats.id is the primary key of your table stats
Thus for each conversions.id you have at most one links.id impacted.
You request is a bit like doing the cartesian product of 2 sets :
[clicks]
SELECT *
FROM links
LEFT OUTER JOIN stats ON links.id = stats.parent_id
[conversions]
SELECT *
FROM links
LEFT OUTER JOIN conversions ON links.id = conversions.link_id
and for each link, you get sizeof([clicks]) x sizeof([conversions]) lines
As you noted the number of unique conversions in your request can be obtained via a
count(distinct conversions.id) = sizeof([conversions])
this distinct manages to remove all the [clicks] lines in the cartesian product
but clearly
sum(conversions.value) = sum([conversions].value) * sizeof([clicks])
In your case, since
count(*) = sizeof([clicks]) x sizeof([conversions])
count(*) = sizeof([clicks]) x count(distinct conversions.id)
you have
sizeof([clicks]) = count(*)/count(distinct conversions.id)
so I would test your request with
SELECT links.id,
count(DISTINCT stats.id) as clicks,
count(DISTINCT conversions.id) as conversions,
sum(conversions.value)*count(DISTINCT conversions.id)/count(*) as conversion_value
FROM links
LEFT OUTER JOIN stats ON links.id = stats.parent_id
LEFT OUTER JOIN conversions ON links.id = conversions.link_id
GROUP BY links.id
ORDER BY links.created desc;
Keep me posted !
Jerome
Jeromes solution is actually wrong and can produce incorrect results!!
sum(conversions.value)*count(DISTINCT conversions.id)/count(*) as conversion_value
let's assume the following table
conversions
id value
1 5
1 5
1 5
2 2
3 1
the correct sum of value for distinct ids would be 8.
Jerome's formula produces:
sum(conversions.value) = 18
count(distinct conversions.id) = 3
count(*) = 5
18*3/5 = 9.6 != 8
For an explanation of why you were seeing incorrect numbers, read this.
I think that Jerome has a handle on what is causing your error. Bryson's query would work, though having that subquery in the SELECT could be inefficient.
Use the following query:
SELECT links.id
, (
SELECT COUNT(*)
FROM stats
WHERE links.id = stats.parent_id
) AS clicks
, conversions.conversions
, conversions.conversion_value
FROM links
LEFT JOIN (
SELECT link_id
, COUNT(id) AS conversions
, SUM(conversions.value) AS conversion_value
FROM conversions
GROUP BY link_id
) AS conversions ON links.id = conversions.link_id
ORDER BY links.created DESC
I use a subquery to do this. It eliminates the problems with grouping.
So the query would be something like:
SELECT COUNT(DISTINCT conversions.id)
...
(SELECT SUM(conversions.value) FROM ....) AS Vals
How about something like this:
select l.id, count(s.id) clicks, count(c.id) clicks, sum(c.value) conversion_value
from (SELECT l.id id, l.created created,
s.id clicks,
c.id conversions,
max(c.value) conversion_value
FROM links l
LEFT JOIN stats s ON l.id = s.parent_id
LEFT JOIN conversions c ON l.id = c.link_id
GROUP BY l.id, l.created, s.id, c.id) t
order by t.created
This will do the trick, just divide the sum with the count of conversation id which are duplicate.
SELECT a.id,
a.clicks,
SUM(a.conversion_value/a.conversions) AS conversion_value,
a.conversions
FROM (SELECT links.id,
COUNT(DISTINCT stats.id) AS clicks,
COUNT(conversions.id) AS conversions,
SUM(conversions.value) AS conversion_value
FROM links
LEFT OUTER JOIN stats ON links.id = stats.parent_id
LEFT OUTER JOIN conversions ON links.id = conversions.link_id
GROUP BY conversions.id,links.id
ORDER BY links.created DESC) AS a
GROUP BY a.id
Select sum(x.value) as conversion_value,count(x.clicks),count(x.conversions)
FROM
(SELECT links.id,
count(DISTINCT stats.id) as clicks,
count(DISTINCT conversions.id) as conversions,
conversions.value,
FROM links
LEFT OUTER JOIN stats ON links.id = stats.parent_id
LEFT OUTER JOIN conversions ON links.id = conversions.link_id
GROUP BY conversions.id) x
GROUP BY x.id
ORDER BY x.created desc;
I believe this will give you the answer that you are looking for.

Mysql - group by result by SUM of 2 columns

I have 3 tables:
user(id, name, id_school)
school(id, name)
result(id_user, stage1, stage2)
Now I would like to get school ranking - is sum of 2 columns: stage1 and stage2 of all users.
seems too simple:
select s.name as school, sum(stage1)+sum(stage2) as rank
from result r
join user u on u.id=r.id_user
join school s on s.id=u.id_school
group by s.id
and i really hope you have indexes.
Try this query. It is only assumption.please provide some data for testing
SELECT
s.id,
s.name AS SchoolName
(r.S1 + r.S2) AS Rank
FROM school as s
LEFT JOIN user as u ON u.id_school = s.id
LEFT JOIN (SELECT id_user , SUM(stage1) as S1 , SUM(stage2) FROM result GROUP BY id_user) as r ON r.id_user = u.id
GROUP BY s.id
ORDER BY Rank DESC
I would do something like this:
SELECT school.name, SUM(result.result1 + result.result2)
FROM school LEFT JOIN user ON (user.id_school = school.id) LEFT JOIN result ON (result.id_user = user.id) GROUP BY school.id
Hope it helps, good luck :)