I have a mysql table "course_marks" which has 3 fields
studentID, courseID, mark.
I want to get the rank of a particular studentID for a particular courseID on the basis of mark(order by mark desc). The query should return a single row that indicates the rank. Ignore the same mark condition. How can I get the result in a single mysql query ?
First of all you should decide how identical marks will be treated when calculating the rank. This query for marks: 5,5,4 will calculate the rank of students with mark 4 as 3rd rank:
SELECT
COUNT(r.mark)+1 as rank
FROM
course_marks cm
LEFT JOIN
course_marks r ON r.courseID = cm.courseID AND r.mark > cm.mark
WHERE
cm.studentID = %STUDENT_ID% AND
cm.courseID = %COURSE_ID%
This query for marks: 5,5,4 will calculate the rank of students with mark 4 as 2nd:
SELECT
COUNT(DISTINCT r.mark)+1 as rank
FROM
course_marks cm
LEFT JOIN
course_marks r ON r.courseID = cm.courseID AND r.mark > cm.mark
WHERE
cm.studentID = %STUDENT_ID% AND
cm.courseID = %COURSE_ID%
Related
in barcodes table i have two column witch they are enter, exit like with student attendance record,
in this table when student entered to class, i add new row in table and enter value is 1 and exit is 0 like with login
after exiting from classroom i check latest row of this student and if enter is 1 then i add new row in table with 1 value for exit like with logout
now i want to calculate counting all students witch enter column is 1 and exit value is 0 to get all present students in class
SAMPLE DATA:
in this DB-FIDDLE i have sample data and into that i have 2 present student that session_id of them is 1, they are 123451,123452 students, but my sql command as second part is incorrect and that return one present student:
(
select count(*)
from barcodes b
where b.session_id = s.id
group by session_id, barcode
having sum(exit) = 0
) as present
for example:
select s.id, s.session_name, s.session_type, s.date_time,
(
select count(*)
from barcodes b where b.session_id = s.id
) as barcode_count ,
(
select count(*)
from barcodes b
where b.session_id = s.id
group by session_id, barcode
having sum(exit) = 0
) as present
from sessions s;
If I get what you want to do right, you can add another, outer aggregation.
...
(SELECT sum(x.count)
FROM (SELECT count(*) count
FROM barcodes b
WHERE b.session_id = s.id
GROUP BY b.session_id,
b.barcode
HAVING sum(b.enter) <> 0
AND sum(b.exit) = 0) x) present
...
But I think there's a much simpler way to get what you want by just taking the sum of enter minus exit for the session. Since any row with exit = 1 also comes with enter = 1, we need to double the exit before subtraction though.
...
(SELECT sum(b.enter - 2 * b.exit)
FROM barcodes b
WHERE b.session_id = s.id) present
...
If you can trust the data, something like this could work:
select
s.id,
s.session_name,
s.session_type,
s.date_time,
count(*) as barcode_count,
sum(exit = 0) - sum(exit = 1) as present
from sessions s
left join barcodes b on b.session_id = s.id
group by s.id, s.session_name, s.session_type, s.date_time
db-fiddle
You can read sum(exit = 0) - sum(exit = 1) as (number of entries) - (number of exits). If 4 students entered a class and 2 left the class, I would expect 2 students (4 - 2) to be still in the class.
My query does return the lowest price from the two columns (price_base, price_special) but it is not returning the correct store_id that corresponds to the lowest price found.
My Query:
SELECT grocery_item.id, grocery_item.category,
grocery_category.name AS cat, grocery_item.name AS itemName,
MIN( if( grocery_price.price_special>0,
grocery_price.price_base)) AS price,
grocery_price.store_id,
grocery_store.name AS storeName
FROM grocery_item
LEFT JOIN grocery_category ON
grocery_category.id=grocery_item.category
LEFT JOIN grocery_price
ON grocery_price.item_id = grocery_item.id
LEFT JOIN grocery_store
ON grocery_store.id=grocery_price.store_id
WHERE grocery_price.selection='no'
AND buy='yes'
GROUP BY grocery_price.item_id
ORDER BY store_id, grocery_item.category, grocery_item.name
Returns this:
ID category cat itemName price store_id storeName
92 3 Bread/Bakery Arnold Bread 2.14 1 Food Lion
But the grocery_price table holds this info:
item_id price_base price_special store_id
92 4.29 2.14 9
92 3.99 0.00 1
so the store_id I need to be returned is 9 (the storeName returned would NOT then be Food Lion)
EDIT: WORKING QUERY based on Uueerdo's comments (thank you!)
SELECT minP.item_id, gi.category, gc.name AS cat,
gi.name as itemName, gp.store_id,
gs.name AS storeName, minP.price
FROM
(SELECT p.item_id, MIN(IF(p.price_special >0,
p.price_special,p.price_base)) AS price
FROM grocery_item AS i
INNER JOIN grocery_price AS p ON (i.id = p.item_id)
WHERE i.buy = 'yes'
GROUP BY p.item_id) AS minP
INNER JOIN grocery_item AS gi ON minP.item_id = gi.id
INNER JOIN grocery_category AS gc on gi.category = gc.id
LEFT JOIN grocery_price AS gp
ON minP.price = IF(gp.price_special > 0,
gp.price_special,gp.price_base)
AND gp.item_id = gi.id
INNER JOIN grocery_store AS gs ON gp.store_id = gs.id
GROUP BY gi.id
ORDER BY gs.id, gi.category,gi.name
The values returned for non-grouped, non-aggregated fields are an (effectively) random selection from the values encountered with the grouped fields' values. Most RDBMS do not even consider such a query valid, and even newer versions of MySQL default to disallowing such queries.
In cases like yours, where you need the non-grouped value(s) associated with the aggregate result (min in this case); the aggregating query must be converted into a subquery, that can be joined back to the aggregated tables to find the source row(s) that correspond to the aggregated value.
Edit: Basically, you need to look at the problem slightly differently. You're currently finding the lowest price for an item and a store that item is listed for; you need to find the lowest price for an item, and use that to find the store(s) that have the item at that price.
This gets you the lowest price for item's marked "buy":
SELECT p.item_id, MIN(IF(p.price_special > 0,p.price_special,p.price_base)) AS price
FROM grocery_item AS i
INNER JOIN grocery_price AS p ON (i.id = p.item_id)
WHERE i.buy = 'yes'
GROUP BY p.item_id
You can then take that to get the rest of the results:
SELECT minP.item_id
, gi.name
, gi.category, gc.category_name AS cat, gi.Name as itemName, gi.buy
, gp.store_id, gp.name AS storeName
, minP.price
FROM ([the query above]) AS minP
INNER JOIN grocery_item AS gi ON minP.item_id = gi.id
INNER JOIN grocery_category AS gc on gi.category = gc.grocery_category_id
/* Guessing on this join since grocery_category_id
was not qualified with it's table name */
INNER JOIN grocery_price AS gp
ON minP.price = IF(gp.price_special > 0,gp.price_special,gp.price_base)
/* Alternatively: ON minP.price IN (gp.price_special, gp.price_base)
... though this could cause false positives if the minP.price is 0
from one store's base price being "free"
*/
INNER JOIN grocery_store AS gs ON gp.store_id = gs.id
;
I am using the following query to retrieve the number of events per state from 2 tables that are linked by a userID.
SELECT state,COUNT(*) AS num
FROM tableUserInfo
WHERE userID IN (SELECT userID
FROM tableEvents
WHERE conditionOne = 1
AND conditionTwo = 2)
GROUP BY state
This query works correctly. My problem is that not all states have user entries, and I need the query to return 0 for those. I was wondering if there was a method such as joining or using an in clause, that would included a set of all states, making the query return 0 for any that didn't have entries in tableEvents?
Do you have a list of states? If not then this would give a list of all the states your database knows about:
SELECT DISTINCT state FROM tableUserInfo
....and enclosing this in brackets it can be dropped in place in the query below:
SELECT s.state, IFNULL(cnt, 0) AS num
FROM list_of_states s
LEFT JOIN (
SELECT state,COUNT(*) AS cnt
FROM tableUserInfo ui
INNER JOIN tableEvents te
ON ui.userId=te.userId
WHERE conditionOne = 1
AND conditionTwo = 2
GROUP BY state
) u
ON s.state=u.state;
Although in the absence of "list_of_states" it would be more efficient to do this:
SELECT ui.state, SUM(IF(te.userId IS NULL, 0, 1)) AS cnt
FROM tableUserInfo ui
LEFT JOIN tableEvents te
ON ui.userId=te.userId
AND te.conditionOne = 1
AND te.conditionTwo = 2
GROUP BY state;
As #raymond-nijland suggested you can use Left Join to include all states.
SELECT tableUserInfo.state,COUNT(tableUserInfo.*) AS num
FROM tableUserInfo Left Join tableEvents on tableUserInfo.userID = tableEvents.userID
WHERE tableEvents.conditionOne = 1 AND tableEvents.conditionTwo = 2
GROUP BY state
I want to get the amount of students who have passed ALL their subjects (mark >= 4).
Ex:
Student 1(OK):
Math: 5
Chemistry: 4
Student 2(OK):
Philosophy: 7
Student 3(NOT OK):
Math: 3
Philosophy: 6
Student 4(NOT OK):
Math: null
Philosophy: 8
DB:
-students(id)
-subjects_students(id_subject, id_student, mark)
SQL (using MySQL):
SELECT count(ss.id_student)
FROM subjects_students ss
WHERE (SELECT count(ss.id_student)
FROM students st
WHERE ss.id_student = st.id)
=
(SELECT count(ss.id_student)
FROM students st
WHERE ss.id_student = st.id
AND ss.mark >= 4)
I can´t seem to get the right amount. I get students who have passed some subjects, but not all of them.
EDIT: mark can be null. Do not count these.
ANSWER:
SELECT COUNT(*)
FROM
(
SELECT COUNT(ss.id_student)
FROM subjects_students ss
GROUP BY ss.id_student
HAVING MIN(ss.mark) >= 4 AND COUNT(ss.mark) = COUNT(*)
) src;
If you want students who have passed all their subjects, then you want to filter out the ones whose mark is too low. Here is one method:
SELECT ss.id_student
FROM subjects_students ss
GROUP BY ss.id_student
HAVING MIN(ss.mark) >= 4;
You can then count the students using this as a subquery.
Note that the join to the students table is unnecessary. All the information you need is in subjects_students.
This should join the students and subject table, then remove any students who have a subject with a failing grade.
SELECT
a.`id`,
c.`id_subject`,
c.`mark`
FROM `students` a
LEFT JOIN `subjects_students` c
ON a.`id` = b.`id_student`
LEFT JOIN `subjects_students` b
ON a.`id` = b.`id_student` AND b.`mark` < 4
WHERE b.`id` IS NULL
GROUP BY a.`id`,c.`id_subject`,c.`mark`;
I do like Gordon's method here is another that might be slightly slower but can get you to the count of students in the same query without 2 steps.
SELECT COUNT(DISTINCT ss.id_student) as CountOfStudents
FROM
subjects_students ss
WHERE
NOT EXISTS (SELECT 1 FROM subjects_students ss2 WHERE ss.id_student = ss2.id_student AND ss2.mark < 4)
i have 5 tables called personal,dailypay,bonuses,iou and loans am trying to write a query that will generate payroll from this table's...my code is
select personal.name as NAME,
(sum(dailypay.pay) + bonuses) - (iou.amount + loans.monthly_due)) as SALARY
from personal
join dailypay on personal.eid = dailypay.eid
left join bonuses on personal.eid = bonuses.eid
left join iou on personal.eid = iou.eid
left join where dailypay.date = 'specified_date'
and bonuses.date_approved = 'specified_date'
and iou.date_approved = 'specified_date'
and loans.date = month(now()
It returns the name and null salary values for staffs that does have records for either bonuses,iou and loans. But i want to sum their dailypay, deduct/add deductions or additions return the values, in the event of no record it should proceed with the summation without any deduction or subtraction.
You missed something when pasting the code, as there is no join for the loans table. Also, you are using the table bonuses as a value, you need a field name also. I added some code for the join and for the field, but used ??? for names that are unknown to me.
When you add or subtract a null value to something else, the result is null, that's why you get null as result when any of the values from the left-joined tables are missing. You can use ifnull(..., 0) to turn a null value into zero.
You need a group by clause, otherwise it would sum up the salary for all persons.
If I get you right, you have several records in the dailypay table for each user, but only one record per user in the other tables? In that case you have the problem that you will be joining the other tables against each row in the dailypay, so if you have 20 payment records for a user, it will count the bonus 20 times. You can use an aggregate like max to get the value only once.
You have put conditions for the left.joined tables in the where clause, but this will turn the joins into inner joins. You should have those conditions in each join clause.
select
personal.name as NAME,
(sum(dailypay.pay) + ifnull(max(bonuses.???), 0)) - (ifnull(max(iou.amount), 0) + ifnull(max(loans.monthly_due), 0)) as SALARY
from
personal
inner join dailypay on personal.eid = dailypay.eid
left join bonuses on personal.eid = bonuses.eid and bonuses.date_approved = 'specified_date'
left join iou on personal.eid = iou.eid and iou.date_approved = 'specified_date'
left join loans on personal.??? = loans.??? and loans.date = month(now())
where
dailypay.date = 'specified_date'
group by
personal.name
There seems to be an extranous left join before the where and a missing closing bracket ) in month(now()
so it should look like:
select personal.name as NAME,
(sum(dailypay.pay) + bonuses) - (iou.amount + loans.monthly_due)) as SALARY
from personal
join dailypay on personal.eid = dailypay.eid
left join bonuses on personal.eid = bonuses.eid
left join iou on personal.eid = iou.eid
where dailypay.date = 'specified_date'
and bonuses.date_approved = 'specified_date'
and iou.date_approved = 'specified_date'
and loans.date = month(now())