You may need some knowledge of cricket to help with this one. Given the following table, would this query produce this result?
Table: batsmen
batsmanname | runsscored | howout
Colin | 10 | bowled
Colin | 20 | caught
Steve | 10 | bowled
Steve | 20 | not out
SELECT batsmanname, SUM(runsscored) / COUNT(howout) AS battingavg
FROM batsmen
WHERE howout <> 'not out'
GROUP BY batsmananme
Result:
batsmanname | battingavg
Colin | 15
Steve | 30
This should produce the results that you want:
SELECT b1.name, SUM(b1.runsscored) / b2.NumOut as bat_avg
FROM batsmen b1
INNER JOIN
(
select name, COUNT(howout) as NumOut
from batsmen
WHERE howout <> 'not out'
GROUP BY name
) b2
ON b1.name = b2.name
GROUP BY b1.name
See SQL Fiddle with demo
Related
I have a single MySQL table with the name 'checkins' and 4 columns.
id | userIDFK | checkin_datetime | shopId
------------------------------------------------
1 | 1 | 2018-01-18 09:44:00 | 3
2 | 2 | 2018-01-18 10:32:00 | 3
3 | 3 | 2018-01-18 11:19:00 | 3
4 | 1 | 2018-01-18 17:57:00 | 3
5 | 1 | 2018-01-18 16:31:00 | 1
6 | 1 | 2018-01-19 08:31:00 | 3
Basically I want to find rows where users have checked-in more than once (>=2) on the same day and the same shop. So for instance if a user checks-in as in rows with ids 1 and 4 (same user, same day, same shop), the query should return a hit with the the entire rows (id, userIDFK, checkin_datetime, shopId). Hope this makes sense.
I already tried using
SELECT id, userIDFK, checkin_datetime, shopId
FROM (
SELECT * FROM 'checkins' WHERE COUNT(userIDFK)>=2 AND COUNT(shopId)>=2
)
The same day part I have no clew how to do it, and I know this query is way off, but this is the best I could.
You can try grouping by userId checkin_date and shopID
SELECT userIDFK, checkin_datetime, shopId,COUNT(SHOPiD)
FROM checkins
GROUP BY userIDFK, DATE(checkin_datetime), shopId
HAVING COUNT(SHOPID)>1
EDIT
You can include a subquery to get all lines:
select b.id,b.userIDFK, b.checkin_datetime, b.shopId
from checkins b
where (SELECT COUNT(SHOPiD)
FROM checkins a
where a.userIDFK=b.userIDFK and date(a.checkin_datetime)=date(b.checkin_datetime) and a.shopId=b.a.shopId
GROUP BY userIDFK, DATE(checkin_datetime), shopId)>1
GROUPBY can be used to get the multiple occurrences.
SELECT id, userIDFK, checkin_datetime, shopId
FROM checkins
GROUP BY userIDFK, DATE(checkin_datetime), shopId
HAVING count(id) > 1;
Hope it helps!
EDIT:
Using inner join you can achieve it. Here is the query:
SELECT c1.* FROM checkins c1 INNER JOIN checkins c2
ON c1.userIDFK = c2.userIDFK
AND date(c1.checkin_datetime) = date(c2.checkin_datetime)
AND c1.shopId = c2.shopId
AND c1.id != c2.id
Cheers!!
I am unable to map the record as my expectation.
Doc Table
-------+-------------------
doc_id | doc_title
-------+-------------------
1 | My book
-------+-------------------
2 | My sec Book
--------------------------
Doc details Table
-----------+--------------+-----------------------
fk_doc_id | doc_version | submit_date
-----------+--------------+-----------------------
1 | 1 | 2015-10-25 14:32:01
-----------+--------------+-----------------------
1 | 2 | 2015-10-26 13:00:01
-----------+--------------+-----------------------
1 | 3 | 2015-10-27 09:00:00
--------------------------+-----------------------
2 | 1 | 2015-10-25 11:15:01
-----------+--------------+-----------------------
2 | 2 | 2015-10-26 10:00:00
--------------------------+-----------------------
Question: How do I join this two tables to get each documents with the latest version doc info? even though I get the latest version but the row info which is not correct.
So far I have tried this query
SELECT *, max(doc_version) AS latest_version
FROM d_doc
JOIN d_doc_dtl ON d_doc.doc_id = d_doc_dtl.fk_doc_id
GROUP BY d_doc.doc_id;
My expected result is
--------+--------------+----------------+--------------------
doc_id | doc_title | latest_version | submit_date
--------+--------------+----------------+--------------------
1 | My book | 3 | 2015-10-27 09:00:00
--------+--------------+----------------+--------------------
2 | My sec book | 2 | 2015-10-26 10:00:00
----------------------------------------+--------------------
but my result is
--------+--------------+----------------+--------------------
doc_id | doc_title | latest_version | submit_date
--------+--------------+----------------+--------------------
1 | My book | 3 | 2015-10-25 14:32:01
--------+--------------+----------------+--------------------
2 | My sec book | 2 | 2015-10-25 11:15:01
----------------------------------------+--------------------
NOTE: the submit_date which is no correct.
SELECT d_doc.doc_id, d_doc.doc_title, max_table.latest_version
FROM d_doc JOIN (
select fk_doc_id, max(doc_version) as latest_version from d_doc_dtl group by fk_doc_id
) as max_table ON d_doc.doc_id = max_table.fk_doc_id
This query should work as you expect. It selects latest document versions in inner subquery and than joins it with documents.
SELECT d.doc_id,
d.doc_title,
dtl.doc_version latest_version,
dtl.submit_date
FROM d_doc d
INNER JOIN (SELECT dt.*
FROM d_doc_dtl dt
INNER JOIN (SELECT fk_doc_id, MAX(doc_version) doc_version
FROM d_doc_dtl
GROUP BY fk_doc_id) dm
ON dt.fk_doc_id = dm.fk_doc_id
AND dt.doc_version = dm.doc_version) dtl
ON d.doc_id = dtl.fk_doc_id
You get wrong results because you selected only max(version), but date as it is not in group by clause can contain any value. First you need to get records containing latest version as shown above.
Easy, instead of
SELECT *, max(doc_version) AS latest_version
Use this
SELECT d_doc.*, max(doc_version) AS latest_version
What you were doing by selecting * is getting all the results after the table is joined and you only wanted the original table results.
select * from doc_table , doc_version where exists( select
max(version_id)
from
doc_version vert
where
(doc_table .DOC_ID = vert.VERSION_DOC_ID) ) group by doc_id;
You can try something like this.
I am in a very complicated problem. Let me explain you first what I am doing right now:
I have a table name feedback in which I am storing grades against course id. The table looks like this:
+-------+-------+-------+-------+-----------+--------------
| id | cid | grade |g_point| workload | easiness
+-------+-------+-------+-------+-----------+--------------
| 1 | 10 | A+ | 1 | 5 | 4
| 2 | 10 | A+ | 1 | 2 | 4
| 3 | 10 | B | 3 | 3 | 3
| 4 | 11 | B+ | 2 | 2 | 3
| 5 | 11 | A+ | 1 | 5 | 4
| 6 | 12 | B | 3 | 3 | 3
| 7 | 11 | B+ | 2 | 7 | 8
| 8 | 11 | A+ | 1 | 1 | 2
g_point has just specific values for the grades, thus I can use these values to show the user courses sorted by grades.
Okay, now first my task is to print out the grade of each course. The grade can be calculated by the maximum occurrence against each course. For example from this table we can see the result of cid = 10 will be A+, because it is present two times there. This is simple. I have already implemented this query which I will write here in the end.
The main problem is when we talk about the course cid = 11 which has two different grades. Now in that situation client asks me to take the average of workload and easiness of both these courses and whichever course has the greater average should be shown. The average would be computed like this:
all workload values of the grade against course
+ all easiness values of the grade against course
/ 2
From this example cid = 11 has four entries,have equal number of grades against a course
B+ grade average
avgworkload(2 + 7)/2=x
avgeasiness(3 + 8)/2 = y
answer x+y/2 = 10
A+ grade average
avgworkload(5 + 1)/2=x
avgeasiness(4 + 2)/2 = y
answer x+y/2 = 3
so the grade should be B+.
This is the query which I am running to get the max occurrence grade
SELECT
f3.coursecodeID cid,
f3.grade_point p,
f3.grade g
FROM (
SELECT
coursecodeID,
MAX(mode_qty) mode_qty
FROM (
SELECT
coursecodeID,
COUNT(grade_point) mode_qty
FROM feedback
GROUP BY
coursecodeID, grade_point
) f1
GROUP BY coursecodeID
) f2
INNER JOIN (
SELECT
coursecodeID,
grade_point,
grade,
COUNT(grade_point) mode_qty
FROM feedback
GROUP BY
coursecodeID, grade_point
) f3
ON
f2.coursecodeID = f3.coursecodeID AND
f2.mode_qty = f3.mode_qty
GROUP BY f3.coursecodeID
ORDER BY f3.grade_point
Here is SQL Fiddle.
I added a table Courses with the list of all course IDs, to make the main idea of the query easier to see. Most likely you have it in the real database. If not, you can generate it on the fly from feedback by grouping by cid.
For each cid we need to find the grade. Group feedback by cid, grade to get a list of all grades for the cid. We need to pick only one grade for a cid, so we use LIMIT 1. To determine which grade to pick we order them. First, by occurrence - simple COUNT. Second, by the average score. Finally, if there are several grades than have same occurrence and same average score, then pick the grade with the smallest g_point. You can adjust the rules by tweaking the ORDER BY clause.
SELECT
courses.cid
,(
SELECT feedback.grade
FROM feedback
WHERE feedback.cid = courses.cid
GROUP BY
cid
,grade
ORDER BY
COUNT(*) DESC
,(AVG(workload) + AVG(easiness))/2 DESC
,g_point
LIMIT 1
) AS CourseGrade
FROM courses
ORDER BY courses.cid
result set
cid CourseGrade
10 A+
11 B+
12 B
UPDATE
MySQL doesn't have lateral joins, so one possible way to get the second column g_point is to repeat the correlated sub-query. SQL Fiddle
SELECT
courses.cid
,(
SELECT feedback.grade
FROM feedback
WHERE feedback.cid = courses.cid
GROUP BY
cid
,grade
ORDER BY
COUNT(*) DESC
,(AVG(workload) + AVG(easiness))/2 DESC
,g_point
LIMIT 1
) AS CourseGrade
,(
SELECT feedback.g_point
FROM feedback
WHERE feedback.cid = courses.cid
GROUP BY
cid
,grade
ORDER BY
COUNT(*) DESC
,(AVG(workload) + AVG(easiness))/2 DESC
,g_point
LIMIT 1
) AS CourseGPoint
FROM courses
ORDER BY CourseGPoint
result set
cid CourseGrade CourseGPoint
10 A+ 1
11 B+ 2
12 B 3
Update 2 Added average score into ORDER BY SQL Fiddle
SELECT
courses.cid
,(
SELECT feedback.grade
FROM feedback
WHERE feedback.cid = courses.cid
GROUP BY
cid
,grade
ORDER BY
COUNT(*) DESC
,(AVG(workload) + AVG(easiness))/2 DESC
,g_point
LIMIT 1
) AS CourseGrade
,(
SELECT feedback.g_point
FROM feedback
WHERE feedback.cid = courses.cid
GROUP BY
cid
,grade
ORDER BY
COUNT(*) DESC
,(AVG(workload) + AVG(easiness))/2 DESC
,g_point
LIMIT 1
) AS CourseGPoint
,(
SELECT (AVG(workload) + AVG(easiness))/2
FROM feedback
WHERE feedback.cid = courses.cid
GROUP BY
cid
,grade
ORDER BY
COUNT(*) DESC
,(AVG(workload) + AVG(easiness))/2 DESC
,g_point
LIMIT 1
) AS AvgScore
FROM courses
ORDER BY CourseGPoint, AvgScore DESC
result
cid CourseGrade CourseGPoint AvgScore
10 A+ 1 3.75
11 B+ 2 5
12 B 3 3
If I understood well you need an inner select to find the average, and a second outer select to find the maximum values of the average
select cid, grade, max(average)/2 from (
select cid, grade, avg(workload + easiness) as average
from feedback
group by cid, grade
) x group by cid, grade
This solution has been tested on your data usign sql fiddle at this link
If you change the previous query to
select cid, max(average)/2 from (
select cid, grade, avg(workload + easiness) as average
from feedback
group by cid, grade
) x group by cid
You will find the max average for each cid.
As mentioned in the comments you have to choose wich strategy use if you have more grades that meets the max average. For example if you have
+-------+-------+-------+-------+-----------+--------------
| id | cid | grade |g_point| workload | easiness
+-------+-------+-------+-------+-----------+--------------
| 1 | 10 | A+ | 1 | 5 | 4
| 2 | 10 | A+ | 1 | 2 | 4
| 3 | 10 | B | 3 | 3 | 3
| 4 | 11 | B+ | 2 | 2 | 3
| 5 | 11 | A+ | 1 | 5 | 4
| 9 | 11 | C | 1 | 3 | 6
You will have grades A+ and C soddisfing the maximum average 4.5
There is a table Remark that contains data as shown below:
SerialNo | RemarkNo | Desp
=============================================
10 | 1 | rainy
10 | 2 | sunny
11 | 1 | sunny
11 | 2 | rainy
11 | 3 | cloudy
12 | 1 | rainy
If I run a query SELECT * FROM remark WHERE remark_no IN (SELECT MAX(remark_no) FROM remark GROUP BY serial_no);, I still get the above result:
What query will return the following result:
10 | 2 | sunny
11 | 3 | cloudy
12 | 1 | rainy
That is, the last record in each group should be returned??
SELECT
r1.*
FROM remark r1
LEFT JOIN remark r2
ON (r1.serial_no = r2.serial_no
AND r1.remark_no < r2.remark_no)
WHERE r2.remark_no IS NULL;
For the top RemarkNo for each SerialNo (along with the other fields from the same row) :
Select all records where there isn't a higher RemarkNo for the same SerialNo
SELECT *
FROM remark r1
WHERE NOT EXISTS
(SELECT SerialNo FROM remark r2
WHERE (r2.RemarkNo>r1.RemarkNo)
AND (r2.SerialNo=r1.SerialNo)
)
http://sqlfiddle.com/#!2/7da1b/21
You can do this by returning your results and using a sort order. For example...
SELECT * from GROUP Order By GROUP.ID DESC
This will return results in the order of last record first. Then if you do not loop through results you will return only one record... the last one recorded.
I have a table say :
id| AccID | Subject | Date
1 | 103 | Open HOuse 1 | 11/24/2011 9:00:00 AM
2 | 103 | Open HOuse 2 | 11/25/2011 10:00:00 AM
3 | 72 | Open House 3 | 11/26/2011 1:10:28 AM
4 | 82 | OPen House 4 | 11/27/2011 5:00:29 PM
5 | 82 | OPen House 5 | 11/22/2011 5:00:29 PM
From the above table, i need all the unique values for the Accid. But say, if there are two or more columns with the same Accid, then i need the one which has the smaller date (among the columns which have the same Accid)
So, from the above table, the o/p should be :
1
3
5
Can any1 please help me in this ? Thanks
SELECT t1.*
FROM [MyTable] t1
INNER JOIN
(
SELECT AccID, MIN(Date) Date
FROM [MyTable]
GROUP BY AccID
) t2 ON t1.AccID = t2.AccID AND t1.Date = t2.Date
More than just the AccID but...
WITH SEL
AS
(
SELECT AccID, MIN(DATE)
FROM table
GROUP BY AccID
)
SELECT table.*
FROM table
JOIN SEL ON SEL.AccID = table.AccID