Read our blogs, tips and tutorials
Try our exercises or test your skills
Watch our tutorial videos or shorts
Take a self-paced course
Read our recent newsletters
License our courseware
Book expert consultancy
Buy our publications
Get help in using our site
551 attributed reviews in the last 3 years
Refreshingly small course sizes
Outstandingly good courseware
Whizzy online classrooms
Wise Owl trainers only (no freelancers)
Almost no cancellations
We have genuine integrity
We invoice after training
Review 30+ years of Wise Owl
View our top 100 clients
Search our website
We also send out useful tips in a monthly email newsletter ...
Software ==> | SQL (203 exercises) |
Topic ==> | Aggregation and grouping (12 exercises) |
Level ==> | Average difficulty |
Subject ==> | SQL training |
This exercise is provided to allow potential course delegates to choose the correct Wise Owl Microsoft training course, and may not be reproduced in whole or in part in any format without the prior written consent of Wise Owl.
If you haven't already done so, run the script in the above folder to generate the Doctor Who database.
The tables you'll need for this exercise are as follows:
You'll need to link each episode to its author and its doctor.
Write a query to list out for each author and doctor the number of episodes made, but restrict your output to show only the author/doctor combinations for which more than 5 episodes have been written. Here's what you should see:
Here we've sorted the rows so the most common combination appears first.
You'll need to use the HAVING clause. Remember that sweaty feet will give horrible odours!
Save this as Favourite combos, then close it down.
You can find other training resources for the subject of this exercise here:
From: | Bhavya59 |
When: | 01 Aug 24 at 12:27 |
SELECT AuthorName
,DoctorName
,COUNT(EpisodeId) AS 'Episodes'
FROM tblEpisode e
INNER JOIN tblAuthor a
ON e.AuthorId=a.AuthorId
INNER JOIN tblDoctor d
ON e.DoctorId=d.DoctorId
GROUP BY AuthorName, DoctorName
HAVING COUNT(EpisodeId)>5
ORDER BY Episodes DESC
From: | Rupchand7 |
When: | 14 Nov 23 at 05:59 |
select AuthorName, DoctorName, count(episodeid) as Episodes from tblauthor join tblepisode
on tblauthor.authorid = tblepisode.authorid join tbldoctor on tblepisode.doctorid = tbldoctor.doctorid
group by authorname, doctorname having count(episodeid) >5 order by Episodes desc
From: | walteragwe |
When: | 21 Jan 21 at 03:09 |
SELECT [AuthorName],[DoctorName],COUNT([EpisodeId]) AS [No of episodes]
FROM [dbo].[tblAuthor] A INNER JOIN [dbo].[tblEpisode] E
ON A.AuthorId = E.AuthorId INNER JOIN [dbo].[tblDoctor] D
ON E.DoctorId = D.DoctorId
GROUP BY [AuthorName],[DoctorName]
HAVING COUNT([EpisodeId]) > 5
ORDER BY [No of episodes] DESC;
GO
From: | smyndlo |
When: | 24 Nov 19 at 13:54 |
SELECT
au.AuthorName Writer, --1st Table
Doc.DoctorName Doctor, --2nd Table
COUNT(Ep.EpisodeId) 'Num Of Episodes' --3rd Table
FROM tblAuthor AS Au
INNER JOIN tblEpisode AS Ep
ON Au.AuthorId = Ep.AuthorId
INNER JOIN tblDoctor Doc
ON Doc.DoctorId = Ep.DoctorId
GROUP BY Au.AuthorName, Doc.DoctorName
HAVING COUNT(Ep.EpisodeId) > 5
ORDER BY [Num Of Episodes] Desc
From: | Andy B |
When: | 25 Nov 19 at 09:33 |
We don't normally post answers to exercises, as it removes the incentive for people to try them, but I've made an exception for this because the query is so well-written and presented!
Kingsmoor House
Railway Street
GLOSSOP
SK13 2AA
Landmark Offices
99 Bishopsgate
LONDON
EC2M 3XD
Holiday Inn
25 Aytoun Street
MANCHESTER
M1 3AE
© Wise Owl Business Solutions Ltd 2024. All Rights Reserved.