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SQL | Scalar functions exercise | Create a scalar function to determine type of each episode
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.
You can learn how to do this exercise if you attend one of more of the courses listed below!
| Software ==> | SQL (198 exercises) |
| Version ==> | Any version of SQL Server |
| Topic ==> | Scalar functions (6 exercises) |
| Level ==> | Average difficulty |
| Classroom ==> | Fast-track SQL / Advanced SQL |
| Online ==> | Fast-track SQL / Advanced SQL |
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First (if you haven't already done so) run the script shown above to generate the Doctor Who training database.
The aim of this exercise is to create a function called fnEpisodeDescription such that you can count how many episodes there are of each type:
-- count number of episodes of each type
SELECT
dbo.fnEpisodeDescription(Title) AS 'Episode type',
COUNT(*) AS 'Number of episodes'
FROM
tblEpisode
GROUP BY
dbo.fnEpisodeDescription(Title)
Here's what this should return:
It's reassuring to see that the number of first and second parts are the same!
To write this function use the CHARINDEX function to determine whether one string of text exists within another:
Your function should look for the word Part 1 (or failing that, Part 2) in the title passed in.
As a guide, CHARINDEX('owl','Wise Owl') would return 6, whereas CHARINDEX('fox','Wise Owl') would return 0.
When you've got your function and query working, save (optionally) the SQL as Part of a couple.sql, then close it down.