Joins and aliases
Overview
Teaching: 30 min
Exercises: 30 minQuestions
Linking tables together and using shorthand
Objectives
to understand how to link tables together
Joins
To combine data from two tables we use the SQL JOIN
command, which comes after
the FROM
command.
We also need to tell the computer which columns provide the link between the two
tables using the word ON
. What we want is to join the data with the same
journal name.
SELECT *
FROM articles
JOIN journals
ON articles.issns = journals.issns;
ON
is like WHERE
, it filters things out according to a test condition. We use
the table.colname
format to tell the manager what column in which table we are
referring to.
Alternatively, we can use the word USING
, as a short-hand. In this case we are
telling the manager that we want to combine articles
with journals
and that
the common column is issns
.
SELECT *
FROM articles
JOIN journals
USING (issns);
We often won’t want all of the fields from both tables, so anywhere we would
have used a field name in a non-join query, we can use table.colname
.
For example, what if we wanted information on published articles in different journals, but instead of their ISSN we wanted the actual journal title.
SELECT articles.issns, journal_title, title, first_author, citation_count, author_count, month, year
FROM articles
JOIN journals
ON articles.issns = journals.issns;
Joins can be combined with sorting, filtering, and aggregation. So, if we wanted average number of authors for articles on different journals, we could do something like
SELECT articles.issns, journal_title, ROUND(AVG(author_count), 2)
FROM articles
JOIN journals
ON articles.issns = journals.issns
GROUP BY articles.issns;
Challenge
Write a query that returns the journal title, total number of articles published and average number of citations for every journal.
Solution
SELECT journal_title, count(*), avg(citation_count) FROM articles JOIN journals ON articles.issns = journals.issns GROUP BY articles.issns;
It is worth mentioning that you can join multiple tables. For example:
SELECT title, first_author, journal_title, publisher
FROM articles
JOIN journals
ON articles.issns = journals.issns
JOIN publishers
ON publishers.id = journals.publisherid;
Challenge:
Write a query that returns the journal title, publisher name, and number of articles published, ordered by number of articles in descending order.
Solution
SELECT journal_title, publisher, count(*) FROM articles JOIN journals ON articles.issns = journals.issns JOIN publishers ON publishers.id = journals.publisherid GROUP BY journal_title ORDER BY count(*) desc;
Aliases
As queries get more complex names can get long and unwieldy. To help make things clearer we can use aliases to assign new names to things in the query.
We can alias both table names:
SELECT ar.title, ar.first_author, jo.journal_title
FROM articles AS ar
JOIN journals AS jo
ON ar.issns = jo.issns;
And column names:
SELECT ar.title AS title, ar.first_author AS author, jo.journal_title AS journal
FROM articles AS ar
JOIN journals AS jo
ON ar.issns = jo.issns;
The AS
isn’t technically required, so you could do
SELECT a.title t
FROM articles a;
but using AS
is much clearer so it is good style to include it.
Extra Challenges (optional)
SQL queries help us ask specific questions which we want to answer about our data. The real skill with SQL is to know how to translate our scientific questions into a sensible SQL query (and subsequently visualize and interpret our results).
Have a look at the following questions; these questions are written in plain English. Can you translate them to SQL queries and give a suitable answer?
Challenge 1
How many articles are there from each first_author? Can you make an alias for the number of articles? Can you order the results by articles?
Solution 1
SELECT first_author, COUNT( * ) AS n_articles FROM articles GROUP BY first_author ORDER BY n_articles DESC;
Challenge 2
How many papers have a single author? How many have 2 authors? How many 3? etc?
Solution 2
SELECT author_count, count( * ) FROM articles GROUP BY author_count;
Challenge 3
How many articles are published for each language? (Ignore articles where language is unknown).
Solution 3
SELECT language, count( * ) FROM articles JOIN languages ON articles.languageid=languages.id WHERE language IS NOT null GROUP BY language;
Challenge 4
How many articles are published for each licence type, and what is the average number of citations for that licence type
Solution 4
SELECT licence, avg( citation_count ), count( * ) FROM articles JOIN licences ON articles.licenceid=licences.id WHERE licence IS NOT null GROUP BY licence;
Challenge 5
Multiple table joins. Select title, first_author, author_count, citation_count, month, year, journal_title and publisher
Solution 5
SELECT title, first_author, author_count, citation_count, month, year, journal_title, publisher FROM articles JOIN journals ON articles.issns=journals.issns JOIN publishers ON publishers.id=journals.publisherid;
Key Points
SQL is ideal for linking tables