This is the second article in my series discussing advanced SQL concepts. I want to describe features that are well supported in popular database management systems for quite some time, but somehow many people still don’t know about their existence. I’d like to explain them with examples, first giving a problem to solve using “plain old” SQL and then showing a better solution using advanced SQL.
You can find the first article about window functions here.
This time I’d like to discuss Common Table Expressions (CTE).
In this post I’ll be using PostgreSQL 10, because it’s the most feature-rich open source database available. Common Table Expressions have been available since Postgres 8.4, so any modern version will be fine. They are also supported by other popular RDBMSes.
This time we’ll be working with three tables:
actors and a linking table
films_actors between them:
Something that may attract your attention are the columns
sequel_id. They are foreign key referencing to the very same table
films and pointing respectively to a prequel or sequel of a given film. To make things clear I prepared a set of sample rows for this table:
As you can see there are two chains of the prequel-sequel relation (1-3-5 and 2-4) and one film that has no connections at all. I don’t think it’s necessary to provide the content of the other two tables - let’s pretend they have some meaningful data.
Here comes the first example:
Example 1. For each film count number of actors starring in it.
Easy. The result looks like this:
Result rows with actors counts
There is no catch here - it’s as simple as it looks. All we need to do is to join
film_actors and count number of rows:
SELECT f.id, COUNT(*) AS actors FROM films f LEFT JOIN films_actors fa ON f.id = fa.film_id GROUP BY f.id
With that solution in mind let’s move on to the next thing.
Example 2. For each film, its prequel and sequel count number of actors starring in them.
This task doesn’t look any more complicated than the previous one. Right?
Result rows with actors counts for prequel and sequel
There is more than one solution, but the most straightforward just uses three identical queries to get information about the film and both its prequel and sequel:
SELECT films.id, films.prequel_id, films.sequel_id, films.actors, prequels.actors AS prequel_actors, sequels.actors AS sequel_actors FROM ( SELECT f.*, COUNT(*) as actors FROM films f JOIN films_actors fa ON f.id = fa.film_id GROUP BY f.id, f.prequel_id ) films LEFT JOIN ( SELECT f.*, COUNT(*) as actors FROM films f JOIN films_actors fa ON f.id = fa.film_id GROUP BY f.id ) prequels ON films.prequel_id = prequels.id LEFT JOIN ( SELECT f.*, COUNT(*) as actors FROM films f JOIN films_actors fa ON f.id = fa.film_id GROUP BY f.id ) sequels ON films.sequel_id = sequels.id
The query works, but has some problems. It’s not easy to read and understand as you have to carefully compare, line by line, all three subqueries to make sure that they do exactly the same thing. Modifying one means also that you need to change others as well. Wouldn’t it be nice to write identical parts once and only refer to them somehow?
Moreover this also shows one general disadvantage of SQL - you need to read queries from inside to outside - because that’s the order in which they are executed. I think it would look much better if we had them one below the other.
And that’s what CTE are mainly about.
Common Table Expressions (CTE)
CTE are a mechanism that allows to define temporary named result sets existing just for one query (you may also think about them as temporary “tables” or “views”). Let’s see how they work in practice by solving Example 1 once again:
WITH film_with_actors_count AS ( SELECT f.*, count(*) AS actors FROM films f JOIN films_actors fa ON f.id = fa.film_id GROUP BY f.id ) SELECT f.id, f.actors FROM film_with_actors_count f
CTE are defined using
WITH … AS clause. Inside them you can put almost any SQL statement you like (not only
SELECT, but also
DELETE). Every CTE has a name, so you can easily refer to it in the main query, just like I did in the example above. Fun fact: there is no comma or semicolon between the last CTE definition and the main query.
Of course you can refer to them as many times you want. As a reference take a look at the new solution to Example 2:
WITH film_with_actors_count AS ( SELECT f.*, count(*) AS actors FROM films f JOIN films_actors fa ON f.id = fa.film_id GROUP BY f.id ) SELECT films.id, films.prequel_id, films.sequel_id, films.actors, prequels.actors AS prequel_actors, sequels.actors AS sequel_actors FROM film_with_actors_count films LEFT JOIN film_with_actors_count prequels on films.prequel_id = prequels.id LEFT JOIN film_with_actors_count sequels on films.sequel_id = sequels.id
Looks better, doesn’t it?
Let’s see what other problems CTE can solve. To visualize the first one I’ll use the example from my previous article about window functions. This time with a little complication:
Example 3. Return a single film with the greatest number of actors for each release year.
So, only one film from each year and only the one with the most actors:
Films with greatest number of actors for each year
To solve this problem we need to use window functions. Adding a new column with a correct values is just a matter of using
RANK() over a correctly partitioned and ordered window:
SELECT f.id, f.release_year, COUNT(*) AS actors, RANK() OVER (PARTITION BY release_year ORDER BY COUNT(*) DESC) AS year_rank FROM films f LEFT JOIN films_actors fa ON f.id = fa.film_id GROUP BY f.id
That’ll give us the following result:
Films with year rank
And now we can simply add a
HAVING clause, right?
SELECT f.id, f.release_year, COUNT(*) AS actors, RANK() OVER (PARTITION BY release_year ORDER BY COUNT(*) DESC) AS year_rank FROM films f LEFT JOIN films_actors fa ON f.id = fa.film_id GROUP BY f.id HAVING year_rank = 1
ERROR: column "year_rank" does not exist LINE 5: HAVING year_col == 1
That’s because window functions are not visible in any other clauses in the same query. To overcome this issue we can simple wrap the above query with another query and add necessary filtering there. Or, to make things clearer and simpler, use CTE:
WITH films_actors_year_rank AS ( SELECT f.id, f.release_year, COUNT(*) AS actors, RANK() OVER (PARTITION BY release_year ORDER BY COUNT(*) DESC) AS year_rank FROM films f LEFT JOIN films_actors fa ON f.id = fa.film_id GROUP BY f.id ) SELECT f.release_year, f.id, f.actors FROM films_actors_year_rank f WHERE year_rank = 1
And now it works just fine. Once more a CTE expression was used to improve readability. This is good, but are they really only about making SQL code nicer?
Well, not exactly. In fact there are problems that simply can’t be solved without CTE.
Example 4. For each film return number of all its prequels and sequels.
And I’m having such thing in mind:
Films with numbers of all their prequels and sequels
We need to count the length of both prequel and sequel chain for each film.
If you think about this problem for a while you may realize that it’s not difficult at all to check if a film has a single prequel or sequel by simply looking at its corresponding foreign key column. It’s not hard to extend it to the second level either. In other words we can check if the prequel’s prequel (or sequel’s sequel) exists by doing a self join. Adding more nesting however requires using more subsequent joins. To solve this problem for any length of the prequel/sequel sequence we’d need something more powerful.
Something like a recursion. Wait, what? In SQL? Yes, it’s possible.
Recursive CTE has an interesting ability to invoke itself. You can put the name of a CTE in its body and therefore make it run recursively. This kind of CTE takes the form of:
WITH RECURSIVE cte AS ( -- [non-recursive term] UNION ALL -- [recursive term] )
A very simple working example from Postgres documentation goes as follows:
WITH RECURSIVE t(n) AS ( VALUES (1) -- non-recursive term UNION ALL SELECT n+1 FROM t WHERE n < 100 -- recursive term ) SELECT * FROM t;
The above query generates numbers from 1 to 100.
As you can see there are few differences between normal and recursive Common Table Expression. First thing is the usage of
RECURSIVE term in the definition, which enables the recursive mode. Second thing is that the query consists of two separate parts connected with a
UNION ALL operator. They are called respectively “non-recursive term” and “recursive term”. You can make sure that the result table will not have any duplicated by using
UNION instead of
UNION ALL. The last thing is the fact that the recursive term query invokes its own CTE - something not possible in the normal mode.
Let’s see how recursive CTE work exactly. Under the hood Postgres uses two temporary tables: working table and result table. The latter is the place that accumulates the final result of a CTE. Technically the whole process is actually iterative, not recursive, but that’s how this operation has been called by the SQL standards committee. Therefore it can be visualized in three steps:
1. Initial step
Initial step is evaluated only once. Executor runs the non-recursive term and puts the result both in working and result tables:
2. Repetitive step
It is evaluated many times, in a loop. Executor runs the recursive term against the content of the working table and then merges its output with the result table. It removes duplicates if the
UNION operator was used. The result is also used to replace the content of the working table and therefore prepare it for the next step. The whole process repeats as long as the working table is not empty.
3. Final step
Database simply return the content of the result table.
Armed with that knowledge let’s get back to Example 4. I asked you to find out how many prequels and sequels each film has.
Let’s start with prequels count and do it step by step. First thing is to write the CTE header:
WITH RECURSIVE films_with_prequels_number(id, sequel_id, prequels_num) AS (
I added the part with column names in brackets here. From now I can skip all aliases, but I need to be careful about putting statements in the right order. The additional column
prequels_num will hold the number of prequels for a particular film.
Now let’s write the non-recursive term, which prepares the first set of rows both for working and result table. Because we’re counting prequels we have to select all films that have zero prequels - their
prequel_id column will be set to
SELECT f.id, f.sequel_id, 0 FROM films f WHERE f.prequel_id IS NULL
Now we have to choose the linking operator. In our case both
UNION ALL work identically, because we’re not expecting any duplicates.
Now the hardest part - the recursive term. We need to take the content of working table by recursively selecting from self and joining it together with the films table, effectively replacing each film with its sequel. We also have to remember about incrementing the
SELECT f.id, f.sequel_id, fr.prequels_num + 1 FROM films_with_prequels_number fr JOIN films f ON fr.sequel_id = f.id )
And that’s it. Using inner
JOIN instead of
LEFT JOIN ensures that the execution will eventually stop, because it effectively discards all rows that have
NULL value in its joining column (in this case
sequel_id). And we don’t have cycles here.
Now let’s see at the whole query with CTE expressions for both prequels and sequels:
WITH RECURSIVE films_with_prequels_number(id, sequel_id, prequels_num) AS ( SELECT f.id, f.sequel_id, 0 FROM films f WHERE f.prequel_id IS NULL UNION ALL SELECT f.id, f.sequel_id, fr.prequels_num + 1 FROM films_with_prequels_number fr JOIN films f ON fr.sequel_id = f.id ), films_with_sequels_number(id, prequel_id, sequels_num) AS ( SELECT f.id, f.prequel_id, 0 FROM films f WHERE f.sequel_id IS NULL UNION ALL SELECT f.id, f.prequel_id, fr.sequels_num + 1 FROM films_with_sequels_number fr JOIN films f ON fr.prequel_id = f.id ) SELECT f.id, f.prequel_id, f.sequel_id, fpn.prequels_num, fsn.sequels_num FROM films f LEFT JOIN films_with_prequels_number fpn ON f.id = fpn.id LEFT JOIN films_with_sequels_number fsn ON f.id = fsn.id
The very last thing we have to do is to write the final query - one that puts everything together. It’s as simple as selecting from
films table and joining it with both CTE.
Bonus example. Find Fibonacci sequence with numbers below 100.
I’ll just post the solution here and leave it for a curious reader to analyze ;)
WITH RECURSIVE fibo(a, b) AS ( VALUES (0,1) UNION ALL SELECT b, a + b FROM fibo WHERE a + b < 100 ) SELECT b FROM fibo
CTE - inconspicuous but powerful
CTE are an interesting SQL feature. They help to organize and simplify complicated queries and also make them easier to maintain by allowing a user to get rid of duplicated parts. In their simplest form however they don’t offer anything more, especially nothing in terms of manipulating data.
You may therefore think that they’re not very useful. But their true potential lies in the recursive mode. It enables you to do a thing otherwise impossible in pure SQL - write a query that invokes itself, which gives you a lot of new possibilities. For example, you can traverse your relational tables like they were graphs. Recursive CTE might seem hard at first glance, but once you get familiar with them, you will appreciate the power they give.