SQL: Structured Query Language (Part 2: Advanced Features)¶
In this lecture, we will learn more advanced features of SQL.
Examples database to be used in this lecture is given in SQL here:
- Remember that while SQL is a standard, there are still differences
in implementations of it.
- Writing queries that do not rely on specific features results in portable applications.
- However, you cannot deny that some constructs may simplify your queries and performance. So, it is important to decide when to use a specific method to write a query.
- Remember: a query is not an algorithm. It is for the most part
a logical statement of what you are interested in.
- Often, there are multiple algorithms to implement it.
- Most DBMSs feature state of the art query optimizers (QOPT) that choose the lowest cost algorithm for a given query and database.
- QOPT engines are very sophisticated, often operate better than even expert human judgment.
- So, instead of trying to optimize your queries, you can try to make your queries easy to optimize: simple queries are better.
- Once you become sophisticated in a specific DBMS, you may learn specific weaknesses and you can develop strategies to adopt for that. We will discuss some.
- Finally, you should still follow some very simple guidelines:
- Do not join with relations unnecessarily.
- Do not sort (order by) or remove duplicates (distinct) unless it is needed.
- A INNER JOIN B: inner join selects tuples that satisfy a join condition, eliminates all tuples that do not satisfy the join condition. A is called the left operand and B is the right operand of the join operation.
- A LEFT OUTER JOIN B returns all tuples in the inner join as well as the tuples in A that do not join with any tuples in in B.
- A RIGHT OUTER JOIN B returns all tuples in the inner join as well as the tuples in B that do not join with any tuples in in A.
- A FULL OUTER JOIN B returns all tuples in the inner join as well as
the tuples from A and B that do not participate in the inner join.
- You can also use terms: JOIN, LEFT JOIN, RIGHT JOIN, FULL JOIN
Inner vs. outer join.¶
- Given R(A,B) and S(B,C) with the following contents:
We get the following results:
SELECT R.A, R.B, S.B, S.C FROM R JOIN S ON R.B=S.B;
A B B C a1 b1 b1 c1
SELECT R.A, R.B, S.B, S.C FROM R LEFT JOIN S ON R.B=S.B;
A B B C a1 b1 b1 c1 a2 b2 null null
SELECT R.A, R.B, S.B, S.C FROM R RIGHT JOIN S ON R.B=S.B;
A B B C a1 b1 b1 c1 null null b3 c3
SELECT R.A, R.B, S.B, S.C FROM R FULL JOIN S ON R.B=S.B;
A B B C a1 b1 b1 c1 a2 b2 null null null null b3 c3
We can use the fact that tuples that do not match have null values for the join.
Outer join examples:¶
Find for each student, total number of classes
SELECT s.id , s.name , count(t.student_id) as numclasses FROM students s left join transcript t on s.id = t.student_id GROUP BY s.id , s.name;
As count(attr) counts the total number of values, students with no classes will have zero classes.
If we used inner join, we would have elimited all students with zero classes as they would not join with transcript.
Find faculty who did not teach any classes
SELECT f.id , f.name FROM faculty f left join classes c on f.id = c.instructor_id WHERE c.instructor_id is null;
Find classes with no students. Return semester, year, section and course name.
SELECT co.crsname , c.course_id , c.semester , c.year , c.section FROM (classes c left join transcript t on c.course_id = t.course_id and c.semester = t.semester and c.year = t.year and c.section = t.section), courses co WHERE t.course_id is null and co.id = c.course_id;
This works because for each class, there is a course tuple. Otherwise, you would need to do a left join with the next table as well.
A query can be treated like a relation in the from clause
It is treated like a virtual relation:
SELECT f.id , f.name , f2class.numclasses , count(*) as numstudents FROM ( SELECT instructor_id AS id , count(*) as numclasses FROM classes GROUP BY instructor_id HAVING count(DISTINCT year) >= 2 ) as f2class , classes c , transcript t , faculty f WHERE f2class.id = c.instructor_id and t.course_id = c.course_id and t.semester = c.semester and t.year = c.year and t.section = c.section and f.id = c.instructor_id GROUP BY f.id , f.name , f2class.numclasses;
The inner query allows us to find faculty who taught in at least two years and total number of classes they taught. We can then use this information in the main query as if it was a real relation.
This query would be very hard to write without an anonymous relation as we cannot count for different types of things with a single group by.
Be careful: Do not use any anonymous relations to make it simpler to write/read the query.
SELECT S.d FROM (SELECT a.* FROM R WHERE b>5) as newR , S WHERE S.c = newR.c;
Anonymous relation is not really necessary here. The same query can be written with a simple join:
SELECT S.d FROM R,S WHERE R.b>5 and S.c=R.c;
When using an anonymous view, query optimizer may miss certain optimizations, especially in older DBMS.
Any query that returns a single number with an aggregate function is called a scalar query.
You can use a scalar query as if it was a number.
SELECT count(*) FROM classes WHERE instructor_id = 4; count ------- 5 (1 row) SELECT instructor_id , count(*) FROM classes c GROUP BY c.instructor_id HAVING count(*) >= 5;
Let’s rewrite this query by substituting the above query for number 5 in the having clause.
SELECT instructor_id , count(*) FROM classes c GROUP BY c.instructor_id HAVING count(*) >= (SELECT count(*) FROM classes WHERE instructor_id = 4);
Comparisons involving sets/bags¶
Many expressions in the WHERE clause (or HAVING) can compare a value against a SET
WHERE grade IN ('A', 'B') WHERE year NOT IN (2015, 2016)
Substitute a query for the set: Find all faculty who never taught a class.
SELECT f.id , f.name FROM faculty f WHERE f.id NOT IN (SELECT instructor_id FROM classes) ;
You can write equivalent queries using EXCEPT and LEFT JOIN.
Set Comparison Operators¶
There are many set comparison operators that can be used in queries. The inner query must return a single column for this to work.
Some useful operations:
value IN (QUERY) value NOT IN (QUERY) value > ANY (QUERY) value >= ALL (QUERY) value > ALL (QUERY) value = ANY (QUERY) --> same as IN value <> ALL (QUERY) --> same as NOT IN
You can also write expressions that check whether a query returns any tuples at all:
EXISTS (QUERY) => True if Query returns at least one tuple NOT EXISTS (QUERY) => True if Query returns no tuples
5 IN (1,2,3,4) FALSE 5 NOT IN (1,2,3,4) TRUE 2 IN (1,2,3,4) TRUE EXISTS (1,2,3,4) TRUE NOT EXISTS (1,2,3,4) FALSE NOT EXISTS () TRUE 5 <ALL (1,2,3,4) FALSE 5 >ALL (1,2,3,4) TRUE
SELECT * FROM students WHERE EXISTS (SELECT 1 FROM classes WHERE semester='Spring' and year=2016);
This is a kind of stupid query: if there is any class offered in Spring 2016, return all students. Otherwise, return no students.
Since it does not matter what we return in EXISTS/NOT EXISTS conditions (we only care whether a tuple is returned or not), we can return something simple like an integer, instead of a relation column.
We will finish section with a few complex queries.
A problem in using the transcript relation is that a student might take a class more than once. However, their grade from the last time they took the class is the one that counts.
When computing credits completed, we need to return a single tuple for each course that corresponds to the last valid grade for that class.
SELECT s.id , s.name , count(t.course_id) as coursescompleted FROM students s LEFT JOIN transcript t ON s.id = t.student_id AND t.grade IS NOT NULL AND t.grade <> 'I' WHERE NOT EXISTS ( SELECT 1 FROM transcript t2 WHERE t2.student_id = t.student_id AND t2.course_id = t.course_id AND (t2.year < t.year OR (t2.year = t.year AND t2.semester = 'Spring' AND t.semester='Fall')) ) GROUP BY s.id , s.name ;
FOR ALL Queries¶
What is we wanted to find students who have taken a class from all the professors in the database.
This is a complex query: we want to check that the set of professors for a student is equivalent to the set of all professors.
In relational algebra, this query would need two set subtractions.
We can represent this query logically as follows:
Find students who took a class from ALL professors in the database. Find students s such that there does not exist a professor f such that s did not take a class from f (or there does not exists a tuple in transcript for s and f)
SQL query will also require two subqueries:
SELECT s.id , s.name FROM students s WHERE NOT EXISTS (SELECT 1 FROM faculty f WHERE NOT EXISTS (SELECT 1 FROM transcript t, classes c WHERE t.student_id = s.id AND c.instructor_id = f.id AND c.course_id = t.course_id AND c.semester = t.semester AND c.year = t.year AND c.section = t.section));
Do we really need this level of complexity? Can we do this using a count?
Return each student if the number of different faculty they took classes with is equal to the number of different faculty in the database.
Let’s write this expression:
SELECT s.id , s.name FROM students s , transcript t , classes c WHERE s.id = t.student_id AND t.course_id = c.course_id AND t.semester = c.semester AND t.year = c.year AND t.section = c.section GROUP BY s.id , s.name HAVING count(DISTINCT c.instructor_id) = (SELECT count(*) FROM faculty) ;
Not only this query is simpler to write, it is likely much more efficient given it has no correlated subqueries.
Most queries that use IN or EXISTS can be rewritten using simple joins. Joins are much easier to optimize.
Set subtraction usually can be expressed using NOT IN or NOT EXISTS.
Using anonymous relations in the from clause may cause the optimizer to miss some optimizations. Simpler the query, the better it is.
There is a subtle difference on the syntax of the two statements:
Attribute NOT IN (select statement) NOT EXISTS (select statement)
For all queries usually require two NOT EXISTS.
SQL aggregates and outer joins are powerful constructs for formulating complex queries, even those involving some sort of negation.