How to create fast database queries

Archive for April, 2010

Groups holding highest ranked items

Comments enabled. I *really* need your comment

Answering questions asked on the site.

Nate asks:

I know you've addressed similar issues related to the greatest-per-group query but this seems to be a different take on that.

Example table:

item_id group_id score
100 1 2
100 2 3
200 1 1
300 1 4
300 2 2

Each item may be in multiple groups. Each instance of an item in that group is given a score (how relevant it is the the group).

So given the data above, when querying for group 1 it should return items 200 and 300 (item 100's highest score is for group 2, so it's excluded).

The classical greatest-n-per-group problem requires selecting a single record from each group holding a group-wise maximum. This case is a little bit different: for a given group, we need to select all records holding an item-wise maximum.

Let's create a sample table:
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Written by Quassnoi

April 22nd, 2010 at 11:00 pm

Posted in MySQL

Hierarchical query in MySQL: limiting parents

with 6 comments

Answering questions asked on the site.

James asks:

Your series on hierarchical queries in MySQL is tremendous! I'm using it to create a series of threaded conversations.

I'm wondering if there is a way to paginate these results.

Specifically, let's say I want to limit the conversations to return 10 root nodes (parent=0) and all of their children in a query.

I can't just limit the final query, because that will clip off children. I've tried to add LIMITs to your stored functions, but I'm not getting the magic just right.

How would you go about doing this?

A quick reminder: MySQL does not support recursion (either CONNECT BY style or recursive CTE style), so using an adjacency list model is a somewhat complicated task.

However, it is still possible. The main idea is storing the recursion state in a session variable and call a user-defined function repeatedly to iterate over the tree, thus emulating recursion. The article mentioned in the question shows how to do that.

Normally, reading and assigning session variables in the same query is discouraged in MySQL, since the order of evaluation is not guaranteed. However, in the case we only use the table as a dummy recordset and no values of the records are actually used in the function, so the actual values returned by the function are completely defined by the function itself. The table is only used to ensure that the function is called enough times, and to present its results in form of a native resultset (which can be returned or joined with).

To do something with the logic of the function (like, imposing a limit on the parent nodes without doing the same on the child nodes), we, therefore, should tweak the function code, not the query that calls the functions. The only thing that matters in such a query is the number of records returned and we don't know it in design time.

Limiting the parent nodes is quite simple: we just use another session variable to track the number of parent branches yet to be returned and stop processing as soon as the limit is hit, that is the variable becomes zero.

Let's create a sample table and see how to do this:
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Written by Quassnoi

April 18th, 2010 at 11:00 pm

Posted in MySQL

Date ranges: overlapping with priority

Comments enabled. I *really* need your comment

Answering questions asked on the site.

Jason Foster asks:

We have a table of student registrations:

student_code course_code course_section session_cd
987654321 ESC102H1 Y 20085
998766543 ELEE203H F 20085

course_code and course_section identify a course, session_cd is an academic session, e. g. 20085, 20091, 20079.

The courses (stored in another table) have associated values for engineering design, complementary studies, etc., like that:

course_code course_section start_session end_session design science studies
ESC102H1 F 20071 20099 10 0 0
AER201Y1 Y 20059 NULL 0 0 30

, or like that:

In-house courses
course_code course_section student_code design science studies
ESC102H1 F 998766543 10 0 0

We are required by an external accreditation body to add up all of the units of engineering design, complementary studies, etc., taken by an individual student.

Where it gets really messy is that we have multiple data feeds for the associated values of courses. For example we have a set from the Registrar's Office, the Civil Department, our In-House version, etc.

The rule is that In-House beats Civil beats the Registrar's Office in the case of any duplication within the overlapping intervals.

The session_cd is of the form YYYY{1,5,9}.

Basically, we have three sets here.

To get the course hours for a given student we should find a record for him in the in-house set, or, failing that, find if the session is within the ranges of one of the external sets (Civil or Registrar). If both ranges contain the academic session the student took the course, Civil should be taken.

The first part is quite simple: we just LEFT JOIN students with the in-house courses and get the hours for the courses which are filled. The real problem is the next part: searching for the ranges containing a given value.

As I already mentioned in the previous posts, relational databases are not in general that efficient for the queries like that. It's easy to use an index to find a value of a column within a given range, but B-Tree indexes are of little help in searching for a range of two columns containing a given value.

However, in this case, the data domain of session_cd is quite a limited set. For each pair of session_start and session_end it is easy to create a set of all possible values between session_start and session_end.

The overlapping parts of the session ranges from the two sets will yields two records for each of the sessions belonging to the range. Of these two records we will need to take the relevant one (that is Civil) by using DISTINCT ON with the additional sorting on the source (Civil goes first).

Then we just join the relevant records to the subset of the students which does not have corresponding records in the in-house version.

Finally, we need to union this with the in-house recordset.
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Written by Quassnoi

April 7th, 2010 at 11:00 pm

Posted in PostgreSQL

Multiple attributes in a EAV table: GROUP BY vs. NOT EXISTS

with 5 comments

Answering questions asked on the site.

Andrew Stillard asks:

I have a store which will hold around 50,000 products in a products table. Each product will have 14 options, giving 700,000 options in total. These are held in an options table which is joined via the product id.

Users search for products based on the options via an Advanced Search menu.

The users need to be able to select multiple options upon which to query. I would normally use a JOIN if it was just the one option to select upon, but because its a variable number i thought it would be best to loop through the WHERE EXISTS statement.

The issue i have currently is that the query is taking a minimum of 18 seconds (And that was a query when the tables only had a fraction of the total products in).
If you can help us speed this up, or suggest an alternative idea that would be greatly appreciated.

The option table mentioned here is in fact an implementation of the EAV model in a relational database.

Each record basically contains 3 things: id of the product it describes; id of the option and the value of the given option for the given product. These fields represent entity, attribute and value, respectively.

This model is very easy to maintain and expand should the need arise: all we have to do to define an extra attribute is to add a record to the EAV table with the name and the value of the attribute. This is a DML operation rather than a DDL one.

However, this model has a serious drawback: we cannot efficiently search for two or more options at once. An index can only be defined on several fields from a single record, so we can only search for a single option using an index.

There are two approaches to writing a query which would search for the entities with the certain conditions on several attributes at once:

  1. For each attribute, find all entities for which the conditions on the given attribute hold, then aggregate the resulting entities, using COUNT(*) as a filter. The number of entity entries should be equal to the number of the attributes.
  2. Takes each entity and for each attribute, check if the condition holds.

The first approach uses a GROUP BY, the second one uses EXISTS.

Let's create a sample table and see which one is more efficient:
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Written by Quassnoi

April 2nd, 2010 at 11:00 pm

Posted in MySQL


with 6 comments

I had a vision tonight.

Image by nataliej

A huge, dark, grim figure approached me, seized me with its long bony arms and made me see all the vanity of the world we are living in.

Bloated database engines, useless ACID requirements, meaningless joins are now in the past for me.

I decided to move to NoSQL.

Where do I begin?

Written by Quassnoi

April 1st, 2010 at 1:00 pm

Posted in Miscellaneous