How to create fast database queries

Archive for July, 2009

MySQL: counting items in pairs

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From Stack Overflow:

I currently have an item table, and a pair table.

The pair table simply contains two columns, which contain the PRIMARY KEY from the item table. There is an index on id1, id2 on pair.

A common query is to find a number of items that are featured in the least number of pairs:

SELECT  id, COUNT(*) AS count
FROM    item i
pair p
ON      (p.id1 = i.id OR p.id2 = i.id)
count, RAND()

, but the query is horible performance wise.

Is there a better query, and/or data structure for this type of thing?

MySQL is not very good in optimizing OR conditions.

It's capable of doing index_merge, which would UNION the results of two indexes, however, as documentation states, this access method works only when comparing the fields against the constants.

This method, therefore, can be used to serve an OR condition neither in a JOIN nor in a subquery.

Let's create sample tables and see the execution plan for the query:

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Written by Quassnoi

July 31st, 2009 at 11:00 pm

Posted in MySQL

PostgreSQL: making duplicate records unique

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From Stack Overflow:

In an existing application, I have a table which has no primary key, which (rarely) has duplicate rows in it.

For integration with another system, I need to add a column to the table that makes these duplicate rows unique.

The table is essentially:

txn# detail# amount

I could just use an incrementing seq# for every existing row, however, the application changes will be to take the existing almost key (basically (txn#, detail#)), and increment the sequence number for this combination

So, e. g., I'll have two rows for (txn#, detail#) = (1, 1), and have seq# 1 for the first and 2 for the second.

If (txn#, detail#) = (513, 44) has 3 duplicate rows, these would have seq# 1, 2 and 3 appropriately.

We just need to update the table with a condition that would distinguish between the rows somehow.

Fortunatly, every PostgreSQL table has an implicit primary key, ctid.

In fact, it's the pointer to the place in the datafile where the row data resides.

In new PostgreSQL 8.4, we can employ a window function to assign a seq to each row.

This script fills the table with duplicate values (two duplicates for each (txn, detail) pair):

CREATE TABLE t_dup (txn INT NOT NULL, detail INT NOT NULL, seq INT);

CREATE INDEX ix_dup_txn_detail ON t_dup (txn, detail);

INTO    t_dup (txn, detail, seq)
FROM    generate_series(1, 100) t,
generate_series(1, 1000) d,
generate_series (1, 2) s;

ANALYZE t_dup;

Here's the query to update it:

SET     seq = rn
FROM    (
SELECT  ctid, ROW_NUMBER() OVER (PARTITION BY txn, detail) AS rn
FROM    t_dup
) d
WHERE   t_dup.ctid = d.ctid

This works perfectly and completes in 9 seconds, but, unfortunately, it's unavailable in older versions of PostgreSQL.

For PostgreSQL 8.3 and below, we need to use a subselect with a COUNT(*).

Fortunately, ctids are comparable, and we can use less than (<) operator on them to build an ordered set.

This is less efficient, but still works (if the keys are almost unique, i. e. there are not many duplicates of each key):

SET	seq =
FROM	t_dup di
WHERE	di.txn = d.txn
AND di.detail = d.detail
AND di.ctid &lt;= d.ctid

This takes a little bit longer (15 seconds), however still works.

Written by Quassnoi

July 30th, 2009 at 11:00 pm

Posted in PostgreSQL

PostgreSQL 8.4: efficient MIN(COUNT(*))

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Answering questions asked on the site.

Frances asks:

I have a table user_views which contains pages viewed by users.

How do I select the user that had the most and the fewest number of page views?

If you need them at the same time, you'll just have to group by myuser, then order by COUNT(*).

With new PostgreSQL 8.4, it's possible to do this in one CTE, to avoid double evaluation:

WITH    q AS
SELECT  myuser, COUNT(*)
FROM    user_pages
FROM    (
SELECT  'max'::TEXT AS which, myuser, count
FROM    q
count DESC
) qmax
FROM    (
SELECT  'min'::TEXT AS which, myuser, count
FROM    q
) qmin

However, it's takes quite a long time if you table is big.

This is probably an answer to the question you asked.

But as a bonus I'd like to tell how to optimize the query for the cases when you need just the user with minimal count of page views.

The main idea here that we should accumulate the least value of COUNT(*) calculated so far, and stop counting when we reach this threshold. This will save us some row lookups and improve the query.

Let's create a sample table:

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Written by Quassnoi

July 29th, 2009 at 11:00 pm

Posted in PostgreSQL

SQL Server: random records avoiding CTE reevaluation

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From Stack Overflow:

I need to get 5 random records from a table plus a further record based on data from the users preferences.

How can I combine the two statements whilst ensuring that no results are repeated (i. e. the favourite is not present in the 5 random records)?

We should select the 6th record so that it would satisfy two conditions: first, it should be preferred by user, second, it should not be among the 5 random records selected earlier.

The problem with random records it that they are, um, random. We usually cannot reproduce 5 random records to apply further conditions on them.

That's where the CTEs would come handy, but there's a little problem.

A CTE, as Microsoft understands them, is just an alias for an online view.

Therefore, each time a CTE is used in a query, it may (or may not) be reevaluated.

It's not a problem for a plain CTEs which are built out of static columns.

However, this is totally a problem when we try to UNION two queries using an undeterministic CTE (like, ordered by NEWID()), as we would do in this case.

Let's create a sample table and see what happens:

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Written by Quassnoi

July 28th, 2009 at 11:00 pm

Posted in SQL Server

Dynamic pivot

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From Stack Overflow:

I have a table and want to transpose its rows to columns, similar to a pivot table but without summarising.

For example I have the following tables:

Question (QuestionID, QuestionText)
Response (ResponseID, ResponseText, QuestionID)

Basically I want to be able to create a dynamic table something like:

Question 1 Question 2 Question 3
Response 1.1 Response 1.2 Response 1.3
Response 2.1 Response 2.2 Response 2.3
Response 3.1 Response 3.2 Response 3.3
Response 4.1 Response 4.2 Response 4.3

The main requirement would be I don't know at design time what the question text will be.

First, let's put it straight: SQL is not a right tool to do this.

SQL operates on sets. An SQL query is a functions that returns a set. And the column layout is an implicit property of a set. It should be defined in design time. This is the way SQL works.

However, there are several approaches that can simplify the task. One of them is using SQL Server's XML abilities.

To arrange the answers in a table, we need to make the following:

  1. Select all questions. This rowset will define the columns of our table.
  2. Select the row numbers for all answers for the question that has the most answers. This rowset will define the rows of our table. The number of rows wil be exactly as the number of answers to the most answered question.
  3. Assign the ordinal number to each answer on any question. This is done using ROW_NUMBER(). This value will uniquely define the position of the answer in a table.
  4. For each row (taken from the rowset made on step 2), take all the questions and LEFT JOIN the answers that need to be put into this row, using the ordinal number calculated on the step 3 as a LEFT JOIN condition. This will put either the answer or an empty string into the corresponding cell.
  5. Transform the rowset retrieved on step above into an XML string made out of <td>'s and enclosed into a <tr>
  6. Transform the resulting rowset into an XML made out of the XML's from the previous step and enclosed into a <table>

Let's create a sample table and see how can we do this:

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Written by Quassnoi

July 27th, 2009 at 11:00 pm

Posted in SQL Server

MySQL: selecting rows before and after filtered one

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From Stack Overflow:

I have a basic MySQL table, terms, comprised of an id and term field.

I want to create an alphabetically sorted dictionary index (in the literal sense), that would list ten 10 terms above the selected term, and 20 below it

An example of this could be found on Urban Dictionary where on the left column you see the current term highlighted, and a number of terms above it, and some below, all sorted alphabetically.

As we all know, MySQL doesn't support a ROW_NUMBER() or a similar function so we end up resorting to user variables and sub-selects.

I also cannot create a view with user defined variables because MySQL doesn't allow that.

MySQL indeed doesn't support ROW_NUMBER() and this function indeed can be emulated using session variables:

However, it will be an overkill in this case, since a more simple and more efficient set-based solution using LIMIT can be used instead.

Let's create a sample table and see how:

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Written by Quassnoi

July 25th, 2009 at 11:00 pm

Posted in MySQL

MySQL: grouping continuous ranges

with 4 comments

From Stack Overflow:

Given the following table:

ID State Date
12 1 2009-07-16 10:00
45 2 2009-07-16 13:00
67 2 2009-07-16 14:40
77 1 2009-07-16 15:00
89 1 2009-07-16 15:30
99 1 2009-07-16 16:00

, how can I group by the field state, while still maintaining the borders between the state changes?

I need MIN(id), MIN(date) and COUNT(*) for each group, and this is expected:

ID State Date Count
12 1 2009-07-16 10:00 1
45 2 2009-07-16 13:00 2
77 1 2009-07-16 15:00 3

Though a pure set-based solution exists, in MySQL, this can be easily done using session variables, which is more efficient.

Let's create a sample table:

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Written by Quassnoi

July 24th, 2009 at 11:00 pm

Posted in MySQL

PostgreSQL: last 12 weeks of a year

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From Stack Overflow:

I have a query like this:

SELECT  EXTRACT(WEEK FROM j.updated_at) as "week",  count(j.id)
FROM    jobs
WHERE   EXTRACT(YEAR FROM j.updated_at)=2009
EXTRACT(WEEK FROM j.updated_at)

, which works fine, but I only want to show the last 12 weeks.

LIMIT 12 works, but only gives me the first twelve and I need the order to be in sequential week order (i. e. not reversed) for charting purposes.

To select last 12 rows in ascending order it's enough to select first 12 rows in descending order and resort them in a subquery:

FROM    (
SELECT  EXTRACT(week FROM updated_at) AS week, COUNT(*) AS cnt
FROM    jobs
WHERE   EXTRACT(year FROM updated_at) = 2008
week DESC
) q

However, this is rather inefficient. This query selects and aggregate all the year data just to fetch 12 last weeks.

More than that, EXTRACT(year FROM updated) is not a sargable predicate, and all table rows (or index rows) need to be examined.

On a sample table of 1,000,000 rows, this query runs for more that 3.5 seconds:

FROM    (
SELECT  EXTRACT(week FROM updated) AS week, COUNT(*) AS cnt
FROM    t_week
WHERE   EXTRACT(year FROM updated) = 2008
week DESC
) q

week cnt
41 3584
42 3467
43 3499
44 3535
45 3511
46 3621
47 3529
48 3500
49 3415
50 3536
51 3504
52 3479
12 rows fetched in 0.0004s (3.5843s)
Sort  (cost=58517.44..58517.47 rows=11 width=16)
  Sort Key: (date_part('week'::text, t_week.updated))
  ->  Limit  (cost=58517.11..58517.14 rows=11 width=8)
        ->  Sort  (cost=58517.11..58517.14 rows=11 width=8)
              Sort Key: (date_part('week'::text, t_week.updated))
              ->  HashAggregate  (cost=58516.75..58516.92 rows=11 width=8)
                    ->  Seq Scan on t_week  (cost=0.00..58491.75 rows=5000 width=8)
                          Filter: (date_part('year'::text, updated) = 2008::double precision)

However, this query can be easily improved. For each year, we can easily calculate the beginning and the end of each of last 12 weeks, and use these values in a more index-friendly query.

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

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Written by Quassnoi

July 23rd, 2009 at 11:00 pm

Posted in PostgreSQL

Hierarchial queries in MySQL: identifying trees

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Continuing on the theme of hierarchical queries in MySQL:

Assume we have a table with hierarchical structure like this:

18 rows fetched in 0.0003s (0.0137s)

We have two trees here: one starting from 1, another one starting from 2.

The problem is: given any item, we should identify the whole tree this item belongs to, and return the whole tree in the hierarchical order.

This also can be easily done using hierarchical queries in MySQL.

In this article: Hierarchical queries in MySQL I shown how to implement a function that returnes tree items in correct order, being called sequentially.

This function is reentrable and keeps its state in session variables, one of which, @start_with, defines the parent element for the tree we want to build.

We have two problems here:

  1. Given an item, define a root of the tree it belongs to
  2. Build a whole tree, starting from the root

The first problem can be solved by iterating the linked list backwards, starting from the variable given as an input.

This article:

describes how to do it in great detail, that's why I'll just put a query here.

Let's create the table described above:

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Written by Quassnoi

July 22nd, 2009 at 11:00 pm

Posted in MySQL

PostgreSQL: round-robin order

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From Stack Overflow:

I have a bunch of URLs stored in a table waiting to be scraped by a script.

However, many of those URLs are from the same site. I would like to return those URLs in a site-friendly order (that is, try to avoid two URLs from the same site in a row) so I won't be accidentally blocked by making too many HTTP requests in a short time.

The database layout is something like this:

create table urls (
site varchar, -- holds e.g. www.example.com or stockoverflow.com
url varchar unique

Example result:

FROM    urls

To solve this task, we just need to assign a number to an URL within each site, and order by this number first, then by site, like this:

number site url
1 example.com http://www.example.com/some/file
1 stackoverflow.com http://stackoverflow.com/questions/ask
1 use.perl.org http://use.perl.org/
2 example.com http://www.example.com/some/other/file
2 stackoverflow.com http://stackoverflow.com/tags

Good news: in new PostgreSQL 8.4, we just can use a window function to do this:

FROM    urls
1000 rows fetched in 0.0169s (0.0178s)

Yes, that simple.

Bad news: in PostgreSQL 8.3 and below, we still need to hack.

Let's create a sample table and see how:

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Written by Quassnoi

July 21st, 2009 at 11:00 pm

Posted in PostgreSQL