CODEWITHSUNDEEP
sqlpostgresqlrankpostgresql-8.3
Arthur
Arthur

Reputation: 3418

Implementing a total order ranking in PostgreSQL 8.3

The issue with 8.3 is that rank() is introduced in 8.4.

Consider the numbers [10,6,6,2].
I wish to achieve a rank of those numbers where the rank is equal to the row number:

rank | score
-----+------
1    | 10
2    | 6
3    | 6
4    | 2

A partial solution is to self-join and count items with a higher or equal, score. This produces:

1    | 10
3    | 6
3    | 6
4    | 2

But that's not what I want.
Is there a way to rank, or even just order by score somehow and then extract that row number?

Upvotes: 0

Views: 1042

Answers (3)

Erwin Brandstetter
Erwin Brandstetter

Reputation: 656724

If you want a row number equivalent to the window function row_number(), you can improvise in version 8.3 (or any version) with a (temporary) SEQUENCE:

CREATE TEMP SEQUENCE foo;
    
SELECT nextval('foo') AS rn, *
FROM   (SELECT score FROM tbl ORDER BY score DESC) s;

db<>fiddle here
Old sqlfiddle

The subquery is necessary to order rows before calling nextval().

Note that the sequence (like any temporary object) ...

  • is only visible in the same session it was created.
  • hides any other table object of the same name.
  • is dropped automatically at the end of the session.

To use the sequence in the same session repeatedly run before each query:

SELECT setval('foo', 1, FALSE);

Upvotes: 2

Daniel V&#233;rit&#233;
Daniel V&#233;rit&#233;

Reputation: 61526

There's a method using an array that works with PG 8.3. It's probably not very efficient, performance-wise, but will do OK if there aren't a lot of values.

The idea is to sort the values in a temporary array, then extract the bounds of the array, then join that with generate_series to extract the values one by one, the index into the array being the row number.

Sample query assuming the table is scores(value int):

SELECT i AS row_number,arr[i] AS score
 FROM (SELECT arr,generate_series(1,nb) AS i
   FROM (SELECT arr,array_upper(arr,1) AS nb
     FROM (SELECT array(SELECT value FROM scores ORDER BY value DESC) AS arr
     ) AS s2
   ) AS s1
 ) AS s0

Upvotes: 1

Ihor Romanchenko
Ihor Romanchenko

Reputation: 28541

Do you have a PK for this table?

Just self join and count items with: a higher or equal score and higher PK.

PK comparison will break ties and give you desired result.

And after you upgrade to 9.1 - use row_number().

Upvotes: 0

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