How to return rows listed in descending order of COUNT(*)?

Question

I have a table called foo with these fields:

- id

- type

- parentId

I want to select a list of parent IDS, in the descending order of their COUNT(*) of how many times they appear in the table. Something like this:

SELECT DISTINCT parentId FROM `foo` 
ORDER BY (COUNT(parentId) DESC where parentId = parentId)

How can this be done in the most efficient way and putting the least load on the server?

There can be thousands-hundreds of thousands of records in the table, so manually going through each record is not acceptable..

Answer 1

Simply by applying a GROUP BY clause, and assuming you have an index , FOREIGN KEY, or PRIMARY KEY on parentId, the performance should be quite good. (parentId looks like it is likely a FORIEGN KEY, so be sure to define the constraint to enforce indexing).

SELECT `parentId`
FROM `foo`
GROUP BY `parentId`
ORDER BY COUNT(*) DESC

Answer 2

How can this be done in the most efficient way and putting the least load on the server?

The key is the the most efficient way.

Not a Count() for sure, but most efficient is... to read a field, which you are storing the Count result. You can update it with a trigger or after insert.

Especially when

There can be thousands-hundreds of thousands of records in the table