Common mistakes in solving SELECT statement exercises

Computer Firm Database

Get pairs of PC models with identical speeds and the same RAM capacity. Each resulting pair should be displayed only once, i.e. (i, j) but not (j, i).Result set: model with the bigger number, model with the smaller number, speed, and RAM.

Here is a solution that occurs quite often:

Solution 1.12.1

Console

Execute

SELECT MAX(model) AS 'model', MIN(model) AS 'model', speed, ram
FROM PC
GROUP BY speed, ram
HAVING MAX(model) > MIN(model);

SELECT MAX(model) AS 'model', MIN(model) AS 'model', speed, ram FROM PC GROUP BY speed, ram HAVING MAX(model) > MIN(model);

    Scheme

I really have no idea why some users output just the maximum and minimum model for each matching pair of speed and ram values. Maybe the correct result returned by such a query for the main database is misleading.

In this task, all models have to be ordered, not just the maximum and the minimum one. Extreme values are mentioned just for the sake of unambiguity, to make it clear a pair of models has to be displayed once, e. g.

Say, if three models 1122, 1121, and 1135 have identical characteristics, the result should look as follows:

The solution presented below is almost correct, albeit somewhat cumbersome.

Solution 1.12.2

Console

Execute

SELECT P.model, L.model, P.speed, P.ram
FROM PC P JOIN 
     (SELECT speed, ram
      FROM PC
      GROUP BY speed, ram
      HAVING SUM(speed)/speed = 2 AND 
             SUM(ram)/ram = 2 
      ) S ON P.speed = S.speed AND 
             P.ram = S.ram JOIN 
      PC L ON L.speed = S.speed AND 
              L.ram = S.ram AND 
              L.model < P.model;

SELECT P.model, L.model, P.speed, P.ram FROM PC P JOIN      (SELECT speed, ram       FROM PC       GROUP BY speed, ram       HAVING SUM(speed)/speed = 2 AND              SUM(ram)/ram = 2       ) S ON P.speed = S.speed AND              P.ram = S.ram JOIN       PC L ON L.speed = S.speed AND               L.ram = S.ram AND               L.model < P.model;

    Scheme

Here, the subquery S gets the unique pairs of characteristics (speed, memory) coinciding for two computers (SUM (speed)/speed = 2) - the sum of identical values divided by this value yields the number of PCs. However, one could just as well use the following HAVING clause in this case:

HAVING COUNT(*) = 2

The subquery is joined to the PC table PC using this pair of characteristics twice. Thereby, the second join is performed just to order the models (L.model <P.model).

The mistake of this solution is, there may be more than two PCs with identical characteristics. In this case, the query under consideration won’t output any of such models.

In spite of the fact this solution can be corrected easily, it’s preferable to rewrite it in a less redundant form.

One more mistake typical for this exercise is caused by the possible presence of identical computer models in the PC table. Thus, when displaying pairs of PCs you need to remove duplicates.

T&S