SQL order_by表达式

Question

Let's say I have a column that contains a float value between 1-100.

I'd like to be able to turn that value into a less precise integer between 1-10 then order the results on this new value.

It may seem odd to want to make the ordering less precise but the SQL statement is ordered by 2 columns and if the first is too precise then the 2nd order column would have no weight.

Essentially I would like to group my first order by into 10 groups and then order each of those groups by another column.

SELECT "sites".* FROM "sites" ORDER BY "sites"."rating" DESC, "sites"."price" ASC LIMIT 24 OFFSET 0

edit: This is a rails app using postgresql

Answer 1

What SQL is that? Use a divide function and, if necessary, round it.

In MySQL look at the DIV command. I don't have the means to test this right now, but it might help point you in the right direction:

SELECT "sites".* FROM "sites" ORDER BY "sites"."rating" DIV 2 DESC, ...

Answer 2

SELECT "sites".* 
FROM "sites" 
ORDER BY FLOOR("sites"."rating"/10) DESC, "sites"."price" ASC 
LIMIT 24 OFFSET 0

Answer 3

Use the round function as follows:

    Select sites.* 
    from sites
    order by round(sites.rating/10, -1) desc, sites.price desc

This will convert a value of 99 to 10.