Question

When I perform the following query

select * from table where c2 = 11 or c3 = 15 or c7 = false

I’m getting this result

| c1 | c2 | c3 | c4  | c5  | c6      | c7     | c8        |
| 24 | 11 | 15 | NNN | NNN |         | true   | false     |
| 28 | 11 | 13 | NNN | NNN |         | true   | false     |
| 26 | 11 | 15 | NNN | NNN | wwwww   | false  | false     |
| 25 | 11 | 2  | NNN | NNN | qqqq    | false  | false     |
| 33 | 23 | 31 | NNN | NNN |         | false  | false     |
| 31 | 23 | 15 | NNN | NNN |         | false  | false     |
| 31 | 23 | 15 | NNN | NNN |         | true   | false     |
| 25 | 11 | 23 | NNN | NNN | qqqqw2  | false  | false     |
| 29 | 11 | 22 | NNN | NNN |         | true   | false     |

I’m trying to order this by search coincidences, like this:

| c1 | c2 | c3 | c4  | c5  | c6      | c7    | c8        |
| 26 |[11]|[15]| NNN | NNN | wwwww   |[false]| false     |
| 24 |[11]|[15]| NNN | NNN |         | true  | false     |
| 25 |[11]| 2  | NNN | NNN | qqqq    |[false]| false     |
| 31 | 23 |[15]| NNN | NNN |         |[false]| false     |
| 25 |[11]| 23 | NNN | NNN | qqqqw2  |[false]| false     |
| 28 |[11]| 13 | NNN | NNN |         | true  | false     |
| 29 |[11]| 22 | NNN | NNN |         | true  | false     |
| 31 | 23 |[15]| NNN | NNN |         | true  | false     |
| 33 | 23 | 31 | NNN | NNN |         |[false]| false     |

So the row that has the 3 options will be on top
then the all the rows that matches at least 2 optiones (in no specific order)
and last all the rows that matches 1 option,

Is there a way to achieve this order? because the other option I have is doing combination of options with and statement and then use union with all the results, but it seems to be a lot more

Thanks in advance

Answer 1

You need to add an order by clause:

order by (case when c2 = 11 then 1 else 0 end) +
         (case when c3 = 15 then 1 else 0 end) +
         (case when c7 = false then 1 else 0 end) desc

(SQL Fiddle example)