SQL - 返回具有最多列匹配的行

Question

Consider this table:

Name   Color1    Color2    Color3   Prize
-----------------------------------------
Bob    Red       Blue      Green    Stapler
Bob    Red       Blue      NA       Pencil
Bob    Red       NA        NA       Lamp
Bob    Red       NA        NA       Chair
Bob    NA        NA        NA       Mouse Pad

Bob has 3 colors. This is what I'm trying to get:

(#1) If Bob has Red, Blue, Green (match 3) ....  Return Stapler
(#2) If Bob has Red, Blue, Purple (match 2) ...  Return Pencil
(#3) If Bob has Red, Orange, Purple (match 1) .  Return Lamp AND Chair rows
(#4) If Bob has Brown, Pink, Black (match 0) ..  Return Mouse Pad

Colors would only appear in their own Columns. So in the example above, Red would only be in Color1 column and never in Color2 or Color3. Black would only be in Color3 and never in Color1 or Color2. Etc...

I only want the row(s) with the most matches.

I would really prefer not to do this with 4 separate SELECT statements and check each time if they return a row. This is how I do it in a stored procedure and it's clunky.

How can I do this in 1 SQL statement? Using Oracle if that matters...

Thanks!!!

Answer 1

7 Years later, I think I'll answer my question with what I came up with. Thanks to Ponder Stibbons for his answer to get me started. Posting here in case anyone else stumbles upon this.

Ponder's answer will return multiple rows. For example, if you match on Red and Blue, it will return both Stapler and Pencil. Only Pencil should be returned.

So, in my solution, exact matches get 2 points for the ranking and "NA" for the column gets you 1. This bumps up the NA rows and creates "catch all" scenarios for the non-matching columns, which was my intent.

select *
  from (
    select t.*, rank() over (order by 
          decode(color1, 'Red', 2, 0) 
        + decode(color2, 'Blue', 2, 0) 
        + decode(color3, 'Purple', 2, 0) 
        + decode(color1, 'NA', 1, 0) 
        + decode(color2, 'NA', 1, 0) 
        + decode(color3, 'NA', 1, 0)         
         desc) rnk
      from t
      where t.name = 'Bob'
  )
  where rnk = 1
  

Answer 2

You can count the matches using case and then fetch the row with the most. In Oracle, this uses a few subqueries:

select name
from (select t.*
      from (select t.*,
                   ((case when color1 in ('Red', 'Blue', 'Green') then 1 else 0 end) +
                    (case when color2 in ('Red', 'Blue', 'Green') then 1 else 0 end) +
                    (case when color3 in ('Red', 'Blue', 'Green') then 1 else 0 end) +
                   ) as numMatches
            from table t
           ) t
      order by nummatches desc
     ) t
where rownum = 1;

Answer 3

I only want the row(s) with the most matches.

You can use function rank() for that:

SQLFiddle

select name, color1, color2, color3, prize 
  from (
    select t.*, rank() over (order by decode(color1, 'Red', 1, 0) 
        + decode(color2, 'Blue', 1, 0) + decode(color3, 'Green', 1, 0) desc) rnk
      from t)
  where rnk = 1

This returns row or rows with most matches.