如何找到最接近的结果？

Question

Lets say I have an array with fields A, B, C, D, E, F whose value can be a random number

I use these fields to make a query WHERE A = 'random number' AND B = 'random number'... AND F = 'random number'

If the result is empty then I remove last field from my array then run query WHERE A = 'random number' AND B = 'random number'... AND E = 'random number'

This process repeats until I get the result. Now, I found a new case where if I remove D field only and run the query, it returns the result.

I am not sure how to handle this case.

Answer 1

One option would be to count how many columns in the row equal to that number, and take the row with the most matches. MySQL has a neat ability to treat boolean values as 0s and 1s in numeric contexts, so this can be done relatively easily (but note the operator precedence and the required parentheses):

SELECT   *
FROM     mytable
ORDER BY (a = random_number) + 
         (b = random_number) + 
         (c = random_number) + 
         (d = random_number) + 
         (e = random_number) DESC
LIMIT    1

Answer 2

If you really want to remove the last column at each step, then:

select t.*
from t
order by (a, b, c, d, e) = ($a, $b, $c, $d, $e) desc,
         (a, b, c, d) = ($a, $b, $c, $d) desc,
         (a, b, c) = ($a, $b, $c) desc,
         (a, b) = ($a, $b) desc,
         a = $a desc
limit 1;

Note: If there is no match then this will still return a row. That seems to follow your pattern.

You can also use:

select t.*
from t
order by (a = $a) desc,
         (b = $b) desc,
         (c = $c) desc,
         (d = $d) desc
         (e = $e) desc
limit 1;