Order by Maximum condition match

Question

Please help me to create a select query which contains 10 \'where\' clause and the order should be like that: the results should be displayed in order of most keywords(where

Answer 1

SELECT *
  FROM (SELECT (CASE WHEN cond1 THEN 1 ELSE 0 END +
                CASE WHEN cond2 THEN 1 ELSE 0 END +
                CASE WHEN cond2 THEN 1 ELSE 0 END +
                ...
                CASE WHEN cond10 THEN 1 ELSE 0 END
               ) AS numMatches,
               other_columns...
          FROM mytable
       ) xxx
 WHERE numMatches > 0
 ORDER BY numMatches DESC

Answer 2

EDIT: This answer was posted before the question was modified with a concrete example. Marcelo's solution addresses the actual problem. On the other hand, my answer was giving priority to matches of specific fields.

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You may want to try something like the following, using the same expressions in the ORDER BY clause as in your WHERE clause:

SELECT    *
FROM      your_table
WHERE     field_1 = 100 OR
          field_2 = 200 OR
          field_3 = 300
ORDER BY  field_1 = 100 DESC,
          field_2 = 200 DESC,
          field_3 = 300 DESC;

I've recently answered a similar question on Stack Overflow which you might be interested in checking out:

• Is there a SQL technique for ordering by matching multiple criteria?

Answer 3

There are many options/answers possible. Best answer depends on size of the data, non-functional requirements, etc.

That said, what I would do is something like this (easy to read / debug):

  Select * from 
    (Select *, iif(condition1 = bla, 1, 0) as match1, ..... , match1+match2...+match10 as totalmatchscore from sourcetable
    where 
      condition1 = bla or
      condition2 = bla2
      ....) as helperquery
    order by helperquery.totalmatchscore desc

Answer 4

I could not get this to work for me on Oracle. If using oracle, then this Order by Maximum condition match is a good solution. Utilizes the case when language feature