Sort by match of LIKE
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16-10-2019 - |
Question
I wondering how I can implement SQL to get results sorted by best match of a like predicate. I have 100K articles in the database and when user call some items by part of name. I want to show the results ordered by the best match of asked query.
I'll try to describe that by pseudo code
select
*
from
articles
where item_nale like '%'+@user_input +'%'
order by "BEST MATCH"
Solution
It isn't possible to calculate relevance with the LIKE predicate. For SQL Server (which from previous questions I believe is your platform?) you'll want to look at full-text search which supports scoring/ranking results by relevance.
OTHER TIPS
Since all your matches have to match the LIKE pattern in order to be included, the simple thing to do is assume that shorter values for the column you're matching are "better" matches, as they're closer to the exact value of the pattern.
ORDER BY LEN(item_nale) ASC
Alternatively, you could assume that values in which the match pattern appear earlier are "better" matches.
ORDER BY PATINDEX('%' + @user_input + '%', item_nale) ASC
Or you could combine the two; look for earliest matches first, and within those, prefer shorter values.
ORDER BY PATINDEX('%' + @user_input + '%', item_nale) ASC, LEN(item_nale) ASC
It's not as sophisticated as what full-text indexing does, and it only accounts for matching on a single column, but it gives decent enough results.
Generate the Levenshtein function, and order by this field.
If you have this function, it could be:
select
field1,
field2,
field3,
fieldN,
item_nale,
MyLevenshteinFunction(item_nale,@user_input) LevenshteinDistance
from (
select
field1,
field2,
field3,
fieldN,
item_nale,
LevenshteinDistance
from
articles
where item_nale like '%'+@user_input +'%'
) as result
order by LevenshteinDistance DESC