Get most approximated value in SQL

Question

I have a table with rows fruit and price for example:

Apple - 1

Melon - 3

Orange - 10

Pear - 5

And I want to make a query to get the fruit with price more approximated to the value I will pass.

Something like:

SELECT * 
FROM fruit
WHERE condition

Any help will be appreciated.

Answer 1

Is this what you want?

SELECT f.* 
FROM fruit f
ORDER BY ABS(f.price - 7)
LIMIT 1;

This gets the fruit whose price is closest to 7. Of course, "7" is a parameter that could be any value.

Answer 2

Well, the comments pretty much have the answer:

SELECT * FROM fruit  
WHERE abs(price-@searchprice) =
  (SELECT abs(price-@searchprice) 
  FROM fruit ORDER BY abs(price-@searchprice) LIMIT 1)

The subquery finds the price difference from the entered price (@searchprice - change the parameter name style according to the needs of your database/driver combination) to the nearest 1 actual price

This is then used to look up all fruit that has a difference equal to that price difference. We do this so that any items tied on price are returned - in your example data if searchprice was 2 then Apple and Melon are both equally near and should be returned