how to get nearest value from database in mysql

select * 
from table 
order by abs(value - $myvalue)
limit 1

Assuming that you have a 10% tolerance (+/-) you could try something like:

select * from table 
where value >= ($myvalue * .9) and value <= ($myvalue * 1.1) 
order by abs(value - $myvalue) limit 1

Slightly updated stealing from others - this should return the nearest result in the assumed tolerance range. (Also, I just noticed the where was incorrect, apologies - now it should work).

(
select   *
from     table
where    value >= $myvalue
order by value asc
limit 1
)
union
(
select   *
from     table
where    value < $myvalue
order by value desc
limit 1
)
order by abs(value - $myvalue)
limit 1

This may look counter-intuitive but the speed will be greater than the other queries shown so far.

This is due to the fact that a greater than and less than query is quicker.

Then doing an ABS on two values is nothing.

This will give you the quickest return in a single query I can think of.

Doing an ABS on a whole table will be slow as it will scan the whole table.

Take the first value from the following:

select * from table order by abs(value - $myvalue);

Get the largest value similar to $val:

SELECT * FROM tab WHERE val <= $val ORDER BY val DESC LIMIT 1

Get the smallest value similar to $val:

SELECT * FROM tab WHERE val >= $val ORDER BY val LIMIT 1

Get the closest value similar to $val in either direction:

SELECT * FROM tab ORDER BY abs(val - $val) LIMIT 1

In my case, I was using the browsers geolocations and trying to find a closest city/state based on the coordinates I had in a table.

table structure:

id    zipcode    city_state   lat    lon
1     12345      Example, GA  85.3   -83.2

Recommend testing this vigorously before using -- probably needs some tweaks, but I came up with this as a start

SELECT city_state, 
   zipcode, 
   ( Abs( lat - -33.867886 ) 
     + Abs( lon - -63.987) ) AS distance
FROM   zipcodes 
ORDER  BY distance 
LIMIT  1;  

For laravel users:

$city = Zipcodes::selectRaw
    ('city_state, zipcode,  ( ABS( lat - ? ) + ABS( lon - ?) ) AS distance', [$lat, $lon])
        ->orderBy('distance')
        ->first();

echo $city->city_state

Hope this helps someone someday.

SELECT * FROM table1 ORDER BY ABS(value - '$myvalue') LIMIT 1 

SELECT number, ABS( number - 2500 ) AS distance
FROM numbers
ORDER BY distance
LIMIT 6

Selecting closest values in MySQL

Try this:

SELECT *,abs((columnname -Yourvalue)) as near
  FROM table
 WHERE order by near limit 0,1

Unfortunately, I think your database will probably do a full table scan for solutions that involve abs, so they will be (very) slow once your table grows. A fast-running solution may be found in this earlier thread specific to PostgreSQL.

Read this page http://dev.mysql.com/doc/refman/5.1/en/mathematical-functions.html#function_round

but your select would look like this

select value from table where ROUND(value) = $myvalue