如何从mysql中的数据库中获取最接近的值

Question

I am using mySQL and CodeIgniter . I have some floating point numbers in my database such as

• 8.3456
• 8.5555
• 4.5556

I want to...

SELECT * FROM table WHERE value = $myvalue

but I can't use value = $myvalue in my SELECT query because $myvalue is not exactly equal to database values. I need to get the nearest value to $myvalue from database.

If $myvalue is 5 I want to select the value 4.5556 .

How can I do this in mySQL?

Answer 1

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

Answer 2

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).

Answer 3

(
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.

Answer 4

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

Answer 5

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.

Answer 6

从以下内容中获取第一个值：

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

Answer 7

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

Selecting closest values in MySQL

Answer 8

Try this:

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

Answer 9

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

Answer 10

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 

Answer 11

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 .