我想在这里给出我的答案，这样人们就可以尝试一下，我认为MySql的解决方案更容易理解:

set @countVal =  (select count(*) from STATION);
/**
499/2 = 249,5 -> 250 -- ceil
499/2 = 249,5 + 1 = 250,5 -- floor 250

500/2 = 250 -- ceil 250
       = 250 + 1 = 251 -- flor 251
**/       
set @ceilVal = ceil(@countVal/2);
set @floorVal = floor( (@countVal/2) + 1);
SELECT ROUND(AVG( latitude ),4) FROM
(SELECT @lineNum:= @lineNum + 1 as id, 
lat_n as latitude
FROM STATION s, ( SELECT @lineNum :=0 ) pivot
ORDER BY lat_n) as a
WHERE id IN ( @ceilVal, @floorVal );

对于任何使用concat_ws函数中的@variable来获得连接值的人，不要忘记用空值重新初始化它。否则，它可以对同一会话使用旧值。

Set @Ids = '';

select 
  @Ids := concat_ws(',',@Ids,tbl.Id),
  tbl.Col1,
  ...
from mytable tbl;

声明: SET @a = 1; 用法: INSERT INTO ' t ' (' c ') VALUES (@a);

设置值

 declare @Regione int;   
 set @Regione=(select  id from users
 where id=1) ;
 select @Regione ;

我想在这里给出我的答案，这样人们就可以尝试一下，我认为MySql的解决方案更容易理解:

set @countVal =  (select count(*) from STATION);
/**
499/2 = 249,5 -> 250 -- ceil
499/2 = 249,5 + 1 = 250,5 -- floor 250

500/2 = 250 -- ceil 250
       = 250 + 1 = 251 -- flor 251
**/       
set @ceilVal = ceil(@countVal/2);
set @floorVal = floor( (@countVal/2) + 1);
SELECT ROUND(AVG( latitude ),4) FROM
(SELECT @lineNum:= @lineNum + 1 as id, 
lat_n as latitude
FROM STATION s, ( SELECT @lineNum :=0 ) pivot
ORDER BY lat_n) as a
WHERE id IN ( @ceilVal, @floorVal );

使用set或select

SET @counter := 100;
SELECT @variable_name := value;

例子:

SELECT @price := MAX(product.price)
FROM product