无法获得MySQL中选择查询

Question

两个领域我已经在我的route表四个字段

busid, routid, position and distance

我想显示busid and distance from a select query。我选择的查询是如下：

$endb = mysql_query("select case when a.position < b.position then a.busid when a.position > b.position then a.busid else null end as busid, a.distance as distance from (select busid,position from route where routid=$result2) a join (select busid,position from route where routid=$end) b on a.busid = b.busid") or die(mysql_error()); 

但是当我使用这个查询然后提示错误：unknown field distance in field list。普莱舍帮助我错过了什么

子查询

Answer 1

失踪距离

select 
    case 
     when a.position < b.position then a.busid 
     when a.position > b.position then a.busid 
     else null 
    end as busid, 
    a.distance as distance 
from (
    select busid, position, distance 
    from route 
    where routid=$result2 
) as a join (
    select busid, position 
    from route 
    where routid=$end 
) as b 
on a.busid = b.busid 

即使是一个更好的版本：

SELECT if (a.position <> b.position, a.busid, null) busid, a.distance 
FROM route a, route b 
WHERE a.busid = b.busid 
AND a.routid= $result2 
AND b.routid= $end 

Answer 2

你不应该在MySQL中from子句中使用子查询，除非必要。它们阻止优化器生成最佳查询计划。

一种更好的方式来编写查询：

select (case when a.position < b.position then a.busid 
      when a.position > b.position then a.busid 
     end) as busid, 
     a.distance 
from route a join 
    route b 
    on a.busid = b.busid and 
     a.routid = $result2 and b.routid = $end; 

您的特定问题，当然，是a.distance没有定义，因为它没有在子查询中定义。