Mysql group by 与 order by 冲突记录

Question

一张表包含经销商的多个地址（纬度和经度），但我想获取最近的经销商

这是一个查询，哪个返回适用于 order by 和 wheredealer=1 的单个经销商，但我需要多个唯一经销商。

检查查询和第一张图片结果

SELECT 

  *, ROUND(
    (
      3959 * ACOS(
        COS(RADIANS(41.355724)) * COS(
          RADIANS(adzip.dealer_zipcode_latitude)
        ) * COS(
          RADIANS(adzip.dealer_zipcode_longitude) - RADIANS(- 87.607332)
        ) + SIN(RADIANS(41.355724)) * SIN(
          RADIANS(adzip.dealer_zipcode_latitude)
        )
      )
    ), 2
  ) AS dealer_distance 
FROM
  ad_dealers_zipcodes AS adzip 
WHERE adzip.dealer_zipcode_dealer_id = 1 #GROUP BY adzip.dealer_zipcode_dealer_id
ORDER BY dealer_distance ASC 

当我通过 adzip.dealer_zipcode_dealer_id 添加组时，查询返回第 4 条记录，但我需要第 1 条结果记录，dealer_distance 为 13.80

Answer 1

实际上，我认为您应该在查询中添加一个级别，以便 group by 语句按您的意愿工作，如下所示：

SELECT *, min(dealer_distance) AS min_dealer_distance  FROM 
(
    SELECT 

      *, ROUND(
        (
          3959 * ACOS(
            COS(RADIANS(41.355724)) * COS(
              RADIANS(adzip.dealer_zipcode_latitude)
            ) * COS(
              RADIANS(adzip.dealer_zipcode_longitude) - RADIANS(- 87.607332)
            ) + SIN(RADIANS(41.355724)) * SIN(
              RADIANS(adzip.dealer_zipcode_latitude)
            )
          )
        ), 2
      ) AS dealer_distance 
    FROM
      ad_dealers_zipcodes AS adzip 
    WHERE adzip.dealer_zipcode_dealer_id = 1 #GROUP BY adzip.dealer_zipcode_dealer_id
    ORDER BY dealer_distance ASC 
) as t1
GROUP BY dealer_zipcode_dealer_id
ORDER BY min_dealer_distance ASC