Question

我有一个问题，比如我有三个表tbldispensaries，tblStates，tblCountries。我将国家和州的ID存储在药房表中，所有与国家和州相关的详细信息分别位于tblCountries和tblStates。

我正在编写一个查询，以便在半径500范围内找到附近的药房，以便在Google地图上显示标记，查询很好，但是当我使用左边连接查询时，它需要花费太多时间响应。我尝试过使用索引，但这对它没有任何影响。所以请建议我克服这个问题。这是我正在使用的查询： -

SELECT
    d.id AS disp,
    d.id AS the_id,
    d. NAME,
    d.address AS addr,
    d.city AS city_name,
    c.country_name AS cntry_name,
    s.state_name AS st_name,
    d.zip AS zipcode,
    d.latitude,
    d.longitude,
    (
        6371 * ACOS(
            COS(RADIANS(33.6119)) * COS(RADIANS(latitude)) * COS(
                RADIANS(longitude) - RADIANS(- 111.8906)
            ) + SIN(RADIANS(33.6119)) * SIN(RADIANS(latitude))
        )
    ) AS distance
FROM
    `tblDispensaries` d
    LEFT JOIN tblCountries c ON (c.country_id = d.country)
    LEFT JOIN tblStates s ON (s.state_id = d.state)
WHERE
    d.id IS NOT NULL
    AND d. STATUS = 1
HAVING distance < 500
LIMIT 0,60

Answer 1

这可能不是一个答案，但是评论是两个大问题。此外，我没有数据，所以我无法测试，但现在是。

SELECT 
    Dispensaries.*,
    c.country_name AS cntry_name,
    s.state_name AS st_name
FROM (
  SELECT
      d.id AS disp,
      d.id AS the_id,
      d. NAME,
      d.address AS addr,
      d.city AS city_name,
      d.zip AS zipcode,
      d.latitude,
      d.longitude,
      (
          6371 * ACOS(
              COS(RADIANS(33.6119)) * COS(RADIANS(latitude)) * COS(
                  RADIANS(longitude) - RADIANS(- 111.8906)
              ) + SIN(RADIANS(33.6119)) * SIN(RADIANS(latitude))
          )
      ) AS distance
  FROM
      `tblDispensaries` d
  WHERE
      d.id IS NOT NULL
      AND d. STATUS = 1
  HAVING distance < 500 
  LIMIT 0,60
) as Dispensaries
LEFT JOIN tblCountries c ON (c.country_id = Dispensaries.country)
LEFT JOIN tblStates s ON (s.state_id = Dispensaries.state)

我认为获取结果并使用JOIN会更快。

MySQL离开连接需要很长时间才能返回

1 个答案: