Question

我＆＃39;有两个表会议和在线

会话

 |   id   |      time    |
 |    1   |   1413170771 |
 |    2   |   1413174398 |
 |    7   |   1413174567 |

在线

 |   id   |   username   |   city   |       lat     |     lon     |
 |    1   |     Jon      |  Toronto |   45.4642700  |  9.1895100  |
 |    2   |     Danny    |  Ottawa  |   46.5645600  |  9.3456883  |
 |    7   |     Martin   |  Calgary |   46.6775339  |  9.5469944  |

我需要从会话和所有在线用户中抽出所有时间并订购它。

例如，如果Martin希望看到所有在线用户并与他保持距离，则结果必须

| id | Users   |  distance |     time   |
|  7 | Martin  |     0     | 1413174567 |
|  1 |  Jon    |    100    | 1413170771 |
|  2 |  Danny  |    250    | 1413174398 |

问题是：

我需要从在线表中提取Martin lat和lon然后计算距离，我想要的距离使用此查询

      $R = 3959;
      $lat= ???
      $lon= ????

      SELECT *, (
      $R * acos( cos( radians($lat) ) * cos( radians( lat ) ) 
      * cos( radians( lon ) - radians($lon) ) + sin( radians($lat) ) * sin(radians(lat)) ) ) AS distance 
      FROM online
      HAVING distance < $rad 
      ORDER BY distance 
      LIMIT 0 , 20"

Answer 1

一个程序可以发挥魔力，检查出来

-- Trigger DDL Statements
DELIMITER //

DROP PROCEDURE IF EXISTS cal_distance;

CREATE PROCEDURE cal_distance (IN var_id INT) 
BEGIN
DECLARE R INT DEFAULT 0;
DECLARE var_lat DOUBLE DEFAULT 0; -- NOT USED?
DECLARE var_lon DOUBLE DEFAULT 0; -- NOT USED?

--GET lon and lat for John
SELECT lat INTO var_lat, lon INTO var_lon 
FROM online 
WHERE id = var_id;

SELECT *, (
R * acos(cos(radians(var_lat)) * cos(radians(var_lat)) 
 * cos(radians(lon) - radians(var_lon)) + sin(radians(var_lat)) * sin(radians(lat))) ) AS distance 
FROM online LEFT JOIN session
USING (id)
HAVING distance >= 0
ORDER BY distance 
END $$

DELIMITER ;

调用程序

CALL cal_distance (1);

如果您想避免产程，请尝试

SET @R = 3959;
SET @var_lon = NULL;
SET @var_lat = NULL;

SELECT lat, lon INTO @var_lat, @var_lon
FROM online 
WHERE id = 1;

SELECT *, (
@R * acos( cos( radians(@var_lat) ) * cos( radians( lat ) ) 
 * cos( radians( lon ) - radians(@var_lon) ) + sin( radians(@var_lat) ) * sin(radians(lat)) ) ) AS distance 
FROM online LEFT JOIN session
USING (id)
HAVING distance >= 0
ORDER BY distance

<强> DEMO

只是使用子查询，但我不喜欢这种方式。

SELECT *, (
3959* acos( cos( radians(var_lat) ) * cos( radians( lat ) ) 
 * cos( radians( lon ) - radians(var_lon) ) + 
sin( radians(var_lat) ) * sin(radians(lat)) ) ) AS distance 
FROM online, (SELECT lat AS var_lat, lon AS var_lon FROM online  WHERE id = 1) AS new_table
HAVING distance >= 0
ORDER BY distance

<强> DEMO

合并两个sql查询

1 个答案: