通过不工作使用组检索最新时间行

时间：2014-09-05 13:24:42

标签： mysql group-by sql-order-by having

表： -

表1： - Person_LatLong 

Person_Id
Lat
Lng
Location_DateTime

Person_LatLong数据： - 

10  23.030211421184454  72.55705489668806   2014-09-02 04:23:42
10  23.03029215373424   72.55710855670746   2014-09-02 04:25:09
10  23.030301667271765  72.55715279051745   2014-09-02 04:27:21
11  19.045563510415214  72.915665750359 2014-09-02 08:22:07
11  19.046567530190785  72.91524330185979   2014-09-02 08:32:36
11  19.04553070382594   72.91621148133476   2014-09-02 08:39:47
11  18.537267778519347  73.83538450865574   2014-09-03 01:44:19
11  18.33554237666039   73.85274219500492   2014-09-03 07:18:02
11  18.331919816746026  73.8525499279805    2014-09-03 07:18:59
11  18.33181875247372   73.85243149060277   2014-09-03 07:19:02
11  18.777939290860722  73.31834934181029   2014-09-04 22:07:31
11  18.790032969638293  73.30265963437363   2014-09-04 22:09:06
11  18.79108238318203   73.29928216416553   2014-09-04 22:09:22
11  18.800857529132163  73.28531940244517   2014-09-04 22:11:22
11  18.812675453346255  73.27794458217039   2014-09-04 22:13:26
11  18.82985965773455   73.25592224937081   2014-09-04 22:15:31
11  18.84531169311457   73.23344887176076   2014-09-04 22:17:49
11  18.869063931831764  73.2185512231118    2014-09-04 22:19:54
11  18.893204517796047  73.20479873759578   2014-09-04 22:22:03
11  18.910161939581506  73.18348844819505   2014-09-04 22:24:30

表2： - LS_For 

Subject_Id -> This is basically Person_Id but I have changed the name.
Watcher_Id
Assistance_Group_Id

LS_For表数据： - 

1   10  1
1   11  1
1   17  1

查询： - 

SELECT *,( 6371 * acos( cos( radians(23.030) ) 
* cos( radians(Lat) ) * cos( radians(Lng) - radians(72.5570) )
 + sin( radians(23.030) ) * sin( radians(Lat) ) ) ) 
AS distance FROM Person_LatLong WHERE Person_Id IN 
(SELECT Watcher_Id FROM LS_For WHERE Subject_Id = 1 AND Assistance_Group_Id = 1)
HAVING distance < 5 ORDER BY Location_DateTime DESC;

输出如下： - ID |拉特| LNG |日期时间| DISTANCE 

10  23.030301667271765  72.55715279051745   2014-09-02 04:27:21 0.037008818510632306
10  23.03029215373424   72.55710855670746   2014-09-02 04:25:09 0.03433299317128307
10  23.030211421184454  72.55705489668806   2014-09-02 04:23:42 0.02417068347133403

11  23.030301667271765  72.55715279051745   2014-09-02 05:27:21 0.037008818510632306
11  23.03029215373424   72.55710855670746   2014-09-02 05:25:09 0.03433299317128307
11  23.030211421184454  72.55705489668806   2014-09-02 05:23:42 0.02417068347133403

添加GROUP BY后： - 

SELECT *,( 6371 * acos( cos( radians(23.030) ) 
* cos( radians(Lat) ) * cos( radians(Lng) - radians(72.5570) )
 + sin( radians(23.030) ) * sin( radians(Lat) ) ) ) 
AS distance FROM Person_LatLong WHERE Person_Id IN 
(SELECT Watcher_Id FROM LS_For WHERE Subject_Id = 1 AND Assistance_Group_Id = 1)
GROUP BY Person_Id HAVING distance < 5 ORDER BY Location_DateTime DESC;

给我以下输出： - 

10  23.030211421184454  72.55705489668806   2014-09-02 04:23:42 0.02417068347133403

但我希望最新的时间行不是最老的。

OUTPUT要求： - 使用2个表，LS_For是我找到Ids的表，我必须根据距离来查找位置距离并根据需要过滤结果。 

10  23.030301667271765  72.55715279051745   2014-09-02 04:27:21 0.037008818510632306
11  23.030301667271765  72.55715279051745   2014-09-02 05:27:21 0.037008818510632306

4 个答案:

答案 0 :(得分：3)

生成一组数据，其中包含每个人的最大日期时间（最近），然后INNER JOIN到它，以便您的基本集仅限于person_LatLong中的最新条目。

(Select person_ID, max(`Location_DateTime`) mldt FROM person_LatLong group by Person_ID)

为每个用户生成最新的latLong ......然后......

可能1 

SELECT a.Person_ID, a.LAT, a.LNG, a.`Location_DateTime`, 
 ( 6371 * acos( cos( radians(23.030) ) 
        * cos( radians(a.Lat) ) * cos( radians(a.Lng) - radians(72.5570) )
        + sin( radians(23.030) ) * sin( radians(a.Lat) ) ) ) AS Distance 
FROM Person_LatLong a
INNER JOIN (SELECT Person_ID, max(`Location_DateTime`) as mldt
            FROM Person_latLong 
            GROUP BY Person_ID) P
  on P.Person_ID = a.Person_Id
 and P.mldt = a.`Location_DateTime` 
LEFT JOIN LS_FOR C
 on a.Person_Id = C.Watcher_ID
  AND C.Subject_Id = 1 
  AND C.Assistance_Group_Id = 1
GROUP BY a.Person_ID, a.LAT, a.LNG, a.`Location_DateTime`
HAVING Distance < 5;

可能2 

SELECT a.Person_ID, a.LAT, a.LNG, a.`Location_DateTime`, 
 ( 6371 * acos( cos( radians(23.030) ) 
        * cos( radians(a.Lat) ) * cos( radians(a.Lng) - radians(72.5570) )
        + sin( radians(23.030) ) * sin( radians(a.Lat) ) ) ) AS Distance 
FROM Person_LatLong a
INNER JOIN (SELECT Person_ID, max(`Location_DateTime`) as mldt
            FROM Person_latLong 
            GROUP BY Person_ID) P
  on P.Person_ID = a.Person_Id
 and P.mldt = a.`Location_DateTime` 
INNER JOIN LS_FOR C
 on a.Person_Id = C.Watcher_ID
WHERE C.Subject_Id = 1 
  AND C.Assistance_Group_Id = 1
GROUP BY a.Person_ID, a.LAT, a.LNG, a.`Location_DateTime`
HAVING Distance < 5;

我修改了将LS_FOR视为左连接的响应，因为LS_FOR中没有每个用户的记录，即使他们没有主题ID或aid_groupID也不会返回该人。 ......

换句话说，上面应该用简单的英语返回：

将为每个唯一的Person_ID返回最新的personLatLong;只要距离<1。 5如果LS_FOR和PersonLatLong条目中有条目，则只返回subjectID为1且asstianceGorupID为1的记录。如果LS_FOR中没有条目，它仍将返回该人（也许你确实希望它作为内部联接...）

答案 1 :(得分：0)

而不是

GROUP BY Person_I

使用

LIMIT 1

OR 

GROUP BY Person_Id HAVING distance < 5 ORDER BY Location_DateTime DESC;

到

GROUP BY Person_Id HAVING distance < 5 ORDER BY Location_DateTime ASC;

答案 2 :(得分：0)

检查此链接。

Retrieving the last record in each group

这里的解决方案比子查询快得多。

   select person1.* from Person_LatLong person1 
    LEFT JOIN Person_LatLong person2 on person1.id = person2.id and 
   person1.dateTiem < person2.dateTime
    where person2.id IS NULL;

这将为您提供最新记录。

答案 3 :(得分：0)

尝试这样做：

SELECT 
  *,( 6371 * acos( cos( radians(23.030) ) * cos( radians(Lat) ) * cos( radians(Lng) - radians(72.5570) ) + sin( radians(23.030) ) * sin( radians(Lat) ) ) ) AS distance
FROM 
  (select * from Person_LatLong ORDER BY Location_DateTime DESC) as t  
WHERE 
  Person_Id IN (SELECT Watcher_Id FROM LS_For WHERE Subject_Id = 1 AND Assistance_Group_Id = 1) 
GROUP BY 
  Person_Id 
HAVING 
  distance < 5 ;
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