我有三张桌子。 一个用户详细信息:
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| Username | Phonenumber |
---------------------------------
| user1 | 0000000000 |
---------------------------------
| user2 | 000000000 |
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表2包含联系信息,这是用户在彼此“了解”时链接在一起的地方:
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| ID | User | Friend |
------------------------
| 1 | 1 | 2 |
------------------------
| 2 | 1 | 3 |
------------------------
| 3 | 3 | 2 |
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最后一个,一个包含位置信息的表:
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| ID | UserID | TimeStamp | Lng | Lat |
----------------------------------------------------------
| 1 | 1 | 2015-01-07 19:54:23 | lngvalue | latvalue |
----------------------------------------------------------
| 2 | 1 | 2015-01-07 19:54:26 | lngvalue | latvalue |
----------------------------------------------------------
| 3 | 2 | 2015-01-07 19:53:52 | lngvalue | latvalue |
----------------------------------------------------------
现在我想从用户1的朋友中选择用户名和电话号码,以及联系人用户的最新位置。
我创建了一个查询,但它显示了用户最早的lng / lat值,而我需要最新的。
SELECT user.Username, user.PhoneNumber, location.Lng, location.Lat
FROM contact, user, location
WHERE User.ID IN (SELECT contact.Friend FROM contact WHERE contact.User = 1)
AND user.ID = location.UserID
GROUP BY user.Username
答案 0 :(得分:1)
我对Randy的查询做了一些小改动,因为我也可以通过检查位置表中的最高ID来选择最新的记录。 该查询对我有用:
SELECT user.Username, user.PhoneNumber, location.Lng, location.Lat
FROM contact, user,
( select max(id) maxid, userid from location group by userid ) loc
, location
WHERE User.ID IN (SELECT contact.Friend FROM contact WHERE contact.User = 1)
AND user.ID = loc.UserID
and location.id = loc.maxid
GROUP BY user.Username
答案 1 :(得分:0)
基本思路:创建所需的lat和日志记录的内联视图..然后加入到那个。
SELECT user.Username, user.PhoneNumber, location.Lng, location.Lat
FROM contact, user,
( select id, userid, max(timestamp) ts from location group by userid ) loc
, location
WHERE User.ID IN (SELECT contact.Friend FROM contact WHERE contact.User = 1)
AND user.ID = loc.UserID
and location.id = loc.id
and location.timestamp = loc.ts
GROUP BY user.Username