从多行读取数据的 MySQL 语句

Question

我需要用一个 mySql 语句来做一个复杂的 mySQL 查询，我有一个这样的表，每个用户的数据分布在不同的行中

 ID  USER_ID FIELD_ID  VALUE
 1      1       2        my name is paul smith
 2      1       3        books
 3      1       4        loggedin
 4      1       5        busy
 5      1       6        lat
 6      1       7        lon
 7      2       2        my name is big boy
 8      2       3        pens
 9      2       4        offline
 10     2       5        idle
 11     2       6        lat
 12     2       7        lon

每一行包含同一个user_id的不同数据

 FIELD_ID=2 contains the user name
 FIELD_ID=3 contains what they bought
 FIELD_ID=4 logged in or offline
 FIELD_ID=5 is busy or idle
 FIELD_ID=6 is their latitude
 FIELD_ID=7 is their longitude

我需要一个 MySQL 语句，它会返回所有已登录且状态为空闲且已购买图书且距离例如 5 英里以内的用户。

所以有四件事必须同时为真：

 1. user has to be LOGGED IN
 2. user has to be idle
 3. user has to have bought books
 4. user has to be within 5 miles of the office (located at lat=12.3, lon=13.3)

所以我从这个开始（做了上面四件事中的三件事）

 SELECT distinct a.*
  FROM table a
   inner join table b on a.user_id = b.user_id and b.field_id = 5 and b.value='idle'
   inner join table c on a.user_id = c.user_id and c.field_id = 4 and c.value = 'loggedin'
   where a.field_id=3 and a.value='BOOK'

我是一个 mysql 新手，似乎上面的语句创建了三个虚拟表并将它们组合起来进行查询，所以我想再创建两个虚拟表（一个用于 lat，另一个用于 long），并将其与我现有的mySQL 函数 calcDist(lat1,lon1,lat2,lon2) 计算 2 个 gps 点之间的距离，如下所示：

 SELECT distinct a.* 
  FROM table a
   inner join table b on a.user_id = b.user_id and b.field_id = 5 and b.value='idle'
   inner join table c on a.user_id = c.user_id and c.field_id = 4 and c.value = 'loggedin'
   inner join table d on a.user_id = d.user_id and d.field_id = 6
   inner join table e on a.user_id = e.user_id and e.field_id = 7
   where a.field_id=3 and a.value='BOOK' and calcDist(12.3,13.3,d.value,e.value)<5.0;

我添加了两行（d和e）并修改了最后一行，其余相同。

对不起，我是一个真正的 mysql 新手，似乎无法同时比较不同行的值。

有人知道这是否是正确的方法吗？

谢谢！

Answer 1

我已经编辑了这个，因为草莓写道，需要一些东西来聚合，这对你的数据不起作用。下面的这个解决方案有效。它与您之前所做的非常相似，只是它包含在子查询中。我已经对此进行了测试并且可以使用

SELECT * FROM
(SELECT DISTINCT
x.USER_ID,
n.value as name,
b.VALUE as bought,
l.VALUE as logged,
s.VALUE as `status`,
lt.VALUE as lat,
ln.VALUE as lon

FROM testing x

LEFT JOIN testing n
ON x.USER_ID = n.USER_ID and n.FIELD_ID = 2

LEFT JOIN testing b
ON x.USER_ID = b.USER_ID and b.FIELD_ID = 3

LEFT JOIN testing l
ON x.USER_ID = l.USER_ID and l.FIELD_ID = 4

LEFT JOIN testing s
ON x.USER_ID = s.USER_ID and s.FIELD_ID = 5

LEFT JOIN testing lt
ON x.USER_ID = lt.USER_ID and lt.FIELD_ID = 6

LEFT JOIN testing ln
ON x.USER_ID = ln.USER_ID and ln.FIELD_ID = 7

)tbl

WHERE tbl.logged = 'loggedin'
AND tbl.status IN ('idle', 'busy')
AND tbl.bought = 'books'
AND calcDist(12.3,13.3, tbl.lat, tbl.lon) > 5

你的 calcDist 函数应该可以做类似的事情

AND (3959 * acos( cos( radians(12.3) ) 
      * cos( radians(x.LAT) ) 
      * cos( radians(x.LON) - radians(13.3)) + sin(radians(12.3))
      * sin( radians(x.LAT) ))) >= 5