计算城镇距离，不包括 SQL 中的时间地点

Question

我有两张桌子

位置： 有城镇名称、位置 latitude/longitude 和城镇开放日期

|--------------|------------|------------|-------------------|
|    Towns     |  latitude  | longitude  |  establish_time   |
|--------------|------------|------------|-------------------|
|  Town_A      |  72.92629  | -12.89272  |    2021-07-21     |
|  Town_B      |  93.62789  | -83.10172  |    2021-08-21     |
|  Town_C      |  68.92612  | -67.17242  |    2021-09-21     |
|--------------|------------|------------|-------------------|

用户：有user_id，用户latitude/longitude，购买电影票的城镇，购买时间

|------------|------------|------------|--------------|-----------------|
|   user_id  |  latitude  | longitude  |    Towns     |  purchase_time  |
|------------|------------|------------|--------------|-----------------|
|    1       |  21.89027  | -53.03772  |   Town_A     |    2021-08-19   |
|    1       |  23.87847  | -41.78172  |   Town_C     |    2021-09-29   |
|    1       |  77.87092  | -96.39242  |   Town_A     |    2021-11-07   |
|    2       |  83.37847  | -62.78278  |   Town_B     |    2021-10-21   |
|    3       |  89.81924  | -80.73892  |   Town_B     |    2021-12-07   |
|    3       |  27.87282  | -18.39183  |   Town_A     |    2021-08-23   |
|------------|------------|------------|--------------|-----------------|

我试图找到购票时从用户位置到所有其他城镇的最短距离 purchase_time

这是我的SQL（雪花）：

SELECT 
        u.*,
        seq4() as seq,
        l.town as nearest_town,
        haversine(u.latitude, u.longitude, l.latitude, l.longitude)
    FROM User AS u, location AS l
    QUALIFY row_number() OVER (PARTION BY u.seq ORDER BY haversine(u.latitude, u.longitude, l.latitude, l.longitude)) = 1

我得到的结果实际上是基于用户位置如下：

|------------|------------|------------|--------------|-----------------|
|   user_id  |  latitude  | longitude  |    Towns     | nearest_town    |
|------------|------------|------------|--------------|-----------------|
|    1       |  21.89027  | -53.03772  |   Town_C     |   Town_C        |
|    1       |  23.87847  | -41.78172  |   Town_C     |   Town_A        | 
|    1       |  77.87092  | -96.39242  |   Town_A     |   Town_B        | <--- This should be Town_A because Town_B was not present at the time of purchase
|    2       |  83.37847  | -62.78278  |   Town_B     |   Town_C        |
|    3       |  89.81924  | -80.73892  |   Town_B     |   Town_A        |
|    3       |  27.87282  | -18.39183  |   Town_A     |   Town_C        | <--- This should be Town_A because Town_B was not present at the time of purchase
|------------|------------|------------|--------------|-----------------|

即使 purchase_time

时城镇不存在，上述查询也会导致 nearest_town

如何在计算nearest_towns

时排除未开放的城镇

Answer 1

将隐式 CROSS JOIN（由 FROM 子句中的逗号分隔表列表暗示）切换为 INNER JOIN，其中 ON 条件是您要添加的日期限制.

类似

FROM Users AS u
    INNER JOIN location AS l
        ON l.establish_time <= u.purchase_time

这用英文说“加入这两张桌子时，只考虑购买这张票之前建立的位置”。

计算城镇距离，不包括 SQL 中的时间地点

Calculating town distance excluding the locations with time in SQL

sql

window-functions

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