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每个人：

我完全被这个问题困住了。我尝试将条件添加为 WHERE 语句、SELECT 语句中的 IF 函数以及加入时的 ON 语句。这些都没有产生我想要的结果。

我需要什么：特定位置的 0.5 度纬度和 0.5 度经度范围内的城市列表。

我得到了什么： 两张表 - 一张包含特定位置列表，另一张包含美国人口普查所规定的城市列表。

表 1 包括：

Location Address
City
State
Location Latitude
Location Longitude

示例位置数据：

Street              City        State  Latitude     Longitude
774 Emerson Street  Palo Alto   CA     37.44239044  -122.15956879

表 2 是美国人口普查数据，包括：

City
State
City Latitude
City Longitude

样本城市数据：

City            State Latitude     Longitude
Palo Alto       CA    37.44188309   -122.14302063
Mountain View   CA    37.38605118   -122.08385468
Sunnyvale       CA    37.36883      -122.0363496

例如，如果我在加利福尼亚州帕洛阿尔托有一个位置，我想要看起来像这样的东西：

Location        City          State
Palo Alto CA    Palo Alto     CA
Palo Alto CA    Mountain View CA
Palo Alto CA    Sunnvyale     CA

我试过了：

SELECT d.`City`
      ,d.`State`
      ,l.`City`
      ,l.`StateAbbreviation`
FROM `LocationDirectoryRevised` AS d
     LEFT JOIN
     `LatLong` AS l
     ON d.`City` = l.`City` AND d.`State` = l.`StateAbbreviation`
WHERE l.`Latitude` BETWEEN (d.`Latitude` + 0.5) AND (d.`Latitude` - 0.5)
      AND
      l.`Longitude` BETWEEN (d.`Longitude` + 0.5) AND (d.`Longitude` - 0.5);

我试过了：

SELECT d.`City`
      ,d.`State`
      ,l.`City`
      ,l.`StateAbbreviation`
FROM `LocationDirectoryRevised` AS d
     LEFT JOIN
     `LatLong` AS l
     ON l.`Latitude` BETWEEN (d.`Latitude` + 0.5) AND (d.`Latitude` - 0.5) &&     
        l.`Longitude` BETWEEN (d.`Longitude` + 0.5) AND (d.`Longitude` - 0.5)

我试过了：

SELECT d.`City`
      ,d.`State`
      ,IF((l.`Latitude` < (d.`Latitude` + 0.5)) && (l.`Latitude` > (d.`Latitude` - 0.5)) &&  (l.`Longitude` < (d.`Longitude` + 0.5)) && (l.`Longitude` > (d.`Longitude` - 0.5)), l.`City`,NULL)
      ,l.`StateAbbreviation`
FROM `LocationDirectoryRevised` AS d
     LEFT JOIN
     `LatLong` AS l
     ON d.`City` = l.`City` AND d.`State` = l.`StateAbbreviation`

但是这些并没有产生我想要的结果。我错过了什么？

谢谢。

4

3 回答 3

0

尝试这个：

WHERE d.`City` = 'Seattle' AND d.`State` = 'WA'
AND
l.`Latitude` < (d.`Latitude` + 0.5) 
AND 
l.`Latitude > (d.`Latitude` - 0.5)
AND
l.`Longitude` < (d.`Longitude` + 0.5) 
AND 
`l.Longitude` > (d.`Longitude` - 0.5);

或者，这也有效：

WHERE d.`City` = 'Seattle' AND d.`State` = 'WA'
AND
l.`Latitude` BETWEEN (d.`Latitude` + 0.5) AND (d.`Latitude` - 0.5)
AND
l.`Longitude` BETWEEN (d.`Longitude` + 0.5) AND (d.`Longitude` - 0.5);
于 2018-01-17T22:44:30.547 回答
0

经过大量工作，我能够使用子查询和稍微调整我的参数来找出主要部分：

SELECT `City`
      ,`StateAbbreviation`
      ,`Latitude`
      ,`Longitude`
FROM `LatLong`
WHERE `Latitude` BETWEEN (SELECT `Latitude` - 0.1
                          FROM `LocationDirectoryRevised`
                          WHERE `ID` = 1)
                      AND (SELECT `Latitude` + 0.1
                          FROM `LocationDirectoryRevised`
                          WHERE `ID` = 1)
      AND
      `Longitude` BETWEEN (SELECT `Longitude` - 0.1
                          FROM `LocationDirectoryRevised`
                          WHERE `ID` = 1)
                      AND (SELECT `Longitude` + 0.1
                          FROM `LocationDirectoryRevised`
                          WHERE `ID` = 1);

现在我必须弄清楚如何将其作为所有 660 个位置的存储过程来执行。如果需要，那将是一个不同的线程。

谢谢。

于 2018-01-21T19:11:17.060 回答
0

你可以使用这个：

drop procedure if exists getNearLocations;
create procedure getNearLocations();
BEGIN
DECLARE n INT DEFAULT 0;
DECLARE i INT DEFAULT 0;
declare e INT default 0;
SELECT COUNT(*) FROM LocationDirectoryRevised INTO n;
SET i=0;
WHILE i<n DO 
select ID from LocationDirectoryRevised into e;
  SELECT `City`
      ,`StateAbbreviation`
      ,`Latitude`
      ,`Longitude`
FROM `LatLong`
WHERE `Latitude` BETWEEN (SELECT `Latitude` - 0.1
                          FROM `LocationDirectoryRevised`
                          WHERE `ID` = e)
                      AND (SELECT `Latitude` + 0.1
                          FROM `LocationDirectoryRevised`
                          WHERE `ID` = e)
      AND
      `Longitude` BETWEEN (SELECT `Longitude` - 0.1
                          FROM `LocationDirectoryRevised`
                          WHERE `ID` = e)
                      AND (SELECT `Longitude` + 0.1
                          FROM `LocationDirectoryRevised`
                          WHERE `ID` = e);
  SET i = i + 1;
END WHILE;
End

这将遍历 660 个位置并生成您正在寻找的数据。

然而，由于子查询的数量，这可能是一个昂贵（缓慢）的查询，但对于 660 个位置，这应该足够了。

于 2018-02-08T19:46:26.260 回答