如何获取一个位置周围的郊区列表,然后使用MySql重复其他位置？
作者：互联网
我使用查询A从一个位置获得指定距离内的郊区列表.
我正在尝试调整查询A以获取location1周围的郊区列表,然后获取location2周围的郊区列表,依此类推(我将称之为查询B).本质上,查询B与查询A的作用相同,但是对于每个单独的位置重复查询.我的问题 – 如何才能使用MySQL.关于如何做到这一点的建议非常感谢.
以下是我正在使用的数据示例. SqlFiddle here
CREATE TABLE `geoname` (
`geonameid` INT(11) NOT NULL,
`asciiname` VARCHAR(200) NULL DEFAULT NULL COLLATE 'utf8_unicode_ci',
`country` VARCHAR(2) NULL DEFAULT NULL COLLATE 'utf8_unicode_ci',
`latitude` DECIMAL(10,7) NULL DEFAULT NULL,
`longitude` DECIMAL(10,7) NULL DEFAULT NULL,
`fcode` VARCHAR(10) NULL DEFAULT NULL COLLATE 'utf8_unicode_ci',
`population` INT(11) NULL DEFAULT NULL,
`area` INT(11) NULL DEFAULT NULL,
PRIMARY KEY (`geonameid`),
INDEX `asciiname` (`asciiname`),
INDEX `country` (`country`),
INDEX `latitude` (`latitude`),
INDEX `longitude` (`longitude`),
INDEX `fcode` (`fcode`),
INDEX `population` (`population`),
INDEX `area` (`area`)
)
COLLATE='utf8_unicode_ci'
ENGINE=InnoDB
;
INSERT INTO geoname(geonameid, asciiname, country, latitude, longitude, fcode, population, area) VALUES
(2147497, 'Tamworth', 'AU', -31.0904800, 150.9290500, 'PPL', 47597, 72),
(8597559, 'Tamworth', 'AU', -21.0457400, 143.6685200, 'PPL', 0, 0),
(8805708, 'Tamworth', 'AU', -21.0471300, 143.6692000, 'HMSD', 0, 0),
(2655603, 'Birmingham', 'GB', 52.4814200, -1.8998300, 'PPL', 984333, 599),
(4782167, 'Roanoke', 'US', 37.2709700, -79.9414300, 'PPL', 97032, 321),
(10114336, 'East Tamworth', 'AU', -31.0854800, 150.9372100, 'PPLX', 2621, 0),
(10114337, 'North Tamworth', 'AU', -31.0786200, 150.9221900, 'PPPL', 0, 0),
(2143940, 'West Tamworth', 'AU', -31.1023600, 150.9144700, 'PPLX', 0, 0),
(2656867, 'Aston', 'GB', 52.5000000, -1.8833300, 'PPLX', 0, 0),
(2646814, 'Hockley', 'GB', 52.5000000, -1.9166700, 'PPLX', 13919, 0),
(2650236, 'Edgbaston', 'GB', 52.4623000, -1.9211500, 'PPLX', 0, 0),
(4754994, 'Cumberland Forest', 'US', 37.1401300, -80.3217100, 'PPLX', 0, 0),
(4774999, 'Mountain Top Estates', 'US', 37.1376300, -80.3247700, 'PPPL', 0, 0),
(4764119, 'Highland Park', 'US', 37.2237400, -80.3917200, 'PPLX', 0, 0);
我尝试了什么
查询A-获取围绕单个兴趣点的郊区
SELECT @lat := latitude, @lng :=longitude FROM geoname WHERE asciiname = 'Tamworth' and country='AU' and population>0 and fcode='PPL';
SELECT
name as suburb, 'Tamworth' as point_of_interest, country,
(
(
ACOS(SIN(@lat * PI() / 180) * SIN(latitude * PI() / 180) + COS(@lat * PI() / 180) * COS(latitude * PI() / 180) * COS((
@lng - longitude
) * PI() / 180)) * 180 / PI()
) * 60 * 1.851999999962112
) AS distance
FROM geoname
WHERE fcode='PPLX' OR fcode='PPPL'
HAVING distance <= '60'
ORDER BY distance ASC;
结果
上面的查询返回兴趣点的一个位置.
+---------------------------------+
| @lat | @lng |
+---------------------------------+
| 52.6339900 | -1.6958700 |
+---------------------------------+
和塔姆沃思周边的郊区列表.
| point_of_interest | suburb | country | distance |
|-------------------|----------------------|---------|--------------------|
| Tamworth | East Tamworth | AU | 0.9548077598752538 |
| Tamworth | North Tamworth | AU | 1.4707125875055387 |
| Tamworth | West Tamworth | AU | 1.915025922482298 |
我尝试使用MySQL用户变量GROUP_CONCAT()和FIND_IN_SET()创建查询B.我的想法是,我可以像使用数组一样循环使用值.如果你愿意,我可以发布我的最后一次尝试,但我甚至不接近解决方案(不是因为没有尝试).
更新：这是我最后一次尝试.
SELECT @lat := GROUP_CONCAT(latitude), @lng :=GROUP_CONCAT(longitude), @city :=GROUP_CONCAT(asciiname), @area :=GROUP_CONCAT(area) FROM geoname WHERE (asciiname = 'Tamworth' or asciiname = 'Birmingham' or asciiname = 'Roanoke') and population>0 and fcode='PPL';
SELECT
FIND_IN_SET(asciiname, @city) as point_of_interest, asciiname as suburb, country,
(
(
ACOS(SIN(FIND_IN_SET(latitude, @lat) * PI() / 180) * SIN(latitude * PI() / 180) + COS(FIND_IN_SET(latitude, @lat) * PI() / 180) * COS(latitude * PI() / 180) * COS((
FIND_IN_SET(longitude, @lng) - longitude
) * PI() / 180)) * 180 / PI()
) * 60 * 1.851999999962112
) AS distance
FROM geoname
HAVING distance <= FIND_IN_SET(distance, @area)
ORDER BY distance ASC;
查询所需的结果B.对于3个兴趣点 – 塔姆沃思,伯明翰和罗阿诺克 – 这是我期望看到的.
| point_of_interest | suburb | country | distance |
|-------------------|----------------------|---------|--------------------|
| Tamworth | East Tamworth | AU | 0.9548077598752538 |
| Tamworth | North Tamworth | AU | 1.4707125875055387 |
| Tamworth | West Tamworth | AU | 1.915025922482298 |
| Birmingham | Aston | GB | 2.347111909955497 |
| Birmingham | Hockley | GB | 2.3581405942861164 |
| Birmingham | Edgbaston | GB | 2.568384753388139 |
| Roanoke | Cumberland Forest | US | 36.66226789588173 |
| Roanoke | Mountain Top Estates | US | 37.02185777044897 |
| Roanoke | Highland Park | US | 40.174566427830094 |
关于如何使用MySQL执行此操作的建议非常感谢.
解决方法:
您只需要执行自我加入. Joining表是SQL的一个非常基础的部分 – 在尝试进一步理解这个答案之前,你真的应该阅读它.
SELECT poi.asciiname,
suburb.asciiname,
suburb.country,
DEGREES(
ACOS(
SIN(RADIANS( poi.latitude))
* SIN(RADIANS(suburb.latitude))
+ COS(RADIANS( poi.latitude))
* COS(RADIANS(suburb.latitude))
* COS(RADIANS(poi.longitude - suburb.longitude))
)
) * 60 * 1.852 AS distance
FROM geoname AS poi
JOIN geoname AS suburb
WHERE poi.asciiname IN ('Tamworth', 'Birmingham', 'Roanoke')
AND poi.population > 0
AND poi.fcode = 'PPL'
AND suburb.fcode IN ('PPLX', 'PPPL')
HAVING distance <= 60
ORDER BY poi.asciiname, distance
在sqlfiddle上看到它.
您会注意到我使用MySQL的IN()运算符作为值的简写= A OR值= B OR ….
您还会注意到我使用了MySQL的DEGREES()和RADIANS()函数,而不是尝试显式执行此类转换.
然后你将纬度分数乘以因子1.851999999962112,这是相当奇怪的：它非常接近1.852,这是海里的精确公里数(历史上定义为纬度的一分钟),但奇怪的是略有不同 – 我以为你打算用它来代替.
最后,您有一个字面值,您将结果集中的距离过滤为字符串,即’60’,而显然这是一个数值,应该是不加引号的.
标签：geonames,mysql,geolocation,user-variables 来源： https://codeday.me/bug/20190928/1825927.html