我想得到那些距离应该在500米左右的经度，纬度（ 25.057550,55.134729 ）。
$ user_long = $ _REQUEST ['user_long']; // 25.057550 $ user_lat = $ _REQUEST ['user_lat']; // 55.134729 从...所在的位置选择*（取500米附近的记录）我已经搜索过它，并找到一些很好的展示 haversine公式答案和谷歌地图与Php / Mysql ，但我没有得到如何开发这种功能。我知道这个问题可能是愚蠢的，因为我没有添加我的代码，但相信我，我不知道我该怎么做，从哪里开始功能。我会很感激，如果有人引导我关于这个功能。
谢谢
解决方案（（19.286558 - 纬度）* pi（）/ 180/2），2）+ COS （）/ 180） * COS（纬度* pi（）/ 180）* POWER（SIN（（ - 99.612494-经度）* pi（）/ 180/2），2））） as距离距表<距离< 10 ORDER BY距离;
这会给你10公里范围内的记录。 在子句中修改500米的查询。
I'm trying to get the longitude and latitude around specific location. Actually I have some records (longitude and latitude) in mysql table. Now i want tofetch those longitude and latitude from table which should have around 500 meter of specific longitude. Please view the code, i hope you will understand my question.
Mysql Location Table ------------------------------------- id | longitude | latitude ------------------------------------- 1 | 25.058036 | 55.133467 ------------------------------------- 2 | 25.056986 | 55.135755 ------------------------------------- 3 | 25.059163 | 55.135616 ------------------------------------- 4 | 25.056860 | 55.132515 ------------------------------------- 5 | 25.055683 | 55.134792I want to get those longitude, latitude which distance should have around 500 meters of specific (25.057550, 55.134729).
$user_long= $_REQUEST['user_long']; // 25.057550 $user_lat= $_REQUEST['user_lat']; // 55.134729 Select * from location where .... (fetch record around 500 meters)I have searched about it and find some well expnained haversine formula answer and Google Map With Php/Mysql but i did not got how can i develop this kind of functionality. I know this question may be stupid for someone because i'm not adding my code, but trust me i don't know how can i do it and from where i should start functionality. I will appreciate if someone guide me regarding this functionality.
Thank You
解决方案 SELECT *,3956*2*ASIN(SQRT(POWER(SIN((19.286558 - latitude)*pi()/180/2),2)+COS(19.286558 * pi()/180) *COS(latitude * pi()/180)*POWER(SIN((-99.612494 -longitude)* pi()/180/2),2))) as distance FROM table having distance < 10 ORDER BY distance;This will give you records within 10 Km range. modify query for 500 meters in having clause.
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