问题描述
我想提出一个Android应用程序。是否有可能来检查，如果用户（给予一定的坐标）正面临着街道，如果这条街是在10-20ft了？
I am making an android app. Is it possible to check if a user (given some coordinates) is facing a street and if that street is within 10-20ft away?
如果不是，是否有一个开放源码的地图，我可以花纹，通过手动使用code /协调，看是否有附近的街头？我想打一个导航系统，为盲人，我需要一些方法来检查用户是否站在马路中间。
If not, is there an open source map which I can "tread" through manually using code/coordinates to see if there is a nearby "street"? I want to make a navigation system for the blind and I need some way to check if the user is standing on the middle of the street.
谢谢，罗希特
推荐答案
您需要阅读上的谷歌地图API地理编码：
You'll want to read up on the Google Maps Geocoding API:
HTTP：//$c$c.google.com/apis /地图/文档/地理编码/
如果你在Android上，有一个特殊的 地理codeR 类，将调用谷歌地图为您提供：
If you're on Android, there's a special Geocoder class that will call Google Maps for you:
Geocoder geocoder = new Geocoder(this, Locale.getDefault());
List<Address> addresses = geocoder.getFromLocation(lat, lng, 1);
（来源：机器人：反向地理编码 - getFromLocation ）
美中不足的是，这只是去给最有可能的地址为用户的当前位置 - 你没有得到指定模糊因数或类似的东西。即便如此，10-20英尺。应该是足够接近此相匹配。
The catch is that this just going to give the most likely address for the user's current location -- you don't get to specify a fuzz factor or anything like that. Even then, 10-20 ft. should be close enough for this to match.
如果这不会得到你所需要的数据，你可能也想看看 OpenStreetMap的，这是一个开放的地图数据库。我没有太多的经验与它自己，但有文档浏览： http://wiki.openstreetmap.org /维基/ Main_Page
If this doesn't get you the data you need, you might also want to look at OpenStreetMap, which is an open maps database. I don't have much experience with it myself, but there's documentation here: http://wiki.openstreetmap.org/wiki/Main_Page
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