我想弄乱我的任务并按城市名称而不是邮政编码搜索天气（我现在如何设置）。使用城市名称输入字符串并从中获取邮政编码最简单的方法是什么？非常感谢帮助！谢谢！
答案 0 :(得分：13)
Google可以帮助你！
https://developers.google.com/maps/documentation/geocoding/
该邮件实际上被Google称为“postal_code”。
"long_name": "94043",
"short_name": "94043",
"types": postal_code
例如，假设你想要获得Clarkston，MI的拉链......
http://maps.googleapis.com/maps/api/geocode/json?address=Clarkston+MI&sensor=true
返回：
{
"results" : [
{
"address_components" : [
{
"long_name" : "Clarkston",
"short_name" : "Clarkston",
"types" : [ "locality", "political" ]
},
{
"long_name" : "Oakland",
"short_name" : "Oakland",
"types" : [ "administrative_area_level_2", "political" ]
},
{
"long_name" : "Michigan",
"short_name" : "MI",
"types" : [ "administrative_area_level_1", "political" ]
},
{
"long_name" : "United States",
"short_name" : "US",
"types" : [ "country", "political" ]
},
{
"long_name" : "48346",
"short_name" : "48346",
"types" : [ "postal_code" ]
}
],
"formatted_address" : "Clarkston, MI 48346, USA",
"geometry" : {
"bounds" : {
"northeast" : {
"lat" : 42.7418310,
"lng" : -83.41402409999999
},
"southwest" : {
"lat" : 42.7252370,
"lng" : -83.42880730000002
}
},
"location" : {
"lat" : 42.73511960,
"lng" : -83.41929410
},
"location_type" : "APPROXIMATE",
"viewport" : {
"northeast" : {
"lat" : 42.74331460,
"lng" : -83.40328670
},
"southwest" : {
"lat" : 42.72692350,
"lng" : -83.43530149999999
}
}
},
"types" : [ "locality", "political" ]
}
],
"status" : "OK"
}
如果您没有收到第一次电话的邮政编码，则必须使用第一次电话的坐标再次拨打同一个网络服务。还是很简单 - 对史蒂文斯角，威斯康星州的呼吁如下：
http://maps.googleapis.com/maps/api/geocode/json?latlng=44.52357920000001,-89.5745630&sensor=true
您可以从“位置”获取lat / lng值。希望这有帮助！
答案 1 :(得分：1)
上面最受欢迎的答案是不完整的。请查看下面的更多最新答案，我在其中描述了我个人经过验证的方法，以便从Google Maps API获得最准确的结果。在拥有超过1亿个独特位置的网站上进行测试。
每个城市都有很多邮政编码。
我之前遇到过类似的问题，因为我必须为包含混合位置+关键字的站点地图生成超过100万个链接组合。
首先，我尝试添加关键字“center”，“central”＆amp;城市名称以及国家的“中心”，它在80％的时间里工作，由于我必须完成的工作量，这还不够好。
所以我一直在寻找更好的解决方案，最后，我找到了2个Google Maps Geocode API的新参数，只需复制/粘贴查询网址中的部分结果即可。
请注意：Google未记录此信息，虽然它刚刚开始工作，但它可能在将来无效。
第一个参数：
&components=country:UK // where "UK" is a country of choice, by utilising this method, rather than adding the Country to the City name, you will avoid clashes and reduce the risk of not getting the postcode.
第二个参数：
&location_type=GEOMETRIC_CENTER& // as is, this will get you a place closest to the central geometrical location of the town/city.
完整示例：
var city_name = 'Edinburgh'; // City/Town/Place Name
var country_code = 'GB'; // Great Britain
var key = 'AIzaSyBk********************cM' // Google API Key
var query = https://maps.googleapis.com/maps/api/geocode/json?address='+city_name+'&components=country:'+country_code+'&location_type=GEOMETRIC_CENTER&key='+key+'&sensor=false
当循环通过JSON时，有时候POSTCODE不在第一层结果中，所以一定要循环第二行，如果第一行缺少POSTCODE。
以下是数组的循环示例：
url = geocode_query;
fetch(url)
.then(res => res.json())
.then((out) => {
result = JSON.parse(out);
postcode = get_postcode(result); // HERE is Your Postcode do what you need with it
})
.catch(err => { throw err });
function get_postcode(results){
city_data = results['results'][0]['address_components'];
for(i=0;i<city_data.length;i++){
var cv = city_data[i];
if(typeof cv['types'][0] != 'undefined')){
if(cv['types'][0] === 'postal_code'){
city['postcode'] = cv['long_name'];
}else if(cv['types'][0] === 'postal_town'){
city['place_name'] = cv['postal_town'];
}
}
}
if(typeof city == 'undefined'){
city_data = results['results'][1]['address_components'];
for(i=0;i<city_data.length;i++){
var cv = city_data[i];
if(typeof cv['types'][0] != 'undefined')){
if(cv['types'][0] === 'postal_code'){
city['postcode'] = cv['long_name'];
}
}
}
}
return city;
}
享受！
答案 2 :(得分：0)
我这样做的方法是打两个电话。
On the first call my query is: geocode('address=' .$cty. ',' .$st, $key);
$cty = city name - $st = state abbreviation - $key is my api key
-
On the second call my query is: geocode('latlng=' .$lat. "," .$lng, $key);
$lat = latitude from first call - $lng = longitude from first call
---------
My function appears below.
--------------------------------------
function geocode($query, $key){
global $lat, $lng, $zipcode, $city, $state;
$url = 'https://maps.googleapis.com/maps/api/geocode/json?'.$query.'&key='.$key;
$json_string = curlfun($url); // this uses CURL to access the url (false on fail)
if($json_string){
$parsed_json = json_decode($json_string, true);
//
if ($parsed_json['status'] == "OK"){
$lat = $parsed_json['results'] [0] ['geometry'] ['location'] ['lat'];
$lng = $parsed_json['results'] [0] ['geometry'] ['location'] ['lng'];
foreach($parsed_json['results'] [0] ['address_components'] as $a){
if($a ['types'] [0] == 'postal_code') $zipcode = $a ['long_name'];
if($a ['types'] [0] == 'locality') $city = $a ['long_name'];
if($a ['types'] [0] == 'administrative_area_level_1') $state = $a ['short_name'];
}
}
else return false;
}
else return false;
if(!$city) return false; // if there is no city just return false.
return true;
}
---------------------------------------------------
在函数调用之后，脚本的其余部分都可以使用全局变量。该函数在失败时返回false，在成功时返回true。应该在主代码中进行适当的错误处理。
答案 3 :(得分：-1)
var res; // store response in res variable
var add_array = res[0].address_components; //add_array = {
"long_name" : "Clarkston",
"short_name" : "Clarkston",
"types" : [ "locality", "political" ]
},
{
"long_name" : "Oakland",
"short_name" : "Oakland",
"types" : [ "administrative_area_level_2", "political" ]
},
{
"long_name" : "Michigan",
"short_name" : "MI",
"types" : [ "administrative_area_level_1", "political" ]
},
{
"long_name" : "United States",
"short_name" : "US",
"types" : [ "country", "political" ]
},
{
"long_name" : "48346",
"short_name" : "48346",
"types" : [ "postal_code" ]
}
var add_array = add_array[add_array.length-1]; //add_array = {
"long_name" : "48346",
"short_name" : "48346",
"types" : [ "postal_code" ]
}
var zip = add_array.long_name; //zip = 48346