使用PHP解析Google地理编码JSON

Question

我正在尝试解析来自Google Geocode API的json响应,但我在理解它时遇到了一些麻烦.

对于那些不熟悉Geocode API的人,请访问以下网址:http://maps.google.com/maps/api/geocode/json？address = alb%20square&_sensor = false

我正在使用以下代码来解析请求

<?php

$address = urlencode($_POST['address']);
$request = file_get_contents("http://maps.google.com/maps/api/geocode/json?address=" . $address . "&sensor=false");
$json = json_decode($request, true);

?>

我正在尝试输出以下内容:

echo $json['results'][0]['formated_address'];

我不确定为什么没有回应.我也试过了$json[0]['results'][0]['formated_address'].我知道这是一个noob问题,但是多维数组让我很困惑.

Answer 1

echo $json['results'][0]['formatted_address'];

如果你正确拼写它会有所帮助;-)

Answer 2

...仅供参考,JSON请求URL必须正确格式化才能返回任何好的内容,否则您可能会收到NULL响应.

例如,这是我从JSON响应中检索经度和纬度值的方法:

// some address values
    $client_address = '123 street';
    $client_city = 'Some City';
    $client_state = 'Some State';
    $client_zip = 'postal code';

// building the JSON URL string for Google API call 
    $g_address = str_replace(' ', '+', trim($client_address)).",";
    $g_city    = '+'.str_replace(' ', '+', trim($client_city)).",";
    $g_state   = '+'.str_replace(' ', '+', trim($client_state));
    $g_zip     = isset($client_zip)? '+'.str_replace(' ', '', trim($client_zip)) : '';

$g_addr_str = $g_address.$g_city.$g_state.$g_zip;       
$url = "http://maps.google.com/maps/api/geocode/json?
        address=$g_addr_str&sensor=false";

// Parsing the JSON response from the Google Geocode API to get exact map coordinates:
// latitude , longitude (see the Google doc. for the complete data return here:
// https://developers.google.com/maps/documentation/geocoding/.)

$jsonData   = file_get_contents($url);

$data = json_decode($jsonData);

$xlat = $data->{'results'}[0]->{'geometry'}->{'location'}->{'lat'};
$xlong = $data->{'results'}[0]->{'geometry'}->{'location'}->{'lng'};

echo $xlat.",".$xlong;

...另外,如果v3 API不起作用,您可以使用相同的代码对Google Map iframe进行硬编码并将其嵌入到您的页面中...请参阅此处的简单教程:http://pmcds.ca/blog/嵌入-谷歌-地图-到最webpage.html

嵌入 不是正确的方法,但有时它成为需要的解决方案.