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如果我理解谷歌错误 OVER_QUERY_LIMIT 正确，那是因为 IP 请求太多。但是如果我去浏览器并加载以下链接： http ://maps.googleapis.com/maps/api/geocode/xml?address=Paris&sensor=true

我可以看到xml！

那为什么呢？

我的代码有问题吗？

这里是：

    $the_map = htmlspecialchars($_POST['map'], ENT_QUOTES);

    error_reporting(0);
    $map_query = file_get_contents('http://maps.google.com/maps/api/geocode/xml?address='.rawurlencode($the_map).'&sensor=false');
    error_reporting(1);

    if ($map_query){
        $geo_code = simplexml_load_string(utf8_encode($map_query));

        if ($geo_code){
            $map_lat = $geo_code->xpath("/GeocodeResponse/result/geometry/location/lat");
            $map_lng = $geo_code->xpath("/GeocodeResponse/result/geometry/location/lng");
            $map_lat = $map_lat[0][0];
            $map_lng = $map_lng[0][0];
        }
    }
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1 回答 1

1

使用curl而不是file_get_contents：

$address = "India+Panchkula";
$url = "http://maps.google.com/maps/api/geocode/json?address=$address&sensor=false&region=India";
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_PROXYPORT, 3128);
curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 0);
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0);
$response = curl_exec($ch);
curl_close($ch);
$response_a = json_decode($response);
echo $lat = $response_a->results[0]->geometry->location->lat;
echo "<br />";
echo $long = $response_a->results[0]->geometry->location->lng;

或查看以下网址

将 file_get_contents 用于 google api 时发出警告

于 2012-09-06T15:32:44.273 回答