从谷歌地图反向地理编码解析json？

Question

如何使用Google Maps JavaScript API v3解析反向地理编码的响应.

geocoder.geocode({'latLng': latlng}, function(results, status) {
                if (status == google.maps.GeocoderStatus.OK) {
                    if (results[0]) {
                        infowindow.setContent(results[0].formatted_address);
                        infowindow.open(map, marker);
                    }
                } 

这会在弹出窗口中显示格式化的地址,但我正在尝试从响应中取出其他位,最好是街道名称或路径(如果找不到街道名称).但是在使用时 obj = JSON.parse(json);我一直在控制台中收到此错误.

SyntaxError:JSON.parse:意外字符

如果它是PHP我会做一堆for each循环.是否可以在JavaScript中执行类似的操作？

这是一个样本

{
   "results" : [
  {
     "address_components" : [
        {
           "long_name" : "131",
           "short_name" : "131",
           "types" : [ "street_number" ]
        },
        {
           "long_name" : "Stubbington Avenue",
           "short_name" : "Stubbington Ave",
           "types" : [ "route" ]
        },
        {
           "long_name" : "Portsmouth",
           "short_name" : "Portsmouth",
           "types" : [ "locality", "political" ]
        },
        {
           "long_name" : "Portsmouth",
           "short_name" : "Portsmouth",
           "types" : [ "administrative_area_level_3", "political" ]
        },
        {
           "long_name" : "Portsmouth",
           "short_name" : "Portsmouth",
           "types" : [ "administrative_area_level_2", "political" ]
        },
        {
           "long_name" : "England",
           "short_name" : "England",
           "types" : [ "administrative_area_level_1", "political" ]
        },
        {
           "long_name" : "United Kingdom",
           "short_name" : "GB",
           "types" : [ "country", "political" ]
        },
        {
           "long_name" : "PO2",
           "short_name" : "PO2",
           "types" : [ "postal_code_prefix", "postal_code" ]
        },
        {
           "long_name" : "Portsmouth",
           "short_name" : "Portsmouth",
           "types" : [ "postal_town" ]
        }
     ],
     "formatted_address" : "131 Stubbington Avenue, Portsmouth PO2, UK",
     "geometry" : {
        "location" : {
           "lat" : 50.8170795,
           "lng" : -1.0709701
        },
        "location_type" : "ROOFTOP",
        "viewport" : {
           "northeast" : {
              "lat" : 50.81842848029149,
              "lng" : -1.069621119708498
           },
           "southwest" : {
              "lat" : 50.8157305197085,
              "lng" : -1.072319080291502
           }
        }
     },
     "types" : [ "street_address" ]
  }
   ],
"status" : "OK"
}

还有一个指向我的开发页面的链接,其中包含我当前的代码

总之,如何从那里弄乱"Stubbington Avenue"？

Answer 1

你不需要JSON.parse那些结果,它已经是json了.

要从那个有效的json中获取"stubbington avenue",你会使用 results[0].address_components[1].short_name

如果你想从这些地址组件中构建实际地址,你可以循环查看打印到控制台的值,如下所示:

for(var i in results[0].address_components){
    console.log(results[0].address_components[i].short_name);
}

而不是记录它们,将它们附加到一个字符串,或将它们附加到一个元素,无论你想用它们做什么.