如何从Json对象中提取正确的值（Google Places Geocoder）

Question

对不起，我需要帮助。

这是我的目标（我无法复制＆amp;＆amp;＆amp; firebug）

与此相同（“Json输出格式”段落）：https://developers.google.com/maps/documentation/geocoding/

{
    "results": [
        {
            "address_components": [
                {
                    "long_name": "1600",
                    "short_name": "1600",
                    "types": ["street_number"]
                },
                {
                    "long_name": "Amphitheatre Pkwy",
                    "short_name": "Amphitheatre Pkwy",
                    "types": ["route"]
                },
                {
                    "long_name": "Mountain View",
                    "short_name": "Mountain View",
                    "types": ["locality", "political"]
                },
                {
                    "long_name": "Santa Clara",
                    "short_name": "Santa Clara",
                    "types": ["administrative_area_level_2", "political"]
                },
                {
                    "long_name": "California",
                    "short_name": "CA",
                    "types": ["administrative_area_level_1", "political"]
                },
                {
                    "long_name": "United States",
                    "short_name": "US",
                    "types": ["country", "political"]
                },
                {
                    "long_name": "94043",
                    "short_name": "94043",
                    "types": ["postal_code"]
                }
            ],
            "formatted_address": "1600 Amphitheatre Pkwy, Mountain View, CA 94043, USA",
            "geometry": {
                "location": {
                    "lat": 37.42291810,
                    "lng": -122.08542120
                },
                "location_type": "ROOFTOP",
                "viewport": {
                    "northeast": {
                        "lat": 37.42426708029149,
                        "lng": -122.0840722197085
                    },
                    "southwest": {
                        "lat": 37.42156911970850,
                        "lng": -122.0867701802915
                    }
                }
            },
            "types": ["street_address"]
        }
    ],
    "status": "OK"
}

我需要迭代address_components并提取administrative_area_level_2，如果不存在administrative_area_level_3则放入变量（并非所有地址都有3级，在这种情况下，在意大利，你可以使用2级）。

所以，我正在尝试这个：

$.each(result.address_components, function(index, types) {
    //FROM HERE PSEUDO CODE... PLEASE HELP
    if (types.type[0]['administrative_area_level_3'] == NULL)
    {
        var administrative_area_level_3 = types.type['administrative_area_level_2'];
    }
});

因此，对于我的图像，行政区域3级和2级将具有相同的值（罗马）。

拜托，你能帮帮我吗？非常感谢你

Answer 1

试试这个，

if (types.type[0]['administrative_area_level_3']==NULL)
                 -------------------------------------^

而不是

if (types.type[0]['administrative_area_level_3'])==NULL

Answer 2

尝试并尝试，这是“解决方案”：

        var area2 = null;
        var area3 = null;
        $.each(result.address_components, function(index, types){
            if (types.types[0]=='administrative_area_level_2')
            {
                area2 = types.long_name;
            }
            if (types.types[0]=='administrative_area_level_3')
            {
                area3 = types.long_name;
            }
        });
        if (area3==null)
        {
            area3 = area2;
            $('input[name=administrative_area_level_3]').val(area3);
        }