javascript - 提取地址组件的更有效方法

Question

目前，我正在使用以下代码获取国家/地区、邮政编码、地区和副地区:

var country, postal_code, locality, sublocality;
for (i = 0; i < results[0].address_components.length; ++i)
{
    for (j = 0; j < results[0].address_components[i].types.length; ++j)
    {
        if (!country && results[0].address_components[i].types[j] == "country")
            country = results[0].address_components[i].long_name;
        else if (!postal_code && results[0].address_components[i].types[j] == "postal_code")
            postal_code = results[0].address_components[i].long_name;
        else if (!locality && results[0].address_components[i].types[j] == "locality")
            locality = results[0].address_components[i].long_name;
        else if (!sublocality && results[0].address_components[i].types[j] == "sublocality")
            sublocality = results[0].address_components[i].long_name;
    }
}

这是不令人满意的。有没有其他方法可以达到同样的效果？

Answer 1

您可以使用以下函数来提取任何地址部分:

function extractFromAdress(components, type){
    for (var i=0; i<components.length; i++)
        for (var j=0; j<components[i].types.length; j++)
            if (components[i].types[j]==type) return components[i].long_name;
    return "";
}

要提取您调用的信息:

var postCode = extractFromAdress(results[0].address_components, "postal_code");
var street = extractFromAdress(results[0].address_components, "route");
var town = extractFromAdress(results[0].address_components, "locality");
var country = extractFromAdress(results[0].address_components, "country");

等...