提取地址组件的更有效方法

Question

目前，我正在使用以下代码来获取国家/地区、邮政编码、地区和次地区：

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;
    }
}

这不令人满意。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组件：

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 "";
}

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");

等等……

Answer 2

我的单线使用函数式方法和map、filter 和 ES2015：

/**
 * Get the value for a given key in address_components
 * 
 * @param {Array} components address_components returned from Google maps autocomplete
 * @param type key for desired address component
 * @returns {String} value, if found, for given type (key)
 */
function extractFromAddress(components, type) {
    return components.filter((component) => component.types.indexOf(type) === 0).map((item) => item.long_name).pop() || null;
}

用法：

const place = autocomplete.getPlace();
const address_components = place["address_components"] || [];

const postal_code = extractFromAddress(address_components, "postal_code");

Answer 3

你可以把它缩短为

var country, postal_code, locality, sublocality;
for (i = 0; i < results[0].address_components.length; ++i) {
    var component = results[0].address_components[i];
    if (!sublocality && component.types.indexOf("sublocality") > -1)
        sublocality = component.long_name;
    else if (!locality && component.types.indexOf("locality") > -1)
        locality = component.long_name;
    else if (!postal_code && component.types.indexOf("postal_code") > -1)
        postal_code = component.long_name;
    else if (!country && component.types.indexOf("country") > -1)
        country = component.long_name;
}

或者您是否想获得更好的格式化结果？那么请向我们展示您的查询。

Answer 4

我是这样做的：

placeParser = function(place){
  result = {};
  for(var i = 0; i < place.address_components.length; i++){
    ac = place.address_components[i];
    result[ac.types[0]] = ac.long_name;
  }
  return result;
 };

那我就用

parsed = placeParser(place)
parsed.route

Answer 5

使用lodash

const result = _.chain(json.results[0].address_components)
  .keyBy('types[0]')
  .mapValues('short_name')
  .value()

Answer 6

我真的相信上面的user1429980 答案值得更多的认可。它真的很好用。我的回答是基于他的功能。我添加了一些示例以更好地说明如何使用提供的代码 user1429980 搜索 JSON 对象：

//searches object for a given key and returns the key's value

extractFromObject (object, key) { return object.filter((component) => component.types.indexOf(key) === 0).map((item)=>item.long_name).pop() || null; }

---

示例 1： Google 的 reverseGeocode API，经度和纬度设置为 43.6532,79.3832（加拿大安大略省多伦多市）：

var jsonData = {} //object contains data returned from reverseGeocode API

var city = extractFromObject(jsonData.json.results[0].address_components, 'locality');

console.log(city); //Output is Toronto

---

示例 2：Google 的 Places API，地点 ID 设置为 ChIJE9on3F3HwoAR9AhGJW_fL-I（美国加利福尼亚州洛杉矶）：

var jsonData = {} //object contains data returned from Google's Places API

var city = extractFromObject(jsonData.json.result.address_components, 'locality');

console.log(city); //Output is Los Angeles

Answer 7

使用 underscore.js 的访问者可以轻松地将地理编码响应中的 address_components 数组转换为对象字面量：

var obj = _.object( 
    _.map(results[0].address_components, function(c){ 
        return  [c.types[0], c.short_name] 
    })
);

Answer 8

if (typeof Object.keys == 'function')
    var length = function(x) { return Object.keys(x).length; };
else
    var length = function() {};

var location = {};      
for (i = 0; i < results[0].address_components.length; ++i)
{
    var component = results[0].address_components[i];
    if (!location.country && component.types.indexOf("country") > -1)
        location.country = component.long_name;
    else if (!location.postal_code && component.types.indexOf("postal_code") > -1)
        location.postal_code = component.long_name;
    else if (location.locality && component.types.indexOf("locality") > -1)
        location.locality = component.long_name;
    else if (location.sublocality && component.types.indexOf("sublocality") > -1)
        location.sublocality = component.long_name;

    // nothing will happen here if `Object.keys` isn't supported!
    if (length(location) == 4)
        break;
}

这是最适合我的解决方案。它也可能对某人有所帮助。

Answer 9

在给定地点类型列表的情况下，我在此之前创建了一个函数：

const getValue = function(data, types=[]){
/* used by results taken from Geocoder.geocode api */
const values = data.reduce((values, address) => {
    return address.address_components.reduce((values2, component) => {
        if(component.types.reduce((result, type) => result || types.indexOf(type) > -1, false))
            values2.push(component.long_name);
        return values2
    }, []);

    if(buff.length)
        return [...values, ...buff];
    return values;
}, []).filter(
    (value, index, self) => {
        return self.indexOf(value) === index;
    }
);

}