Question

我正在使用Google地图返回位置名称以及用户所在位置的状态。这是我感兴趣的部分：

"address_components" : [
        {
           "long_name" : "Salem",
           "short_name" : "Salem",
           "types" : [ "locality", "political" ]
        },
        {
           "long_name" : "Rockingham",
           "short_name" : "Rockingham",
           "types" : [ "administrative_area_level_2", "political" ]
        },
        {
           "long_name" : "New Hampshire",
           "short_name" : "NH",
           "types" : [ "administrative_area_level_1", "political" ]
        },
        {
           "long_name" : "United States",
           "short_name" : "US",
           "types" : [ "country", "political" ]
        }
     ]

有没有办法找到哪些对象包含“locality”和“administrative_area_level_1”值，并使用JavaScript或jQuery返回相应对象的“long_name”字符串？我该怎么做？

Answer 1

你去吧。请找小提琴here

使用如下功能：

function getLocality(){
    for(var i = 0 ; i< address_components.length; i++){
        var obj = address_components[i];
        var arr = obj["types"];
        for(var j = 0; j<arr.length;j++ ){
            if(arr[j] == "locality"){
                return obj;
            }
        }

    }
}

或者更确切地说编写相同的Array原型

Array.prototype.getByType = function(type){
    for(var i = 0 ; i< address_components.length; i++){
        var obj = address_components[i];
        var arr = obj["types"];
        for(var j = 0; j<arr.length;j++ ){
            if(arr[j] == type){
                return obj;
            }
        }

    }
}

使用原型如下：

address_components.getByType("administrative_area_level_1").long_name // Note that you pass the type to the prototype function that does the job for you

其中address_components是Google地图返回的数组

Answer 2

您可以使用此JS库，而不是污染您的代码; DefiantJS（http://defiantjs.com），它使用“搜索”方法扩展全局对象JSON。此方法使您可以使用XPath查询语法搜索JSON结构。结果是您的代码变得更加清晰和可读。

这是工作小提琴;
http://jsfiddle.net/hbi99/SjvZ9/

var data = {
       ...
    },
    res = JSON.search( data, '//*[types="administrative_area_level_1" or types="locality"]' ),
    str = '';

    for (var i=0; i<res.length; i++) {
        str += res[i].long_name +'<br/>';
    }

    document.getElementById('output').innerHTML = str;

不熟悉XPath？查看这个XPath评估器，给你一个简单的提示; http://www.defiantjs.com/#xpath_evaluator

Answer 3

试试这个：

var data = {"address_components" : [
        {
           "long_name" : "Salem",
           "short_name" : "Salem",
           "types" : [ "locality", "political" ]
        },
        {
           "long_name" : "Rockingham",
           "short_name" : "Rockingham",
           "types" : [ "administrative_area_level_2", "political" ]
        },
        {
           "long_name" : "New Hampshire",
           "short_name" : "NH",
           "types" : [ "administrative_area_level_1", "political" ]
        },
        {
           "long_name" : "United States",
           "short_name" : "US",
           "types" : [ "country", "political" ]
        }
]};
for(var i = 0; i < data.address_components.length; i++){
    if(data[i].long_name == 'something'){
      // do something
    }
    //or
    if(data[i].types[0] == 'administrative_area_level_1'){
      // do something
    }
}

Answer 4

试试这个，

var addr_comp = [{"long_name": "Salem","short_name": "Salem","types": ["locality", "political"]}, {"long_name": "Rockingham","short_name": "Rockingham","types": ["administrative_area_level_2", "political"]}, {"long_name": "New Hampshire","short_name": "NH","types": ["administrative_area_level_1", "political"]}, {"long_name": "United States","short_name": "US","types": ["country", "political"]}];
function getlocalityAdminLevel(addrType){
    for(var a=0,len=addr_comp.length;a<len;a++){
        ac=addr_comp[a];
        if(ac.types){
            for (var t = 0, tl = ac.types.length; t < tl; t++) {
                ty = ac.types[t];
                if (ty == addrType) {
                    return ac;break;
                } 
            }
        }
    }
}
console.log(getlocalityAdminLevel('locality'));
console.log(getlocalityAdminLevel('administrative_area_level_1'));

Working demo

您可以像使用

一样使用long_name和short_name

locality=getlocalityAdminLevel('locality');
if(locality && locality.long_name){
    alert(locality.long_name);
}

Demo 1

使用JavaScript / jQuery搜索JSON数组

4 个答案: