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我是使用 Json 的新手，我想从 Json array 获取 Particular Id 数据。
这是我正在使用的 api 网址。

http://example/plapi/areas?jsoncallback=?

这个返回 json 数组的 url 像

([{"area_id":"25","ca_id":"0","area_name":"Durgapuram","alias_area":"","area_sp‌​l":"No","area_lat":"16.52069","area_lng":"80.64067","area_desc":"","area_image":"‌​","map_image":"","area_status":"1","city_id":"27","position":"0","m_keyword":""},‌​{"area_id":"26","ca_id":"0","area_name":"Arundalpet","alias_area":"","area_spl":"‌​No","area_lat":"16.30520","area_lng":"80.44169","area_desc":"","area_image":"","m‌​ap_image":"","area_status":"1","city_id":"31","position":"0","m_keyword":""})]

在那我只想获取数据city_id = 31。

要获得 id 为 31 的数据，我正在这样做

var data = {
"id": "31",
};
$.getJSON("http://example/plapi/areas?jsoncallback=?",JSON.stringify(data) ,function(result){console.log(result);
});

但它什么也没返回。

我只想获得 31 个 id 相关数据而不是其他数据。请帮我怎么办？

4

1 回答 1

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查找 JS 代码City_ID仅从您的JSON:-

var jsonData = [
    {
       
        "area_id": "25",
        "ca_id": "0",
        "area_name": "Durgapuram",
        "alias_area": "",
        "area_spl": "No",
        "area_lat": "16.52069",
        "area_lng": "80.64067",
        "area_desc": "",
        "area_image": "1",
        "map_image": "",
        "area_status": "1",
         "city_id": "27",
        "position": "0",
        "m_keyword": ""
    },
   {
        
       "area_id": "26",
        "ca_id": "0",
        "area_name": "Arundalpet",
        "alias_area": "",
        "area_spl": "No",
        "area_lat": "16.30520",
        "area_lng": "80.44169",
        "area_desc": "",
        "area_image": "",
        "map_image": "",
        "area_status": "1",
       "city_id": "31",
        "position": "0",
        "m_keyword": ""
    }
];

for(var obj in jsonData){
    if(jsonData.hasOwnProperty(obj)){
    for(var prop in jsonData[obj]){
        if(jsonData[obj].hasOwnProperty(prop)){
            if(prop=="city_id"){
                prop;
                alert(prop);
                break;
            }

        }
    }
}
}

很高兴能帮助你

于 2015-08-27T06:16:59.817 回答