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我将如何解析 node.js 中的以下 JSON 以提取 temp 和 city 的值

{
"message":"",
"cod":"200",
"type":"base",
"calctime":"",
"units":"internal",
"count":1,
"list":
    [
        {"id":2823368,
        "coord":{"lat":47.666672,"lon":9.6},
        "name":"London",
        "main":{"temp":275.79,"pressure":1020,"humidity":74,"temp_min":272.59,"temp_max":281.48},
        "dt":1362137169,
        "date":"2013-03-01 11:26:09",
        "wind":{"speed":1.5,"deg":0},
        "clouds":{"all":90},
        "weather":[{"id":804,"main":"Clouds","description":"overcast clouds","icon":"04d"}],
        "sys":{"country":"DE","population":18135},
        "url":"http:\/\/openweathermap.org\/city\/2823368"
        }
        ]
}

我通过以下方式获得上述 JSON:

var response = JSON.parse(body);

console.log(response);

任何帮助将非常感激。

4

2 回答 2

1

使用以下获取临时和城市(来自 url 的城市)或使用名称

var temp = response.list[0].main.temp,
url = response.list[0].url,
city = url.split('/')[3],
name = response.list[0].name;
于 2013-03-01T14:25:30.667 回答
0
var temp = response.list[0].main.temp;
var city = response.list[0].name;

至于“城市”,我不确定您在寻找什么,因为您的输入中没有该名称的键,但我猜了一下。

于 2013-03-01T12:27:44.870 回答