解析JSON在node.js中

我将如何解析以下JSON在node.js中提取温度和城市价值解析JSON在node.js中

{ 
"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" 
     } 
     ] 
}

我通过获取上述JSON：

var response = JSON.parse(body); 

console.log(response);

任何帮助将非常感激。

来源

2013-03-01 spaniard89

使用下面进入温度和城（从URL）或者使用名称

var temp = response.list[0].main.temp, 
url = response.list[0].url, 
city = url.split('/')[3], 
name = response.list[0].name;

来源

2013-03-01 14:25:30 user568109

var temp = response.list[0].main.temp; 
var city = response.list[0].name;

至于“城市”我不知道你在找什么，因为有在你输入该名称的关键，但我花了猜测。

来源

2013-03-01 12:27:44

解析JSON在node.js中

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