Question

我需要访问温度并将其附加到我的页面如果这些都是对象，那么我想我不能使用索引，因为顺序无关紧要，这让我失望。

{
"response": {
"version": "0.1",
"termsofService": "http://www.wunderground.com/weather/api/d/terms.html",
"features": {
  "conditions": 1
 }
},
"current_observation": {
"image": {
  "url": "http://icons-ak.wxug.com/graphics/wu2/logo_130x80.png",
  "title": "Weather Underground",
  "link": "http://www.wunderground.com"
},
"display_location": {
  "full": "Toms River, NJ",
  "city": "Toms River"

},
"observation_location": {
  "full": "Stonehedge, Toms River, New Jersey",
  "city": "Stonehedge, Toms River",
  "state": "New Jersey",
  "country": "US"

},
"estimated": {},
"station_id": "KNJTOMSR5",
"observation_time": "Last Updated on April 11, 10:56 PM EDT",
"observation_time_rfc822": "Fri, 11 Apr 2014 22:56:39 -0400",
"observation_epoch": "1397271399",
"local_time_rfc822": "Fri, 11 Apr 2014 22:56:40 -0400",
"local_epoch": "1397271400",
"local_tz_short": "EDT",
"local_tz_long": "America/New_York",
"local_tz_offset": "-0400",
"weather": "Clear",
**"temperature_string": "59.6 F (15.3 C)"**

 }
}

我的javascript现在显然是因为我还没弄明白如何访问这个项目

$(function() {
$("#getzip").submit(function() {
var zip_data =$(this).serialize();
    $.getJSON("get_weather.php",null, function(data); {

我需要将它附加到我的页面的dom，我认为它应该看起来像这样吗？

   (#output).append(data.temperature_string);

Answer 1

将json对象转换为javascript。

像这样使用。

在Php中

 $json ='{"temperature":"36 c"}';

在Javascript中

$.post("get_weather.php",null, function(data){
 var temperature = data.temperature; // here you'll get temperature from json into variable
 console.log(temperature);
           },"json");

全部

Answer 2

console.log（data [＆＃34; temperature_string＆＃34;]）应该访问该数据。

访问json对象中的特定属性对

2 个答案: