Question

我有一个$ .getJSON请求，从API中提取数据。我需要从该请求中提取数据以在其外部使用。这是代码：

  var weatherData = "http://api.openweathermap.org/data/2.5/weather?q=Location,us&appid=APIkey";

  $.getJSON(weatherData,function(data){
    var icon = data.weather[0].icon;
  });


  var skycons = new Skycons();
  var conditions  = [
        "clear-day", "clear-night", "partly-cloudy-day",
        "partly-cloudy-night", "cloudy", "rain", "sleet", "snow", "wind",
        "fog"], i;
  var icons = ["01d", "01n", "02d", "02n", "03d", "09d", "13d", "50d"];

  if (icon == "03n" || icon == "04d" || icon == "04n")
    icon = "03d";
  if (icon == "09n" || icon == "10d" || icon == "10n" || icon == "11d" || icon == "11n")
    icon = "09d";
  if (icon == "13n")
    icon = "13d";
  if (icon == "50n")
    icon = "50d";


  for(i = conditions.length; i--; )
    skycons.set(conditions[i], conditions[i]);

  skycons.play();

显然，问题是我需要能够访问“ var icon = data.weather [0] .icon;”。在$ .getJSON请求之外。

我了解jQuery是异步的。我已经阅读了这里的所有相关问题，但我无法理解答案。请使其尽可能简单。

Answer 1

万一有人碰到了我的代码并且也在使用Skycons，这就是我结合Openweathermap的API来根据我所在位置的当前天气情况显示相应的Skycon的方式。

      var weatherData = "http://api.openweathermap.org/data/2.5/weather?q=Location,us&appid=APIkey";

      $.getJSON(weatherData,function(data){
        var icon = data.weather[0].icon;

        var skycons = new Skycons();
        var conditions  = [
              "clear-day", "clear-night", "partly-cloudy-day",
              "partly-cloudy-night", "cloudy", "rain", "sleet", "snow", "wind",
              "fog"], i;
        var icons = ["01d", "01n", "02d", "02n", "03d", "09d", "13d", "50d"];

        if (icon == "03n" || icon == "04d" || icon == "04n")
          icon = "03d";
        if (icon == "09n" || icon == "10d" || icon == "10n" || icon == "11d" || icon == "11n")
          icon = "09d";
        if (icon == "13n")
          icon = "13d";
        if (icon == "50n")
          icon = "50d";

        icon = icons.indexOf(icon);
        icon = conditions[icon];


        document.getElementById("test").innerHTML = "<canvas id=\"" + icon + "\"></canvas>";
        
        for(i = conditions.length; i--; )
          skycons.set(conditions[i], conditions[i]);

        skycons.play();
      });

我简化了可能的图标代码，并删除了2个Skycon条件（雨夹雪和风），以更好地匹配Openweathermap API的图标代码。

在Javascript函数中从$ .getJSON请求中传递数据

1 个答案: