Question

这适用于天气应用。我试图从json中获取一个值，将其赋值给一个变量，然后在json回调中的函数调用中使用该变量作为参数。如果我使用switch语句，这是有效的，但我试图避免这种情况，所以我构造了一个带有对象和html值的函数。运行时，它应该选择正确的html并将其附加到div。

问题是它只有在明确定义＆＃39; icon＆＃39;的值时才有效。在iconPicker函数中。否则，最后的警报会一直给我“未定义的”＃。感觉我在这里错过了一些基本的范围问题。救命啊！

const iconPicker = (icon) => {
  const rainy = '<div class="icon rainy"><div class="cloud"></div><div class="rain"></div></div>';
  const sunny = '<div class="icon sunny"><div class="sun"><div class="rays"></div></div></div>';
  const overcast = '<div class="icon cloudy"><div class="cloud"></div><div class="cloud"></div></div>';
  const partlySunny = '<div class="cloud"></div><div class="sun"><div class="rays"></div></div>';
  const stormy = '<div class="icon thunder-storm"><div class="cloud"></div><div class="lightning"><div class="bolt"></div><div class="bolt"></div></div></div>';
  const snowy = '<div class="icon flurries"><div class="cloud"></div><div class="snow"><div class="flake"></div><div class="flake"></div></div></div>';

  const iconObj = {
    'rain': rainy,
    'chancerain': rainy,
    'clear': sunny,
    'sunny': sunny,
    'mostlysunny': sunny,
    'fog': overcast,
    'cloudly': overcast,
    'hazy': partlySunny,
    'mostlycloudy': partlySunny,
    'partlycloudy': partlySunny,
    'partlysunny': partlySunny,
    'chancetstorms': stormy,
    'tstorms': stormy,
    'chanceflurries': snowy,
    'chancesleet': snowy,
    'chancesnow': snowy,
    'flurries': snowy,
    'sleet': snowy,
    'snow': snowy
  }
  $('.icon').append(iconObj[icon]);
  alert(iconObj[icon]);
}

const currentConditions = () => {
  let icon;
  $.getJSON('http://api.wunderground.com/api/e95fb12f6c69ae61/geolookup/conditions/q/autoip.json', function(json) {
    console.log(json);
    tempf = json.current_observation.temp_f;
    tempc = json.current_observation.temp_c;
    icon = json.current_observation.icon;
    city = json.location.city;
    country = json.location.country_name;
    $('.location').html(city + ', ' + country);
    $('.weather').html(tempf + '&deg' + '<a href=# class="scale scaleF">F</a>');
    $('.weather').on('click', 'a', function(e) {
      e.preventDefault();
      let element = $(e.target);
      if(element.hasClass('scaleF')) {
        $('.weather').html(tempc + '&deg' + '<a href=# class="scale scaleC">C</a>');
      } else if (element.hasClass('scaleC')){
        $('.weather').html(tempf + '&deg' + '<a href=# class="scale scaleF">F</a>');
      }
    });
    iconPicker(icon);
  });
};

使用json回调中的值作为另一个函数中的参数

0 个答案: