我想知道是否有另一种方法可以用更少的代码来提高此Switch语句的效率?
我听说过对象更好,更清洁,因此不再需要使用中断。如何正确使用?
谢谢。
代码在这里-https://jsfiddle.net/lmanhaes/cq1g5dyt/4/
$.each(response.weather, function(index) { //retrieve data
let icon;
switch (response.weather[index].currentConditions) { //switch case for icons
case "Cloud":
icon = '<img src="./weather_icons/cloud.png" alt="cloud" width="22px"/>';
break;
case "Hail":
icon = '<img src="./weather_icons/hail.png" alt="hail" width="22px"/>';
break;
case "Heavy Cloud":
icon = '<img src="./weather_icons/heavy cloud.png" alt="heavy-clouds" width="22px"/>';
break;
case "Heavy Rain":
icon = '<img src="./weather_icons/heavy rain.png" alt="heavy-rain" width="22px"/>';
break;
case "Rain":
icon = '<img src="./weather_icons/rain.png" alt="rain" width="22px"/>';
break;
case "Sleet":
icon = '<img src="./weather_icons/sleet.png" alt="sleet" width="22px"/>';
break;
case "Snow":
icon = '<img src="./weather_icons/snow.png" alt="snow" width="22px"/>';
break;
case "Sun":
icon = '<img src="./weather_icons/sun.png" alt="sun" width="22px"/>';
break;
case "Sun and Clouds":
icon = '<img src="./weather_icons/sun and cloud.png" alt="sun-clouds" width="22px"/>';
break
case "Thunderstorm":
icon = '<img src="./weather_icons/thunderstorm.png" alt="thunderstorm" width="22px"/>';
break;
}
当然。您可以建立天气类型字典。
const weather = {
"Cloud": {
"file": "cloud.png",
"label": "cloud"
},
// ...
};
然后使用:
const entry = weather[response.weather[index].currentConditions];
let icon = `<img src="./weather_icons/${entry.file}" alt="${entry.label}" width="22px" />`;
每个天气状况都有三个与之关联的唯一字符串:API返回的字符串,图像文件名和图像alt
。您可以使用由API响应字符串索引的对象来简洁地执行此操作,其中每个值可以是另一个对象或数组:
const weatherStrings = {
Cloud: ['cloud', 'cloud',
Hail: ['hail', 'hail'],
'Heavy Cloud': ['heavy cloud', 'heavy-clouds'],
// ...
}
// ...
success: function(response) {
for (const item of response.weather) {
const [filename, alt] = weatherStrings[item.currentConditions];
const icon = `<img src="./weather_icons/${filename}.png" alt="${alt}" width="22px">`;
// ...
}