如何更改此 JavaScript 以设置背景颜色
how can I change this JavaScript to set the background colour
所以我得到了这段代码,其中 returns 您输入的位置的天气。
与其只返回一个图标,我希望它也能改变背景的颜色。
我只是无法理解这一切。
function getLocation(){
var location = document.getElementById("location").value;
location = location.replace(" ", "%20");
if (location == ""){
document.getElementById("location").classList.add("error");
}
else {
document.getElementById("location").classList.remove("error");
getWeather(location);
}
}
function getWeather(location){
var ajax = new XMLHttpRequest();
var json;
var apiKEY = "3521a940efd69dc5b6f3dd982d18c618";
var url = "http://api.openweathermap.org/data/2.5/weather?q=" + location + " ,uk&appid=" + apiKEY;
ajax.open("GET", url, true);
ajax.send();
ajax.onreadystatechange = function(){
if (ajax.readyState == 4 && ajax.status == 200){
json = JSON.parse(ajax.responseText);
document.getElementById("locationForm").style.display = "none";
document.getElementById("weather").style.display = "block";
if (json != undefined){
var weather = json.weather[0].main
setIconAndDescription(weather, location)
}
else {
description = "Oops, I couldn't find the weather in " + location;
document.getElementById("description").innerHTML = description;
}
}
}
}
function setIconAndDescription(weather, location){
var icon;
var description;
weather = weather.toLowerCase();
if (weather == "clear sky"
|| weather == "clear"){
icon = "clear.svg";
description = "Yay, sunshine!";
}
else if (weather == "few clouds"){
icon = "few-clouds.svg";
description = "It's a little cloudy.";
}
这个想法是使用颜色理论,即红色表示热,蓝色表示红色等等。
任何帮助都会很棒,
谢谢,
扎克
你的意思是改变页面的背景颜色?在那种情况下,只需使用 document.body.style.backgroundColor = "red";
我想你想在这个阶段做:
函数 setIconAndDescription(天气、位置){
var icon;
var description;
weather = weather.toLowerCase();
if (weather == "clear sky"
|| weather == "clear"){
icon = "clear.svg";
description = "Yay, sunshine!";
document.getElementById("description").style.backgroundColor = 'your-color';
}
else if (weather == "few clouds"){
icon = "few-clouds.svg";
description = "It's a little cloudy.";
document.getElementById("description").style.backgroundColor = 'your-color';
}
}
所以我得到了这段代码,其中 returns 您输入的位置的天气。
与其只返回一个图标,我希望它也能改变背景的颜色。
我只是无法理解这一切。
function getLocation(){
var location = document.getElementById("location").value;
location = location.replace(" ", "%20");
if (location == ""){
document.getElementById("location").classList.add("error");
}
else {
document.getElementById("location").classList.remove("error");
getWeather(location);
}
}
function getWeather(location){
var ajax = new XMLHttpRequest();
var json;
var apiKEY = "3521a940efd69dc5b6f3dd982d18c618";
var url = "http://api.openweathermap.org/data/2.5/weather?q=" + location + " ,uk&appid=" + apiKEY;
ajax.open("GET", url, true);
ajax.send();
ajax.onreadystatechange = function(){
if (ajax.readyState == 4 && ajax.status == 200){
json = JSON.parse(ajax.responseText);
document.getElementById("locationForm").style.display = "none";
document.getElementById("weather").style.display = "block";
if (json != undefined){
var weather = json.weather[0].main
setIconAndDescription(weather, location)
}
else {
description = "Oops, I couldn't find the weather in " + location;
document.getElementById("description").innerHTML = description;
}
}
}
}
function setIconAndDescription(weather, location){
var icon;
var description;
weather = weather.toLowerCase();
if (weather == "clear sky"
|| weather == "clear"){
icon = "clear.svg";
description = "Yay, sunshine!";
}
else if (weather == "few clouds"){
icon = "few-clouds.svg";
description = "It's a little cloudy.";
}
这个想法是使用颜色理论,即红色表示热,蓝色表示红色等等。
任何帮助都会很棒,
谢谢,
扎克
你的意思是改变页面的背景颜色?在那种情况下,只需使用 document.body.style.backgroundColor = "red";
我想你想在这个阶段做:
函数 setIconAndDescription(天气、位置){
var icon;
var description;
weather = weather.toLowerCase();
if (weather == "clear sky"
|| weather == "clear"){
icon = "clear.svg";
description = "Yay, sunshine!";
document.getElementById("description").style.backgroundColor = 'your-color';
}
else if (weather == "few clouds"){
icon = "few-clouds.svg";
description = "It's a little cloudy.";
document.getElementById("description").style.backgroundColor = 'your-color';
}
}