Javascript 如何使用AJAX更改完整内容
我希望当我按下从Words.html到SelectNumber.html的按钮时,整个页面的内容都会发生变化Javascript 如何使用AJAX更改完整内容,javascript,html,jquery,css,ajax,Javascript,Html,Jquery,Css,Ajax,我希望当我按下从Words.html到SelectNumber.html的按钮时,整个页面的内容都会发生变化 <html> <head> <meta charset="UTF-8"> <title>Number Game</title> <link rel = "stylesheet" href = "stylesheet.css"> </head
<html>
<head>
<meta charset="UTF-8">
<title>Number Game</title>
<link rel = "stylesheet" href = "stylesheet.css">
</head>
<body>
<div id="Screen2">
<p> Hello World</p>
</div>
<script src = "Main.js"></script>
</body>
</html>
这是Words.html
<html>
<head>
<meta charset="UTF-8">
<title>Number Game</title>
<link rel = "stylesheet" href = "stylesheet.css">
</head>
<body>
<div id = "firstScreen">
<h1 id ="Title" class = "title">
The<br>Number Game
</h1>
<input type = "image" src = "button.png" class = "button1" onclick = "loadScreen">
<h3 class = "start">START</h3>
</div>
</body>
<script src = "Main.js"> </script>
</html>
这是SelectNumber.html
<html>
<head>
<meta charset="UTF-8">
<title>Number Game</title>
<link rel = "stylesheet" href = "stylesheet.css">
</head>
<body>
<div id="Screen2">
<p> Hello World</p>
</div>
<script src = "Main.js"></script>
</body>
</html>
数字游戏
你好,世界
当我按下输入按钮时,我希望整个内容从Words.html更改为NumberSelect.html。
函数加载屏幕(){
var load=new XMLHttpRequest();
load.onreadystatechange=函数(){
document.getElementById(“firstScreen”).innerHTML=this.responseText;
}
load.open(“GET”、“select.html”、true);
load.send();
}
函数myFunction(加载){
document.getElementById(“firstScreen”).innerHTML=this.responseText;
}
加载屏幕()代码>//这是正确的代码//您犯了一些类型和结束标记错误//而且您没有声明“loadscreen()”这是有效的解决方案-我认为loadscreen();使事情出错。它使Words.html上的内容消失(我的意思是什么也不显示);