最简单的ajax实现
程序员文章站
2022-06-15 14:18:07
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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>测试ajax</title>
<script type="text/javascript" charset="utf-8">
var request = false;
function createRequest(){
try{
request = new XMLHttpRequest();
}catch (e1)
{
try{
request = new ActiveXObject("Msxml2.XMLHTTP");
}catch (e2)
{
try{
request = new ActiveXObject("Microsoft.XMLHTTP");
}
catch (e3)
{
request = false;
}
}
}
}
function updatePage()
{
if(request.readyState == 4)
{
if(request.status == 200)
{
//alert("succeed");
var test = document.getElementById("test");
test.value = request.responseText;
}
else if(request.status == 404)
{
alert("url is not exist");
}
else
{
alert("requset.status="+request.status);
}
}
}
function getCustomerInfo()
{
//createRequest();
var phone = document.getElementById("phone").value;
var url = "11.txt";
createRequest();
if (request) {
request.open("GET", url, true);
request.onreadystatechange = updatePage;
request.send(null);
} else {
alert("XMLGttpRequest init failed");
}
}
</script>
</head>
<body>
<h1>It works!</h1>
<form action="">
<div>
<input type="text" class="phone" id="phone" name="phone" onchange="getCustomerInfo()"/>
<input type="text" id="test" />
</div>
</form>
</body>
</html>转载于:https://my.oschina.net/u/574191/blog/495685