一个最简单的ajax问题
来源:百度知道 编辑:UC知道 时间:2024/09/24 04:54:23
html页
<script language="javascript">
function funphp100(url){
xmlHttp = new XMLHttpRequest();
xmlHttp.open("GET","php.php?id="+url,true);
xmlHttp.onreadystatechange =byphp();
xmlHttp.send(null);
}
function byphp(){
var byphp100 = xmlHttp.responseText;
document.getElementById('php100').innerHTML = byphp100
}
</script>
<a href="#"onclick="funphp100('o')">o</a>
<a href="#"onclick="funphp100('t')">t</a>
<a href="#"onclick="funphp100('x')">x</a>
<div id="php100"></div>
php页
<?php
$id=$_GET['id'];
for($i=0;$i<10;$i++){
echo $id;
}
?>
byphp100 没有得到值
哪里错了?
<script language="javascript">
function funphp100(url){
xmlHttp = new XMLHttpRequest();
xmlHttp.open("GET","php.php?id="+url,true);
xmlHttp.onreadystatechange =byphp();
xmlHttp.send(null);
}
function byphp(){
var byphp100 = xmlHttp.responseText;
document.getElementById('php100').innerHTML = byphp100
}
</script>
<a href="#"onclick="funphp100('o')">o</a>
<a href="#"onclick="funphp100('t')">t</a>
<a href="#"onclick="funphp100('x')">x</a>
<div id="php100"></div>
php页
<?php
$id=$_GET['id'];
for($i=0;$i<10;$i++){
echo $id;
}
?>
byphp100 没有得到值
哪里错了?
xmlHttp.onreadystatechange =byphp(); ----byphp后面没有()
应该为xmlHttp.onreadystatechange =byphp;
再者建议你装个火狐浏览器,安装个firebug-1.3.3-fx.xpi装个插件,能够显示出你Ajax里面的错误,