使用onsubmit事件触发ajax时返回值怎么返回
这个是表单
<br /> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><br /> <html xmlns="http://www.w3.org/1999/xhtml"><br /> <head><br /> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /><br /> <title>无标题文档</title><br /> <br /> <br /> </head><br /> <br /> <body><br /> <br /> <form action="abc.php" method="post" name="form" onSubmit="return check_form()"/><br /> 用户名<input type="text" name="username" id="username" /><br /> 验证码<input type="text" name="identifying_code" id="identifying_code"/><br /> <input type="submit" name="login" value="登录"><br /> </form><br /> <div id="tishi"></div><br /> <script language="javascript" src="js/xmlhttprequest.js"></script><br /> <script language="javascript" src="js/js2.js"></script><br /> </body><br /> </html><br /> <br />
这个是xmlhttprequest.js页面
<br />
var xmlhttp = false;<br />
if (window.XMLHttpRequest) { //Mozilla、Safari等浏览器<br />
xmlhttp=new XMLHttpRequest();<br />
} <br />
else if (window.ActiveXObject) { //IE浏览器<br />
try {<br />
xmlhttp = new ActiveXObject("Msxml2.XMLHTTP");<br />
} catch (e) {<br />
try {<br />
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");<br />
} catch (e) {}<br />
}<br />
}<br />
这个是js2.js页面里面有我标记的路线可以看到整个执行的过程。
<br />
var flag;<br />
function check_form(){<br />
<br />
var username=document.getElementById('username').value;<br />
var identifying_code=document.getElementById('identifying_code').value;<br />
url='login_chk.php?username='+username+'&identifying_code='+identifying_code;<br />
alert('这里第1步');<br />
xmlhttp.open('get',url,true);<br />
alert('这里第2步');<br />
xmlhttp.onreadystatechange = function(){<br />
alert('这里第5步');<br />
if(xmlhttp.readyState == 4){<br />
alert('这里第6步');<br />
if(xmlhttp.status == 200){<br />
alert('这里第7步');<br />
msg = xmlhttp.responseText;<br />
alert('这里第8步');<br />
<br />
if(msg==1){<br />
document.getElementById("tishi").innerHTML="<font color='#FF0000'>输入正确正在跳转</font>"; <br />
flag=true;<br />
<br />
}<br />
else{<br />
document.getElementById("tishi").innerHTML="<font color='#FF0000'>用户名或者密码错误</font>";<br />
flag=false;<br />
}<br />
alert('这里第9步');<br />
<br />
}<br />
}<br />
<br />
//alert(flag);<br />
}<br />
xmlhttp.send(null);<br />
alert('这里第3步');<br />
<br />
<br />
if(flag==true){<br />
alert('这里是true');<br />
alert(flag);<br />
return true;<br />
}<br />
else {<br />
alert('这里第4步');<br />
alert(flag);<br />
return false;<br />
}<br />
<br />
} <br />
这个是login_chk.php页面
<br />
<?php<br />
if(strcmp($_GET['username'],'abcd')==0&&strcmp($_GET['identifying_code'],'abcd')==0){<br />
$msg=1;<br />
<br />
}<br />
else{<br />
$msg=0;<br />
}<br />
<br />
echo $msg;<br />
<br />
?><br />
<br />
最重要的是要取得check_form的返回值，才能决定表单是否能提交。求高手更改代码取到check_form的返回值
------解决方案--------------------
xmlhttp.open('get',url,true);
这是异步通讯，所以你要定义接受返回数据的回调函数
xmlhttp.onreadystatechange = function(){
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