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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