日期:2014-05-17  浏览次数:20779 次

求助:表单数据不能提交到javascript函数里。不知道哪里错了?
<html>

<head>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312">
<meta name="GENERATOR" content="Microsoft FrontPage 4.0">
<meta name="ProgId" content="FrontPage.Editor.Document">
<title>系统</title>
<link rel=stylesheet href="style.css" type="text/css">

<script type="text/javascript">

function jumpbj()
{
var bj=document.getElementById("banji");
alert(bj);  if(bj.value>=1&&bj.value<=40)
  {var myurl='teacher-'+bj+'.asp';
  var str = 'window.open("' + myurl + '")';
  eval(str);
  return true;}
  else
  {alert("你输入的班级代码不对,请从新输入!");
  return false;}
}
//-->
</script>

<style type="text/css">
<!--
.STYLE1 {
font-size: large;
color: #FF0000;
font-weight: bold;
font-family: "宋体";
}
body {
background-image: url(bg2003.gif);
}
.STYLE2 {font-family: "宋体"}
-->
</style>
</head>

<body topmargin="0">
<p>&nbsp;</p>
<p>&nbsp;</p>
<div align="center">
  <!--#include file="top.asp"--><div align="center">
  <center> </div>
  
  <table border="0" width="700" cellspacing="0" cellpadding="0">
  <tr> 
  <td width="100%"> <p>&nbsp;</p>
  <table width="90%" border="1" align="center" cellpadding="3" cellspacing="3" bordercolorlight="#008080" bordercolordark="#FFFFFF" id="b1">
  <!--DWLayoutTable-->
  <form action="" method="post" name="form" onsubmit="return jumpbj()">
  <tr> 
  <td width="457" height="27" valign="top"><span class="STYLE3">请输入你的班级代码:</span> 
  <input name="banji" type="VarNumeric" id="banji" size="6" >
    
  <label> 
  <input type="submit" name="Submit" value="提交">
  </form>
</body>
</html>
================================================
代码中红色的 alert(bj);显示为[object],asp表单中的bj没有被提取或提交?

------解决方案--------------------
var myurl='teacher-'+bj.value+'.asp'
另外少用eval 直接window.open(myurl);