关于FORM


<input   type= "submit "   value= "提交 "   name= "sub "> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<input   type= "submit "   value= "修改 "   name= "sub0 ">

这个是一个预览的页面，就先前的一个页面输入数据，提交到这个页面，显示我想实现的是,如果我发现这个页面的信息正确的，我就点提交按钮发送到sub.asp，如果我发现信息错误，就点修改按钮，跳到edit.asp，请问该怎么办？

------解决方案--------------------
<input type= "submit " value= "提交 " name= "sub " onclick= "location.href= 'sub.asp ' "> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<input type= "submit " value= "修改 " name= "sub0 " onclick= "location.href= 'edit.asp ' ">

------解决方案--------------------
<form action= "sub.asp " method= "post ">
<input type= "submit " value= "提交 " name= "sub "> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<input type= "button " value= "修改 " name= "sub0 " onclick= "location.href= 'edit.asp ' ">
</form>