Question

button调用servlet ，页面发生了跳转，但是没有执行servlet中doPost（）方法？求教！

<Script type="text/javascript">
function Submit(){
window.location="/UserLogin/servlet/success";
}
</Script>
<input type="button" name="seek" value="查询功能" id="seek" onclick="Submit()"/>

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
String seek=(String)request.getParameter("seek");
String alter=(String)request.getParameter("alter");
if(seek!=null||alter!=null){
if(seek=="seek" && alter==null){
request.getRequestDispatcher("/seek.jsp").forward(request,response);
}
if(alter=="alter" && seek==null){
request.getRequestDispatcher("/alter.jsp").forward(request,response);
}
}else{
request.getRequestDispatcher("/success.jsp").forward(request,response);
}

}

Answer 1

应该是执行了doget

window.location="/UserLogin/servlet/success";//这句话会执行doget，并且传递不了参数。严格来说应该是
window.location.href="";
要加个form表单，
<Script type="text/javascript">
function Submit(){
document.form1.submit();
}
</Script>
<form action="/UserLogin/servlet/success" method="post" name="form1">
<input type="button" name="seek" value="查询功能" id="seek" onclick="Submit()"/>

</form>

Answer 2

首先看你url的地址是否争取。
其次看看 你doget()里面是不是写了 dopost()