![](/img/trans.png)
[英]HTTP Status 405 - HTTP method POST is not supported by this URL java servlet
[英]HTTP status 405 - HTTP method POST is not supported by this URL
当我运行我的项目时,出现“此 URL 不支持 HTTP 方法 POST”错误。 有趣的是,两天前它运行得非常好。 在我对我的代码进行了一些更改但随后恢复了我的原始代码并且它给了我这个错误之后。 请你帮助我好吗?
这是我的 index.html:
<form method="post" action="login.do">
<div>
<table>
<tr><td>Username: </td><td><input type="text" name="e_name"/>
</td> </tr>
<tr><td> Password: </td><td><input type="password" name="e_pass"/>
</td> </tr>
<tr><td></td><td><input type="submit" name ="e_submit" value="Submit"/>
这是我的登录 servlet:
public class Login extends HttpServlet {
/**
* Processes requests for both HTTP
* <code>GET</code> and
* <code>POST</code> methods.
*
* @param request servlet request
* @param response servlet response
* @throws ServletException if a servlet-specific error occurs
* @throws IOException if an I/O error occurs
*/
protected void processRequest(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException, SQLException {
response.setContentType("text/html;charset=UTF-8");
PrintWriter out = response.getWriter();
try {
/*
* TODO output your page here. You may use following sample code.
*/
int status;
String submit = request.getParameter("e_submit");
String submit2 = request.getParameter("a_submit");
out.println("Here1");
String e_name = request.getParameter("e_name");
String e_password = request.getParameter("e_pass");
String a_name = request.getParameter("a_name");
String a_password = request.getParameter("a_pass");
out.println(e_name+e_password+a_name+a_password);
Author author = new Author(a_name,a_password);
Editor editor = new Editor(e_name,e_password);
// If it is an AUTHOR login:
if(submit==null){
status = author.login(author);
out.println("Author Login");
//Incorrect login details
if(status==0) {
out.println("Incorrect");
RequestDispatcher view = request.getRequestDispatcher("index_F.html");
view.forward(request, response);
}
//Correct login details --- AUTHOR
else {
out.println("Correct login details");
HttpSession session = request.getSession();
session.setAttribute(a_name, "a_name");
RequestDispatcher view = request.getRequestDispatcher("index_S.jsp");
view.forward(request, response);
}
}
//If it is an EDITOR login
else if (submit2==null){
status = editor.login(editor);
//Incorrect login details
if(status==0) {
RequestDispatcher view = request.getRequestDispatcher("index_F.html");
view.forward(request, response);
}
//Correct login details --- EDITOR
else {
out.println("correct");
HttpSession session = request.getSession();
session.setAttribute(e_name, "e_name");
session.setAttribute(e_password, "e_pass");
RequestDispatcher view = request.getRequestDispatcher("index_S_1.html");
view.forward(request, response);
} }
out.println("</body>");
out.println("</html>");
} finally {
out.close();
}
}
@Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
super.doPost(req, resp);
}
@Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
super.doGet(req, resp);
}}
我的 web.xml 看起来像这样:
<?xml version="1.0" encoding="UTF-8"?>
<servlet>
<servlet-name>action</servlet-name>
<servlet-class>org.apache.struts.action.ActionServlet</servlet-class>
<init-param>
<param-name>config</param-name>
<param-value>/WEB-INF/struts-config.xml</param-value>
</init-param>
<init-param>
<param-name>debug</param-name>
<param-value>2</param-value>
</init-param>
<init-param>
<param-name>detail</param-name>
<param-value>2</param-value>
</init-param>
<load-on-startup>2</load-on-startup>
</servlet>
<servlet>
<servlet-name>Login</servlet-name>
<servlet-class>controller.Login</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>action</servlet-name>
<url-pattern>*.do</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>Login</servlet-name>
<url-pattern>/login.do</url-pattern>
</servlet-mapping>
<session-config>
<session-timeout>
30
</session-timeout>
</session-config>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
我使用 Glassfish v3 服务器 - 让我知道您需要知道的任何其他信息
这是因为在您的doGet()
和doPost()
方法上,您将其称为super
方法。 而是在上述各个方法中调用processRequest()
。
默认情况下, super.doGet()
和super.doPost()
方法返回HTTP 405,因此您不必调用超类doGet()
和doPost()
。
为什么您的代码中有一个processRequest
方法? 谁将调用该方法?
您无法通过调用super.doGet()
或super.doPost()
来使用该方法,而需要显式调用该方法。
@Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
processRequest(req,resp)
}
@Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
processRequest(req,resp)
}
编辑
做这个
response.sendRedirect("index_F.html");
return;
代替
RequestDispatcher view = request.getRequestDispatcher("index_F.html");
view.forward(request, response);
在doPost()
内部,您必须调用processRequest()
。
您在调用doGet()
和doPost()
方法时并未实际实现(使用super
)。
HttpServlet基本上遵循模板方法模式,其中所有未覆盖的HTTP方法都返回HTTP 405
错误“ Method not support”。 覆盖此类方法时,不应调用super
方法,因为否则您仍然会收到HTTP 405
错误。 您的doPost()方法也是如此。
在上述各个方法中调用processRequest(req,resp)
。
编辑:
第二,
不要使用调度程序将请求转发到HTML。 如果只想显示html,请使用重定向。
response.sendRedirect("index_F.html");
return;
另外,优良作法是在注销或发送无效凭据时使用redirect
。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.