Question

我有主页，当我点击参考Servlet不工作时我得到错误404.我认为问题在web.xml映射，但不知道在哪里。请帮我纠正这个问题。谢谢。

我的web.xml

<!--Homepage.-->
<servlet>
    <servlet-name>HomePageServlet</servlet-name>
    <servlet-class>ru.pravvich.servlets.HomePageServlet</servlet-class>
</servlet>

<servlet-mapping>
    <servlet-name>HomePageServlet</servlet-name>
    <url-pattern>/</url-pattern>
</servlet-mapping>

<!--Add user in database.-->
<servlet>
    <servlet-name>AddUserServlet</servlet-name>
    <servlet-class>ru.pravvich.servlets.AddUserServlet</servlet-class>
</servlet>

<servlet-mapping>
    <servlet-name>AddUserServlet</servlet-name>
    <url-pattern>/addition</url-pattern>
</servlet-mapping>

我的jsp主页：

<body>
    <ul>
        <li><a href="addition.jsp">addition</a></li>
    </ul>
</body>

使用doGet方法的servlet：

public class HomePageServlet extends HttpServlet {
    @Override
    protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
        req.setCharacterEncoding("UTF8");
        req.getRequestDispatcher("/WEB-INF/views/index.jsp").forward(req,resp);
    }
}

通过http://localhost:8080/items/，我得到了我的主页。

但是，当我点击index.jsp的引荐时，请返回：HTTP Status [404] – [Not Found]

addition.jsp

中我的/WEB-INF/views/addition.jsp同样谎言

我的Servlet用于addition.jsp：

public class AddUserServlet extends HttpServlet {

    private DBJoint db;

    @Override
    protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {

        db = (DBJoint) getServletContext().getAttribute("db");

        db.getDBExecutor().addUser(
            new User(req.getParameter("name"), 
                     req.getParameter("login"), 
                     req.getParameter("email")));

        req.setAttribute("serverAnswer", "Add ok!");
        req.getRequestDispatcher("/WEB-INF/views/answer.jsp").forward(req, resp);

    }
}

addition.jsp：

<body>
    <form method="post" action="addition">
        <input type="text" required placeholder="name" name="name"><br>
        <input type="text" required placeholder="login" name="login"><br>
        <input type="text" required placeholder="email" name="email"><br>
        <input type="submit" value="add">
    </form>
</body>

Answer 1

我建议使用try/catch和调试器模式。

尝试使用这样的getRequestDispatcher

request.getRequestDispatcher("answer.jsp").forward(request, response);

或

req.getRequestDispatcher("~/WEB-INF/views/answer.jsp").forward(req, resp);

我认为您需要获取每个参数的参数，而不是设置参数。试试这个;

public class AddUserServlet extends HttpServlet {

private DBJoint db;

@Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {

    db = (DBJoint) getServletContext().getAttribute("db");

    String Name = request.getParameter("name");
    String Login= request.getParameter("login");
    String Email= request.getParameter("email");

    db.getDBExecutor().addUser(
        new User(Name, Login, Email);

    //And you need to 'serverAnswer' item in your 'answer.jsp' you know.
    request.setAttribute("serverAnswer", "Add ok!");
    request.getRequestDispatcher("answer.jsp").forward(req, resp);
    }    
}

然后您可以在getAttribute，

中使用answer.jsp这样的内容

<%String Answer= (String)request.getAttribute("serverAnswer"); %><%= Answer%>

不要怪我，只是我想帮助你，我希望它对你有所帮助，如果你想要你可以看看我的试验项目; https://github.com/anymaa/GNOHesap

有一个很好的编码：）

不要工作servlet。 web.xml中的servlet映射问题

1 个答案: