Question

xml文件位于我项目的WebContent/WEB-INF/web.xml中。我正在使用Eclipse并运行Tomcat（它不是通过Eclipse安装的。我更喜欢它是一个单独的安装。）

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
  <display-name>EmployeeManagement</display-name>
  <welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
  </welcome-file-list>
  <context-param>
    <param-name>name</param-name>
    <param-value>Pramod</param-value>
  </context-param>
  <servlet-mapping>
        <servlet-name>Registration</servlet-name>
        <url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
   </servlet-mapping>
</web-app>

当表单页面提交给servlet时，它不起作用。我每次都会收到404错误。我一直遇到这个问题。有人请帮助我。

Answer 1

您缺少对映射很重要的<servlet>...</servlet>标记。所以使用以下内容：

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
<display-name>EmployeeManagement</display-name>
<welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<context-param>
    <param-name>name</param-name>
    <param-value>Pramod</param-value>
</context-param>
<servlet>
    <servlet-name>Registration</servlet-name>
    <servlet-class>com.yourPackageName.yourServletName</servlet-class>
</servlet>
<servlet-mapping>
    <servlet-name>Registration</servlet-name>
    <url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
</web-app>

并且您应该在表单上给出action值，如下所示：

<form action="/EmployeeManagement/WebContent/Registration" method="post">

      //Some code here

</form>

并注意到以下所有值都区分大小写：

<servlet>
    <servlet-name>Registration</servlet-name>
    <servlet-class>com.yourPackageName.yourServletName</servlet-class>
</servlet>
<servlet-mapping>
    <servlet-name>Registration</servlet-name>
    <url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>

您的servlet名称Registration在<servlet>...</servlet>和<servlet-mapping>...</servlet-mapping>标记上应该相同，并且package名称应与servlet类所在的名称相同。

Answer 2

你没有将servlet名称映射到servlet类，它应该如下所示，

在<servlet-class>中给出servlet的路径

    <servlet>
         <servlet-name>Registration</servlet-name>
         <servlet-class>com.Registration<servlet-class>
    </servlet>
    <servlet-mapping>
         <servlet-name>Registration</servlet-name>
         <url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
    </servlet-mapping>

Answer 3

检查您的表单操作。那条路是

/EmployeeManagement/WebContent/Registration

或

YOURAPPCONTEXT/EmployeeManagement/WebContent/Registration

或

YOURAPPNAME/EmployeeManagement/WebContent/Registration

Answer 4

您忘记了配置的重要部分。您应该在web.xml代码前添加servlet-mapping：

<servlet>
    <servlet-name>Registration</servlet-name>
    <servlet-class>com.name.of.your.servlet.class</servlet-class>
</servlet>

Answer 5

您已指定servlet-mapping并在Registration中使用了名称servlet-name，但未在之前定义。

在servlet映射中使用它之前，需要定义servlet

<servlet>
    <servlet-name>Registration</servlet-name>
    <servlet-class>[fully qualifyied name of your servlet]</servlet-class>
</servlet>

Answer 6

您缺少在web.xml中定义servlet的另一部分

<servlet>
   <servlet-name>Registration</servlet-name>
   <servlet-class>
      package.path.to.RegistrationServlet
  </servlet-class>
</servlet>

在web.xml中映射servlet

6 个答案: