javaweb项目已经发布起来了，可是输入controller对应的地址，总是自己跳转到login？

Question

web.xml中没看出异常啊？

比如输入http://localhost:8080/dyn2/dy...
会自动调整到http://localhost:8080/dyn2/login

<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee"
 xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
 <description>Dynamic Matrix powered by Spring</description>
 <display-name>Dynamic Matrix</display-name>
 <context-param>
  <param-name>webAppRootKey</param-name>
  <param-value>dyn2.root</param-value>
 </context-param>
 <context-param>
  <param-name>log4jConfigLocation</param-name>
  <param-value>/WEB-INF/log4j.properties</param-value>
 </context-param>
 <context-param>
  <param-name>contextConfigLocation</param-name>
  <param-value>/WEB-INF/applicationContext.xml</param-value>
 </context-param>
 <filter>
  <filter-name>characterEncodingFilter</filter-name>
  <filter-class>org.springframework.web.filter.CharacterEncodingFilter</filter-class>
  <init-param>
   <param-name>encoding</param-name>
   <param-value>utf8</param-value>
  </init-param>
 </filter>
 <!--<filter>
  <filter-name>ContentTypeFilter</filter-name>
  <filter-class>net.counters.dynm.web.ContentTypeFilter</filter-class>
  <init-param>
   <param-name>charset</param-name>
   <param-value>utf8</param-value>
  </init-param>
 </filter>
 -->
 <filter>
  <filter-name>springSecurityFilterChain</filter-name>
  <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
 </filter>
 <!-- <filter>
  <filter-name>UrlRewriteFilter</filter-name>
  <filter-class>org.tuckey.web.filters.urlrewrite.UrlRewriteFilter</filter-class>
 </filter> -->
 <filter-mapping>
  <filter-name>characterEncodingFilter</filter-name>
  <url-pattern>/*</url-pattern>
 </filter-mapping>
 <!--<filter-mapping>
  <filter-name>ContentTypeFilter</filter-name>
  <url-pattern>/*</url-pattern>
 </filter-mapping>
 -->
 <filter-mapping>
  <filter-name>springSecurityFilterChain</filter-name>
  <url-pattern>/*</url-pattern>
 </filter-mapping>
 <!-- <filter-mapping>
  <filter-name>UrlRewriteFilter</filter-name>
  <url-pattern>/*</url-pattern>
 </filter-mapping> -->
 <listener>
  <listener-class>org.springframework.web.util.Log4jConfigListener</listener-class>
 </listener>
 <listener>
  <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
 </listener>
 <listener>
     <listener-class>org.springframework.web.util.HttpSessionMutexListener</listener-class>
 </listener>
 <servlet>
  <servlet-name>dynMatrix</servlet-name>
  <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
  <load-on-startup>1</load-on-startup>
 </servlet>
 <!-- <servlet-mapping>
  <servlet-name>dynMatrix</servlet-name>
  <url-pattern>/app/*</url-pattern>
 </servlet-mapping> -->
 <servlet-mapping>
  <servlet-name>dynMatrix</servlet-name>
  <url-pattern>/</url-pattern>
 </servlet-mapping>
 <session-config>
     <session-timeout>60</session-timeout>
 </session-config>
 <error-page>
  <error-code>403</error-code>
  <location>/errors/403.html</location>
 </error-page>
</web-app>

这是controller

    @RequestMapping(value="/create",method=RequestMethod.GET)
    public String getCreateForm(Model model){
        this.addEnableMap(model);
        model.addAttribute(new WebUser());
        //加载分类
        String cwhere = " where type_parent_id=0";
        List<CheckListType> checkListType = checkLMDao.getCheckListTypeByWhere(cwhere, 0, 0);
        
        model.addAttribute("checkListType", checkListType);
        return "createUserForm";
    }

Answer 1

Requesting this address may require some login information for verification. The request in ajax will pass this information.
If you directly enter the browser without this information, it will be intercepted by the security filter and forwarded to the login page. .

Answer 2

Maybe this filter intercepted your request
Are you responsible for the project alone? If it’s not a question like this, it’s best to ask colleagues in the same project team