Question

我想创建一个登录页面，在注销后我希望用户显示登录页面而不是上一页

如何防止用户在注销后返回上一页。我已经清除了缓存....但是通过按下后退按钮用户将转到上一页。我希望在退出后用户按下后按钮登录页面刷新并显示

    <s:form action="Login" >
    <s:textfield label="username" name="userName"/>
    <s:password label="password" name="password"/>
    <s:submit name="login" value="login"></s:submit>
    </s:form>

如何管理会议也可以。任何人都可以帮助我登录.java

  package action;

 import com.opensymphony.xwork2.ActionSupport;


public class Login extends ActionSupport {

private String userName;
private String password;

public Login() {
}

@Override
  public String execute() {



  Map  session = ActionContext.getContext().getSession();
  session.put("logged-in","yes");
  return SUCCESS;


}
    @Override
       public void validate()
    {
    if(getUserName().length()==0)
    {
         addFieldError("userName", "User Name is required");
    }
   else if (!getUserName().equals("prerna"))
   {
       addFieldError("userName", "Invalid User");
   }

     if(getPassword().length()==0)
    {
         addFieldError("password", "password is required");
    }

     else   if (!getPassword().equals("prerna")) {
        addFieldError("password", getText("password.required"));
    }



   }


      public String getUserName() {
       return userName;
      }

/**
 * @param userName the userName to set
 */
public void setUserName(String userName) {
    this.userName = userName;
}

/**
 * @return the password
 */
public String getPassword() {
    return password;
}

/**
 * @param password the password to set
 */
public void setPassword(String password) {
    this.password = password;
}
 }
Logout.java

   public class Logout {

     public Logout() {
       }

       public String execute() throws Exception {

     Map session = ActionContext.getContext().getSession();
     session.remove("logged-in");

    return "success";
}

}

logout.jsp

   <s:property value="userName"/>
     <s:property value="password"/>
    <s:url action="Logout.action" var="urlTag">

      </s:url>
      <s:a href="%{urlTag}">URL Tag Action (via %)</s:a>

拦截 logintest

  package interceptor;

    import action.Login;
    import com.opensymphony.xwork2.ActionContext;
    import com.opensymphony.xwork2.ActionInvocation;
    import com.opensymphony.xwork2.interceptor.Interceptor;
    import java.util.Map;



 public class logintest implements Interceptor {

   public logintest() {
    }

public void destroy() {
    throw new UnsupportedOperationException("Not supported yet.");
}

public void init() {
    throw new UnsupportedOperationException("Not supported yet.");
}

public String intercept(ActionInvocation actionInvocation) throws Exception {
   Map<String, Object> session = ActionContext.getContext().getSession();

    // sb: feel free to change this to some other type of an object which
    // represents that the user is logged in. for this example, I am using
    // an integer which would probably represent a primary key that I would
    // look the user up by with Hibernate or some other mechanism.
    String userId = (String) session.get("logged-in");

    // sb: if the user is already signed-in, then let the request through.
    if (userId != null) {
        return actionInvocation.invoke();
    }

    Object action = actionInvocation.getAction();

    // sb: if the action doesn't require sign-in, then let it through.


    // sb: if this request does require login and the current action is
    // not the login action, then redirect the user
    if (!(action instanceof Login)) {
        return "loginRedirect";
    }

    // sb: they either requested the login page or are submitting their
    // login now, let it through
    return actionInvocation.invoke();

   }

}

struts.xml中

              <!DOCTYPE struts PUBLIC
     "-//Apache Software Foundation//DTD Struts Configuration 2.1//EN"
         "http://struts.apache.org/dtds/struts-2.1.dtd">

 <struts>
      <!-- Configuration for the default package. -->
<package name="default" extends="struts-default">

    <interceptors>

        <interceptor name="logintest"
class="interceptor.logintest"></interceptor>

        <interceptor-stack name="newStack">
            <interceptor-ref name="logintest"/>
            <interceptor-ref name="defaultStack" />
        </interceptor-stack>
    </interceptors>
    <global-results  >
        <result name="loginRedirect" type="redirect" >/login.jsp</result>
    </global-results>
    <action class="action.Login" name="Login">
        <interceptor-ref name="newStack"></interceptor-ref>
        <result name="input">/login.jsp</result>

        <result name="success">/loginsuccess.jsp</result>

    </action>

    <action class="action.Logout" name="Logout">

        <interceptor-ref name="newStack"></interceptor-ref>

        <result name="success">/login.jsp</result>
    </action>
</package>

Answer 1

作为Liviu。提到，该行为由客户端的浏览器控制。您可以做的最多是为每个登录页面的请求发送无缓存和可能的无存储标头，以便浏览器不存储这些标题，当用户按下时，浏览器必须重新请求页面，结果登录页面。

您想要设置的特定标题是：

response.setHeader("Cache-Control", "no-cache, no-store");
response.setDateHeader("Expires", 0);
response.setHeader("Vary", "*");

Answer 2

试试这个

<html>
<head>
<title>Back Button Demo: Page One</title>
<script>
function backButtonOverride()
{
  // Work around a Safari bug
  // that sometimes produces a blank page
  setTimeout("backButtonOverrideBody()", 1);

}

function backButtonOverrideBody()
{
  // Works if we backed up to get here
  try {
    history.forward();
  } catch (e) {
    // OK to ignore
  }
  // Every quarter-second, try again. The only
  // guaranteed method for Opera, Firefox,
  // and Safari, which don't always call
  // onLoad but *do* resume any timers when
  // returning to a page
  setTimeout("backButtonOverrideBody()", 500);
}
</script>
</head>
<body onLoad="backButtonOverride()">
<h1>Back Button Demo: Page One</h1>
<a href="page2.html">Advance to Page Two</a>
</body>
</html>

Source Link

注销后避免访问历史记录

2 个答案: