java - 获取 HttpSession|来自简单 java 类而不是 servlet 类的请求

Question

我希望 session 对象不在 servlet 类中，而是来 self 们应用程序的普通对象。

WEB.XML

<listener>
        <listener-class>com.abc.web.ApplicationManager</listener-class>
</listener>
<listener>
        <listener-class>com.abc.web.SessionManager</listener-class>
</listener>

ViewPrices.java

public class ViewPrices implements Cloneable, Serializable {

 Session session = request.getSession();
                   servletContext.getSession()
                   anyWay.getSession();
}

Answer 1

调用这个:

RequestFilter.getSession();
RequestFilter.getRequest();

在您的自定义过滤器上:

public class RequestFilter implements Filter {

    private static ThreadLocal<HttpServletRequest> localRequest = new ThreadLocal<HttpServletRequest>();


    public static HttpServletRequest getRequest() {
        return localRequest.get();
    }

    public static HttpSession getSession() {
        HttpServletRequest request = localRequest.get();
        return (request != null) ? request.getSession() : null;
    }


    @Override
    public void doFilter(ServletRequest servletRequest, ServletResponse servletResponse, FilterChain filterChain) throws IOException, ServletException {
        if (servletRequest instanceof HttpServletRequest) {
            localRequest.set((HttpServletRequest) servletRequest);
        }

        try {
            filterChain.doFilter(servletRequest, servletResponse);
        } finally {
            localRequest.remove();
        }
    }

    @Override
    public void init(FilterConfig filterConfig) throws ServletException {
    }

    @Override
    public void destroy() {
    }
}

您将把它注册到您的 web.xml 文件中:

<filter>
    <filter-name>RequestFilter</filter-name>
    <filter-class>your.package.RequestFilter</filter-class>
</filter>

<filter-mapping>
    <filter-name>RequestFilter</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>