我怎么才能从ServletRequest request 中的到URL,参数我能得到。

2024-10-31 15:19:26
有5个网友回答
网友(1):

给一个例子你,希望对你能有帮助
package dao;

import javax.servlet.Filter;

import javax.servlet.FilterConfig;

import javax.servlet.ServletException;

import javax.servlet.ServletRequest;

import javax.servlet.ServletResponse;

import javax.servlet.FilterChain;

import java.io.IOException;

import javax.servlet.http.HttpServletRequest;

import javax.servlet.http.HttpSession;

import javax.servlet.http.HttpServletResponse;

import po.login;

public class permissionFilter implements Filter {

private FilterConfig filterConfig;

private FilterChain chain;

private HttpServletRequest request;

private HttpServletResponse response;

public void destroy() {

this.filterConfig = null;

}

public void init(FilterConfig filterConfig) throws ServletException {

this.filterConfig = filterConfig;

}

public void doFilter(ServletRequest servletRequest,

ServletResponse servletResponse, FilterChain chain) {

this.chain = chain;

this.request = (HttpServletRequest) servletRequest;

this.response = ((HttpServletResponse) servletResponse);

// 获取当前页面文件名此处url为:/Gzlkh/login.jsp

String url = request.getRequestURI();

// 此处截取的url为:login.jsp

url = url.substring(url.lastIndexOf("/") + 1, url.length());

try {

HttpSession session = request.getSession();

// 获取网站访问根目录

String accessPath = request.getContextPath();

// 获取用户登录验证信息

login st = (login) session.getAttribute("st");

if (noFileUrl(url, request)) {

// 不需要判断权限的请求如登录页面,则跳过

chain.doFilter(request, response);// 继续执行请求

} else if (st == null) {

response.sendRedirect(accessPath + "/login.jsp");

// 未登录或超时,返回登陆页面

} else {

verifyUrl(url, st);// 判断当前user是否拥有访问此url的权限

}

} catch (Exception sx) {

sx.printStackTrace();

}

}

网友(2):

HttpServletRequest httpServletRequest = (HttpServletRequest)request;
HttpServletResponse httpServletResponse = (HttpServletResponse)response;
转化一下,然后用httpServletRequest调用getparameter("参数"),getRequestURL() 方法

网友(3):

request.getRequestURL()

网友(4):

request.getRequestURL()、request.getRequestURI()
楼主派分吧 :)

网友(5):

request.getparameter("参数名");