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我正在尝试将 BASIC 身份验证添加到我的 RESTful Web 服务。目前我有一个 Apache Tomcat 6.0 服务器的 BASIC 身份验证,但我需要在 WebSphere 应用程序服务器版本上部署我的 Web 服务。6.1 以及我在 WebSphere 上运行 BASIC 身份验证时遇到问题。

Java 中有没有办法检查 HTTP 请求的身份验证标头,如果提供的用户名/密码(以 Base64 编码)与已知帐户不匹配,是否会强制用户输入新的用户名/密码?

我曾尝试实现 Spring Security,但由于我的项目完全没有 Spring,因此试图让它工作是一个巨大的痛苦,我试图找到一个简单的解决方案来解决我相当简单的问题。

我目前使用的技术包括:Java、Jersey/JAX-RS、带有 Maven 插件的 Eclipse。

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2 回答 2

9

您应该能够设置一个servlet 过滤器,该过滤器在您的 REST 处理程序之前执行、检查“授权”请求标头、base 64 对其进行解码、提取用户名和密码并进行验证。像这样的东西:

public void doFilter(ServletRequest req,
                     ServletResponse res,
                     FilterChain chain) {
  if (request instanceof HttpServletRequest) {
    HttpServletRequest request = (HttpServletRequest) req;
    String authHeader = Base64.decode(request.getHeader("Authorization"));
    String creds[] = authHeader.split(":");
    String username = creds[0], password = creds[1];
    // Verify the credentials here...
    if (authorized) {
      chain.doFilter(req, res, chain);
    } else {
      // Respond 401 Authorization Required.
    }
  }
  doFilter(req, res, chain);
}

所有 servlet 容器都有一个标准的方式来配置过滤器链。

于 2011-06-15T16:32:49.840 回答
5

基于maerics答案的完整实现。

import java.io.IOException;

import javax.servlet.Filter;
import javax.servlet.FilterChain;
import javax.servlet.FilterConfig;
import javax.servlet.ServletException;
import javax.servlet.ServletRequest;
import javax.servlet.ServletResponse;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import org.apache.commons.lang.StringUtils;

import sun.misc.BASE64Decoder;

public class AuthenticationFilter implements Filter {

    private static final String AUTHORIZATION_HEADER_NAME = "Authorization";
    private static final String WWW_AUTHENTICATE_HEADER_NAME = "WWW-Authenticate";
    private static final String WWW_AUTHENTICATE_HEADER_VALUE = "Basic realm=\"Default realm\"";
    private static final String BASIC_AUTHENTICATION_REGEX = "Basic\\s";
    private static final String EMPTY_STRING = "";
    private static final String USERNAME_PASSWORD_SEPARATOR = ":";
    private static final BASE64Decoder DECODER = new BASE64Decoder();

    public void init(FilterConfig arg0) throws ServletException {
    }

    public void doFilter(ServletRequest req, ServletResponse res, FilterChain chain) throws IOException, ServletException {

        HttpServletRequest httpReq = (HttpServletRequest) req;
        HttpServletResponse httpRes = (HttpServletResponse) res;

        String authHeader = httpReq.getHeader(AUTHORIZATION_HEADER_NAME);

        if (authHeader == null) {
            this.requestAuthentication(httpRes);
            return;
        }

        authHeader = authHeader.replaceFirst(BASIC_AUTHENTICATION_REGEX, EMPTY_STRING);
        authHeader = new String(DECODER.decodeBuffer(authHeader));      

        if (StringUtils.countMatches(authHeader, USERNAME_PASSWORD_SEPARATOR) != 1) {
            this.requestAuthentication(httpRes);
            return;         
        }

        String[] creds = authHeader.split(USERNAME_PASSWORD_SEPARATOR);
        String username = creds[0];
        String password = creds[1];         

        //TODO: implement this method
        if (!authenticatedUser(username, password)) {
            this.requestAuthentication(httpRes);
            return;
        }

         chain.doFilter(req, res);
    }

    private void requestAuthentication(HttpServletResponse httpRes) {

        httpRes.setHeader(WWW_AUTHENTICATE_HEADER_NAME, WWW_AUTHENTICATE_HEADER_VALUE);
        httpRes.setStatus(HttpServletResponse.SC_UNAUTHORIZED);
    }

    public void destroy() {
    }
}
于 2013-06-15T01:40:08.860 回答