twitter - 在 Vaadin 14.2 应用程序中使用 pac4j 通过 Twitter 进行身份验证：“code”：32，“message”：“无法对您进行身份验证。”

问题描述

我正在尝试将 Twitter 身份验证（最好是 SSO）集成到 Vaadin 应用程序中。为此，我从头开始创建了一个 Vaadin 应用程序并尝试集成 pac4j（参见以下步骤）。不幸的是，我收到错误“代码”：32，“消息”：“无法验证您。” 尽管有效的电子邮件/密码组合。任何想法如何让它发挥作用？

从https://vaadin.com/start/v14下载 Vaadin 14.2 项目
运行 Application.java 并访问 localhost:8080 工作正常。

使用 pac4j 为 Spring Boot 扩展 pom.xml：

 <!-- https://mvnrepository.com/artifact/org.pac4j/spring-security-pac4j -->
 <dependency>
     <groupId>org.pac4j</groupId>
     <artifactId>spring-security-pac4j</artifactId>
     <version>5.1.0</version>
 </dependency>

使用 pac4j 为 Twitter 客户端扩展 pom.xml：

 <!-- https://mvnrepository.com/artifact/org.pac4j/pac4j-oauth -->
 <dependency>
     <groupId>org.pac4j</groupId>
     <artifactId>pac4j-oauth</artifactId>
     <version>4.0.3</version>
 </dependency>

使用 Spring Security 扩展 pom.xml

 <!-- https://mvnrepository.com/artifact/org.springframework.security/spring-security-config -->
 <dependency>
     <groupId>org.springframework.security</groupId>
     <artifactId>spring-security-config</artifactId>
 </dependency>

创建类 Pac4jConfig 就像本演示文稿中的 Spring Security 示例一样http://www.pac4j.org/gettingstarted.html

 import org.pac4j.core.config.Config;
 import org.pac4j.oauth.client.TwitterClient;
 import org.springframework.context.annotation.Bean;
 import org.springframework.context.annotation.Configuration;
 @Configuration
 public class Pac4jConfig {
     @Bean
     public Config config() {
         TwitterClient twitterClient = new TwitterClient();
         Config config = new Config("http://localhost:8080", twitterClient);
         return config;
     }
 }

像本演示文稿中的 Spring Security 示例一样创建类 SecurityConfig http://www.pac4j.org/gettingstarted.html

 import org.pac4j.core.config.Config;
 import org.pac4j.springframework.security.web.CallbackFilter;
 import org.pac4j.springframework.security.web.SecurityFilter;
 import org.springframework.beans.factory.annotation.Autowired;
 import org.springframework.context.annotation.Configuration;
 import org.springframework.core.annotation.Order;
 import org.springframework.security.config.annotation.web.builders.HttpSecurity;
 import org.springframework.security.config.annotation.web.configuration.EnableWebSecurity;
 import org.springframework.security.config.annotation.web.configuration.WebSecurityConfigurerAdapter;
 import org.springframework.security.web.authentication.www.BasicAuthenticationFilter;

 @EnableWebSecurity
 public class SecurityConfig {

     @Configuration
     @Order(1)
     public static class TwitterWebSecurityConfigurationAdapter
             extends WebSecurityConfigurerAdapter {
         @Autowired
         private Config config;

         protected void configure(final HttpSecurity http) throws Exception {
             final SecurityFilter filter = new SecurityFilter(config, "TwitterClient");
             http.antMatcher("/twitter/**").addFilterBefore(filter, BasicAuthenticationFilter.class);
         }
     }

     @Configuration
     public static class DefaultWebSecurityConfigurationAdapter
             extends WebSecurityConfigurerAdapter {
         @Autowired
         private Config config;

         protected void configure(final HttpSecurity http) throws Exception {
             final CallbackFilter callbackFilter = new CallbackFilter(config);
             http.authorizeRequests().anyRequest().permitAll().and().addFilterBefore(callbackFilter,
                     BasicAuthenticationFilter.class);
         }
     }
 }

在本演示文稿http://www.pac4j.org/gettingstarted.html的 Spring Security 示例中创建类 TwitterTestApplication，如类“Application”

 import java.util.Map;
 import javax.servlet.http.HttpServletRequest;
 import javax.servlet.http.HttpServletResponse;
 import org.springframework.stereotype.Controller;
 import org.springframework.web.bind.annotation.RequestMapping;
 @Controller
 public class TwitterTestApplication {
     @RequestMapping("/twitter/index.html")
     public String twitter(HttpServletRequest request, HttpServletResponse response,
             Map<String, Object> map) {
         return "Hello world";
     }
 }

调用 http://localhost:8080/twitter/index.html 会导致这个错误：

 There was an unexpected error (type=Internal Server Error, status=500).
 key cannot be blank
 org.pac4j.core.exception.TechnicalException: key cannot be blank
     at org.pac4j.core.util.CommonHelper.assertTrue(CommonHelper.java:107)
     [...]

通过我的凭据扩展 TwitterClient 的构造函数调用（当我在 twitter.com 上复制/粘贴它们时工作正常）并允许电子邮件：
```
TwitterClient twitterClient = new TwitterClient("[my-email]", "[my-password]", true);
```

调用 http://localhost:8080/twitter/index.html 会导致这个错误：

There was an unexpected error (type=Internal Server Error, status=500).
com.github.scribejava.core.exceptions.OAuthException: Response body is incorrect. Can't extract token and secret from this: '{"errors":[{"code":32,"message":"Could not authenticate you."}]}'
org.pac4j.core.exception.TechnicalException: com.github.scribejava.core.exceptions.OAuthException: Response body is incorrect. Can't extract token and secret from this: '{"errors":[{"code":32,"message":"Could not authenticate you."}]}'
    at org.pac4j.oauth.redirect.OAuth10RedirectionActionBuilder.getRedirectionAction(OAuth10RedirectionActionBuilder.java:62)
    at org.pac4j.core.client.IndirectClient.getRedirectionAction(IndirectClient.java:109)
    at org.pac4j.core.engine.DefaultSecurityLogic.redirectToIdentityProvider(DefaultSecurityLogic.java:224)
    at org.pac4j.core.engine.DefaultSecurityLogic.perform(DefaultSecurityLogic.java:157)
    at org.pac4j.springframework.security.web.SecurityFilter.doFilter(SecurityFilter.java:73)
    at org.springframework.security.web.FilterChainProxy$VirtualFilterChain.doFilter(FilterChainProxy.java:334)
    at org.springframework.security.web.authentication.logout.LogoutFilter.doFilter(LogoutFilter.java:116)

标签： twitterspring-securityvaadinpac4j

twitter - 在 Vaadin 14.2 应用程序中使用 pac4j 通过 Twitter 进行身份验证：“code”：32，“message”：“无法对您进行身份验证。”

问题描述

解决方案

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