小编Nuñ*_*ada的帖子

我的 Spring Boot appl 中有这个配置类。v1.5.3. 发布

@Configuration
@Profile("dev && cub")
@PropertySource("file:///${user.home}/.cub/application-dev.properties")
public class CubDevelopmentConfig {
..
}

这个属性在我的 application.properties 中定义

spring.profiles.active=dev, cub

但未加载配置类

我也试过 @Profile("{dev && cub}")

我有一个基本的 SpringBoot 应用程序。使用 Spring Initializer、JPA、嵌入式 Tomcat、Thymeleaf 模板引擎，并打包为可执行 JAR 文件。我在扩展自的存储库中定义了此方法

CrudRepository<HotelPrice, Long>, PagingAndSortingRepository<HotelPrice, Long> {

这是方法

List<HotelPrice> getByHotelAndUpdateDateGreaterThan (Hotel hotel, Date date, PageRequest pageRequest);

和服务：

public List<HotelPrice> getMonthlyMinPriceDate(Hotel hotel) {
        return hotelPriceRepository.getByHotelAndUpdateDateGreaterThan
                (hotel, DateUtils.monthlyDate(), new PageRequest(1, 1,new Sort(Sort.Direction.DESC, "price")));
    }

但是当我运行 Junit 测试时，我收到了这个错误：

 java.util.NoSuchElementException
at java.util.ArrayList$Itr.next(ArrayList.java:860)
at java.util.Collections$UnmodifiableCollection$1.next(Collections.java:1042)
at org.springframework.data.jpa.repository.query.CriteriaQueryParameterBinder.bind(CriteriaQueryParameterBinder.java:65)
at org.springframework.data.jpa.repository.query.ParameterBinder.bind(ParameterBinder.java:101)
at org.springframework.data.jpa.repository.query.ParameterBinder.bindAndPrepare(ParameterBinder.java:161)
at org.springframework.data.jpa.repository.query.ParameterBinder.bindAndPrepare(ParameterBinder.java:152)
at org.springframework.data.jpa.repository.query.PartTreeJpaQuery$QueryPreparer.invokeBinding(PartTreeJpaQuery.java:236)
at org.springframework.data.jpa.repository.query.PartTreeJpaQuery$QueryPreparer.createQuery(PartTreeJpaQuery.java:157)
at org.springframework.data.jpa.repository.query.PartTreeJpaQuery.doCreateQuery(PartTreeJpaQuery.java:86)
at org.springframework.data.jpa.repository.query.AbstractJpaQuery.createQuery(AbstractJpaQuery.java:190)
at org.springframework.data.jpa.repository.query.JpaQueryExecution$CollectionExecution.doExecute(JpaQueryExecution.java:123)
at org.springframework.data.jpa.repository.query.JpaQueryExecution.execute(JpaQueryExecution.java:87)
at org.springframework.data.jpa.repository.query.AbstractJpaQuery.doExecute(AbstractJpaQuery.java:116)
at org.springframework.data.jpa.repository.query.AbstractJpaQuery.execute(AbstractJpaQuery.java:106)
at org.springframework.data.repository.core.support.RepositoryFactorySupport$QueryExecutorMethodInterceptor.doInvoke(RepositoryFactorySupport.java:499)
at org.springframework.data.repository.core.support.RepositoryFactorySupport$QueryExecutorMethodInterceptor.invoke(RepositoryFactorySupport.java:477)
at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:179)
at org.springframework.data.projection.DefaultMethodInvokingMethodInterceptor.invoke(DefaultMethodInvokingMethodInterceptor.java:56)
at …

我想基于LocalDateTime对一个Collection进行分组,但我只想得到小时,而不是分钟,秒......

.collect(Collectors.groupingBy(cp -> getUpdateLocalDate())


public LocalDate getUpdateLocalDate() {
    return getUpdateDate().toInstant().atZone(ZoneId.systemDefault()).toLocalDate();
}

但是我还需要Date,比如:2018-10-12T09:00:00不仅仅是小时

阅读文档。的deploy:deploy-file（https://maven.apache.org/plugins/maven-deploy-plugin/deploy-file-mojo.html），似乎只有必需的参数是必需参数<file>, <repositoryId> and <url>但是当我运行，：

mvn deploy:deploy-file -Durl={url} -DrepositoryId={repoId} -Dfile=D:\Users\nunito\IdeaProjects\calzada\target\calzada.zip

我收到此错误：

[ERROR] Failed to execute goal org.apache.maven.plugins:maven-deploy-plugin:2.8.2:deploy-file (default-cli) on project oib-kw-guards-web: The artifact i
nformation is incomplete or not valid:
[ERROR]   [0]  'groupId' is missing.
[ERROR]   [1]  'artifactId' is missing.
[ERROR]   [2]  'version' is missing.

我有一个 twiiter 应用程序：

// 不记名令牌

// API 密钥

// API 密钥秘密

// 访问令牌：

// 访问令牌秘密：

和 Read, Write, and Direct Messages 权限

我使用 ConfigurationBuilder 类在 Java 中以编程方式配置 Twitter4J：

ConfigurationBuilder cb = new ConfigurationBuilder();
    cb.setDebugEnabled(true)
            .setOAuthConsumerKey("IX9qalH5azhSk61ZoFMiL83Jg")   // API key
            .setOAuthConsumerSecret("tidoT2GgVZE3B3txDb64tuinlv0wZHbKesX5bNIcjPWpxmcYcs") // API key secret
            .setOAuthAccessToken("1345093646195010258-goY3W4oyqSltGFSUNPwJy8pSjSmfI5") // Access token:
            .setOAuthAccessTokenSecret("ijkpYB8sacVcBdvKHOnh95Y3QRITzD3c4Uq8nUUMLTN8O"); //Access token secret:
    TwitterFactory tf = new TwitterFactory(cb.build());
    return tf.getInstance();

但是当我尝试发布时出现此错误：

401:Authentication credentials (https://dev.twitter.com/pages/auth) were missing or incorrect. Ensure that you have set valid consumer key/secret, access token/secret, and the system clock …

我有这门课：

@Configuration\n@EnableWebSecurity\n@EnableGlobalMethodSecurity(prePostEnabled = true)\npublic class ApiWebSecurityConfig extends WebSecurityConfigurerAdapter {\n\n    private static final String SALT = "fd&lkj\xc2\xa7isfs23#$1*(_)nof";\n\n    private final JwtAuthenticationEntryPoint unauthorizedHandler;\n    private final JwtTokenUtil jwtTokenUtil;\n    private final UserSecurityService userSecurityService;\n\n    @Value("${jwt.header}")\n    private String tokenHeader;\n\n\n    public ApiWebSecurityConfig(JwtAuthenticationEntryPoint unauthorizedHandler, JwtTokenUtil jwtTokenUtil,\n            UserSecurityService userSecurityService) {\n        this.unauthorizedHandler = unauthorizedHandler;\n        this.jwtTokenUtil = jwtTokenUtil;\n        this.userSecurityService = userSecurityService;\n    }\n\n    @Autowired\n    public void configureGlobal(AuthenticationManagerBuilder auth) throws Exception {\n        auth\n                .userDetailsService(userSecurityService)\n                .passwordEncoder(passwordEncoder());\n    }\n\n    @Bean\n    public BCryptPasswordEncoder passwordEncoder() {\n        return new BCryptPasswordEncoder(12, new SecureRandom(SALT.getBytes()));\n    }\n\n    @Bean\n    @Override\n    public AuthenticationManager authenticationManagerBean() throws …

我刚刚为WebLogic Server版本12.1.3.0.0创建了一个身份验证提供程序,(身份验证提供程序通过在许多可配置的JAAS LoginModule之上构建身份验证序列来遵守标准JAAS框架.)但是当我启动Wl I时有这个错误:

这里的步骤:

1)设置ENV

%WL_HOME%/server/bin/setWLSEnv.cmd

2)生成MBean和stubs:

java -cp %WL_HOME%/server/lib/* -verbose -DcreateStubs="true" \
weblogic.management.commo.WebLogicMBeanMaker -MDF WSAuthentication.xml \
-files C:\Development\Workspaces\Eclipse\WLAuthenticationProvider\src

3)使用生成的存根和MBI文件打包认证提供程序和登录模块.

java -DMJF=C:\Development\Workspaces\Eclipse\WLAuthenticationProvider\jar\WSAuthentication.jar \
-Dfiles=C:\Development\Workspaces\Eclipse\WLAuthenticationProvider\src weblogic.management.commo.WebLogicMBeanMaker

4)在startWebLogic.cmd中添加-DUseSunHttpHandler = true

weblogic.security.service.SecurityServiceRuntimeException: [Security:090877]Service Common JAASAuthenticationService unavailable, see exception text: com.bea.common.engine.ServiceInitializationException: com.bea.common.engine.SecurityServiceRuntimeException: [Security:097533]SecurityProvider service class name for MyAuthentication is not specified.
        at weblogic.security.service.PrincipalAuthenticator.initialize(PrincipalAuthenticator.java:155)
        at weblogic.security.service.PrincipalAuthenticator.<init>(PrincipalAuthenticator.java:315)
        at weblogic.security.service.CommonSecurityServiceManagerDelegateImpl.doATN(CommonSecurityServiceManagerDelegateImpl.java:731)
        at weblogic.security.service.CommonSecurityServiceManagerDelegateImpl.postInitializeRealm(CommonSecurityServiceManagerDelegateImpl.java:515)
        at weblogic.security.service.CommonSecurityServiceManagerDelegateImpl.postLoadRealm(CommonSecurityServiceManagerDelegateImpl.java:861)
        at weblogic.security.service.CommonSecurityServiceManagerDelegateImpl.postInitializeRealms(CommonSecurityServiceManagerDelegateImpl.java:927)
        at weblogic.security.service.CommonSecurityServiceManagerDelegateImpl.postInitialize(CommonSecurityServiceManagerDelegateImpl.java:1109)
        at weblogic.security.service.SecurityServiceManager.postInitialize(SecurityServiceManager.java:943)
        at weblogic.security.SecurityService.start(SecurityService.java:159)
        at weblogic.server.AbstractServerService.postConstruct(AbstractServerService.java:78)
        at sun.reflect.GeneratedMethodAccessor6.invoke(Unknown Source)
        at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
        at java.lang.reflect.Method.invoke(Method.java:497)
        at org.glassfish.hk2.utilities.reflection.ReflectionHelper.invoke(ReflectionHelper.java:1017)
        at org.jvnet.hk2.internal.ClazzCreator.postConstructMe(ClazzCreator.java:388)
        at …

我使用 Spring Initializer、嵌入式 Tomcat、Thymeleaf 模板引擎生成了一个 Spring Boot Web 应用程序，并将其打包为可执行 JAR 文件。

使用的技术：

Spring Boot 1.4.2.RELEASE、Spring 4.3.4.RELEASE、Thymeleaf 2.1.5.RELEASE、Tomcat Embed 8.5.6、Maven 3、Java 8

这是我的安全配置类：

@Configuration
@EnableWebSecurity
@PropertySource("classpath:/com/tdk/iot/config/app-${APP-KEY}.properties")
public class SecurityConfig extends WebSecurityConfigurerAdapter {

    @Value("${securityConfig.formLogin.loginPage}")
    private String loginPage;

    @Override
    protected void configure(HttpSecurity http) throws Exception {

        http
            .formLogin()
                .loginPage(loginPage)
                .permitAll()
                .loginProcessingUrl("/login")
                .failureUrl("/login.html?error=true")
                .defaultSuccessUrl("/books/list")
                .and()
            .exceptionHandling()
                .accessDeniedPage("/denied")
                .and()
            .authorizeRequests()
                .antMatchers("/mockup/**").permitAll()
                .antMatchers("/books/**").permitAll()
                .antMatchers("/welcome/**").authenticated()
                .and()
            .logout()
                .permitAll()
                .logoutSuccessUrl("/index.html");
    }

    @Autowired
    public void configureGlobal(AuthenticationManagerBuilder auth) throws Exception {
        auth
            .inMemoryAuthentication()
                .passwordEncoder(new StandardPasswordEncoder())
                .withUser("test1").password("test1").roles("ADMIN").and()
                .withUser("test2").password("test2").roles("USER").and()
                .withUser("test3").password("test3").roles("SUPERADMIN");
    }

    @Bean …

我使用Spring Initializer,嵌入式Tomcat,Thymeleaf模板引擎和包作为可执行JAR文件生成了一个Spring Boot Web应用程序.

使用的技术:

Spring Boot 1.4.2.RELEASE,Spring 4.3.4.RELEASE,Thymeleaf 2.1.5.RELEASE,Tomcat Embed 8.5.6,Maven 3,Java 8

这是我在启动数据库时调用的bean

@SpringBootApplication
@EnableAutoConfiguration
@Import({SecurityConfig.class})
public class BookApplication {

    public static void main(String[] args) {
        SpringApplication.run(BookApplication.class, args);
    }
}



@Configuration
public class PersistenceConfig {

...

    /**
         * Creates an in-memory "books" database populated 
         * with test data for fast testing
         */
        @Bean
        public DataSource dataSource(){
            return
                (new EmbeddedDatabaseBuilder())
                .addScript("classpath:db/H2.schema.sql")
                .addScript("classpath:db/H2.data.sql")
                .build();
        }

当我执行此插入时

CREATE TABLE IF NOT EXISTS t_time_lapse (
      id          bigint  PRIMARY KEY,
      name        varchar(50) NOT …

我有一个基本的 SpringBoot 2.0.4.RELEASE 应用程序。使用 Spring Initializer、JPA、嵌入式 Tomcat、Thymeleaf 模板引擎，并打包为可执行 JAR 文件。

我有一个具有角色的用户对象：

@Entity
@Table(name="t_user")
public class User implements Serializable, UserDetails {


 @ManyToMany(cascade = CascadeType.MERGE, fetch = FetchType.EAGER)
    @JoinTable(
        name="t_user_role",
        joinColumns=@JoinColumn(name="user_id", referencedColumnName="id"),
        inverseJoinColumns=@JoinColumn(name="role_id", referencedColumnName="id"))
    private Set<Role> roles = new HashSet<>();
..
}

当我启动应用程序时。我创建了所有角色：

roleService.save(new Role(RolesEnum.USER.getRoleName()));
roleService.save(new Role(RolesEnum.ADMIN.getRoleName()));

然后我创建一个具有 USER 角色的用户：

User user1 = new User();


         Role role = roleService.findByName(RolesEnum.USER.getRoleName());

         user.getRoles().add(role);
          userService.save(user);

但是当我创建另一个具有相同角色的用户时：

User user2 = new User();


         Role role = roleService.findByName(RolesEnum.USER.getRoleName());

         user2.getRoles().add(role);
          user2Service.save(user);

我收到这个错误：

Multiple representations of the same entity [com.tdk.backend.persistence.domain.backend.Role#1] are …