小编wth*_*ira的帖子

我已经创建了一个登录页面angular2.它有两个输入字段as userName和password.登录中有一个按钮.当我打算为此编写测试用例时loginComponent,不得不predicate用来识别button.这是我的测试用例.

import { async, ComponentFixture, TestBed } from '@angular/core/testing';
import { By } from '@angular/platform-browser';
import { DebugElement } from '@angular/core';
import { LoginComponent } from './login.component';


export class LoginComponentTest {
  constructor(public loginComponent: LoginComponent){
    this.loginComponent.logIn('Chathura', '1234');

  }
}

describe('LoginComponent', () => {
  beforeEach(() => {
    TestBed.configureTestingModule({
      declarations: [
        LoginComponent
      ],
    })
  });

  const fixture = TestBed.createComponent(LoginComponent);

  const button = fixture.query(predicate: Predicate<DebugElement>, scope?: Function) : DebugElement

});

我将通过提供输入值进行测试,然后单击登录按钮,看看接下来会发生什么.

我无法找到正确使用谓词的正确教程.

我尝试使用创建s3桶

aws s3api create-bucket —bucket kubernetes-aws-wthamira-io

它会犯这个错误

调用CreateBucket操作时发生错误(IllegalLocationConstraintException):未指定的位置约束与此请求发送到的区域特定端点不兼容.

我使用aws configure区域设置并将区域设置为eu-west-1

An error occurred (IllegalLocationConstraintException) when calling
the CreateBucket operation: The unspecified location constraint is
incompatible for the region specific endpoint this request was sent
to.

但是犯同样的错误.我怎么解决这个问题.

我使用osx终端连接aws

我正在尝试更新Spring Boot中的数据源,当数据库名称,密码或主机名等数据库属性在spring配置文件或自定义数据库属性文件中更改时.当属性更改时,应用程序必须通过侦听属性更改来自行更新.

一旦数据库配置发生变化,我就会使用Spring执行器来重启bean.但是用户必须明确地发布重新启动的请求.必须通过监听更改并更新数据源来避免此步骤.

你能告诉我在Spring启动时做到这一点的最佳方法吗？

我将使用 axon 制作 cqrs 应用程序。我尝试将 axon RabbitMq 配置到我的应用程序。如果添加 spring-boot-starter-amqp 和 axon-amqp 会导致此错误。我该如何解决这个问题。

[ERROR] Tests run: 1, Failures: 0, Errors: 1, Skipped: 0, Time elapsed: 2.931 s <<< FAILURE! - in com.thamira.research.ProductServiceApplicationTests
[ERROR] contextLoads(com.thamira.research.ProductServiceApplicationTests)  Time elapsed: 0.007 s  <<< ERROR!
java.lang.NoClassDefFoundError: org/springframework/messaging/handler/annotation/support/MessageHandlerMethodFactory
Caused by: java.lang.ClassNotFoundException: org.springframework.messaging.handler.annotation.support.MessageHandlerMethodFactory

[INFO] 
[INFO] Results:
[INFO] 
[ERROR] Errors: 
[ERROR]   ProductServiceApplicationTests.contextLoads » NoClassDefFound org/springframew...
[INFO] 
[ERROR] Tests run: 1, Failures: 0, Errors: 1, Skipped: 0
[INFO] 
[INFO] ------------------------------------------------------------------------
[INFO] BUILD FAILURE
[INFO] ------------------------------------------------------------------------
[INFO] Total time: 15.136 s …

我使用@Audited annotation作为我的基本模型.我为所有实体扩展了它.但它不起作用.有什么方法我可以使用它

这是我的基本模型

@MappedSuperclass
@Getter
@Setter
@Audited 
public abstract class BaseModelObject implements Serializable {

    /**
     * 
     */
    private static final long serialVersionUID = 4194525198831057382L;


    @Id     
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "id")
    protected Long id;
}

这是我的模特课

@Entity
public class City extends BaseModelObject {

    private static final long serialVersionUID = 1L;

    @Column
    private String name;
 }

我试图将我的注册服务器从"Eureka"更改为"Consul",并将Consul更改为我的配置服务器.使用Consul进行服务发现是成功的.但我无法理解如何获取键/值对选项来引导我的应用程序.我有可能做到这一点吗？

我使用具有以下依赖性的弹簧靴

<dependency>
    <groupId>org.springframework.cloud</groupId>
    <artifactId>spring-cloud-starter-consul-config</artifactId>
</dependency>
<dependency>
        <groupId>org.springframework.cloud</groupId>
    <artifactId>spring-cloud-starter-consul-discovery</artifactId>
</dependency>

这是我的春季启动应用程序ConsuleDemoApplication.class

@EnableDiscoveryClient
@SpringBootApplication
@RestController
public class ConsuleDemoApplication {

    @RequestMapping("/")
    public String home() {
        return "Hello world";
    }

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

和我的bootstrap.yml是

spring:
  cloud:
    consul:
      discovery:
        healthCheckInterval: 15s
        instanceId: ${spring.application.name}:${spring.application.instance_id:${random.value}}

  application:
    name: consul_demo

我尝试使用axon 配置cqrs和事件源. SeatReseveCreateCommand工作正常.但SeatReserveUpadateCommand无法正常工作.

这是我的SeatReserve 聚合

@Aggregate
public class SeatReserve {
    @AggregateIdentifier
    private String id;
    private String seatid;
    private Date date;

    @SuppressWarnings("unused")
    private SeatReserve() {
    }

    @CommandHandler
    public SeatReserve(SeatReseveCreateCommand seatReseveCreateCommand) {
        apply(new SeatReseveCreateEvent(seatReseveCreateCommand.getMyid(), seatReseveCreateCommand.getSeatId(),
                seatReseveCreateCommand.getDate()));
    }

    @CommandHandler
    public SeatReserve(SeatReserveUpadateCommand upadateCommand) {
        apply(new SeatReserveUpadateEvent(id, upadateCommand.getSeatId()));
    }

    @EventSourcingHandler
    public void on(SeatReseveCreateEvent seatReseveCreateEvent) {
        this.id = seatReseveCreateEvent.getId();
        this.seatid = seatReseveCreateEvent.getSeatId();
        this.date = seatReseveCreateEvent.getDate();
    }

    @EventSourcingHandler
    public void on(SeatReserveChangeEvent upadateEvent) {
        seatid = upadateEvent.getSeatId();
    }

}

这是我的控制器

@RestController …

我需要对我的后端实施 CSRF 保护。我正在使用以下配置。但应用程序允许没有 CSRF 令牌的 Post 和 Get 请求。

@Slf4j
@EnableWebFluxSecurity
@EnableReactiveMethodSecurity
public class SecurityConfig {

    @Bean
    SecurityWebFilterChain springSecurityFilterChain(ServerHttpSecurity http) {
        http
                .csrf(csrf -> csrf.csrfTokenRepository(CookieServerCsrfTokenRepository.withHttpOnlyFalse()))
                .authorizeExchange()
                .anyExchange().authenticated()
                .and().oauth2ResourceServer().jwt();
        return http.build();
    }
}

在 HTTP 请求中包含实际的 CSRF 令牌

 @ControllerAdvice
    public class SecurityControllerAdvice {
        @ModelAttribute
        Mono<CsrfToken> csrfToken(ServerWebExchange exchange) {
            Mono<CsrfToken> csrfToken = exchange.getAttribute(CsrfToken.class.getName());
            return csrfToken.doOnSuccess(token -> {
                exchange.getAttributes()
                        .put(CsrfRequestDataValueProcessor.DEFAULT_CSRF_ATTR_NAME, token);
            });
        }
    }

我尝试使用邮递员的API。但这对我不起作用。

春季版

 <parent>
        <groupId>org.springframework.boot</groupId>
        <artifactId>spring-boot-starter-parent</artifactId>
        <version>2.5.3</version>
        <relativePath/>
   </parent> 

依赖项：

<dependency>
            <groupId>org.springframework.security</groupId>
            <artifactId>spring-security-oauth2-resource-server</artifactId>
            <version>RELEASE</version>
        </dependency>
        <dependency>
            <groupId>org.springframework.security</groupId>
            <artifactId>spring-security-oauth2-jose</artifactId>
            <version>RELEASE</version>
        </dependency> …