小编unk*_*own的帖子

我将 Localstack 与 Testcontainers((testcontainers:localstack:1.15.2 )) 一起使用进行集成测试，并在测试设置中设置秘密，如下所示：代码示例

 import com.amazonaws.services.secretsmanager.AWSSecretsManager;
import com.amazonaws.services.secretsmanager.AWSSecretsManagerClientBuilder;
import com.amazonaws.services.secretsmanager.model.CreateSecretRequest; 
import org.junit.Rule;
import org.junit.Test;
import org.testcontainers.containers.localstack.LocalStackContainer;
import org.testcontainers.utility.DockerImageName; 
import static org.testcontainers.containers.localstack.LocalStackContainer.Service.SECRETSMANAGER;

public class QueueServiceTest {

    DockerImageName localstackImage = DockerImageName.parse("localstack/localstack:0.11.3");
    @Rule
    public LocalStackContainer localstack = new LocalStackContainer(localstackImage)
            .withServices(SECRETSMANAGER).withEnv("LOCALSTACK_HOSTNAME", "localhost").withEnv("HOSTNAME", "localhost");
    @Test
    public void someTestMethod() {
        AWSSecretsManager secretsManager = AWSSecretsManagerClientBuilder.standard()
                .withCredentials(localstack.getDefaultCredentialsProvider()).withRegion(localstack.getRegion())
                .build();

        String secretString = "usrnme";
        CreateSecretRequest request = new CreateSecretRequest().withName("test")
                .withSecretString(secretString)
     .withRequestCredentialsProvider(localstack.getDefaultCredentialsProvider());
        secretsManager.createSecret(request);
    }

}

现在测试崩溃并出现错误：

com.amazonaws.services.secretsmanager.model.AWSSecretsManagerException：请求中包含的安全令牌无效。（服务：AWSSecretsManager；状态代码：400；错误代码：UnrecognizedClientException；请求 ID：314b0dee-69ed-4b08-9cd0-2618b8e14b25；代理：null）

在 com.amazonaws.http.AmazonHttpClient$RequestExecutor.handleErrorResponse(AmazonHttpClient.java:1819) 在 com.amazonaws.http.AmazonHttpClient$RequestExecutor.handleServiceErrorResponse(AmazonHttpClient.java:1403) 在 com.amazonaws.http.AmazonHttpClient$RequestExecutor.executeOneRequest (AmazonHttpClient.java:1372) 在 com.amazonaws.http.AmazonHttpClient$RequestExecutor.executeHelper(AmazonHttpClient.java:1145) 在 com.amazonaws.http.AmazonHttpClient$RequestExecutor.doExecute(AmazonHttpClient.java:802) …

我在apache tomcat5.5上使用Spring 2.0.6和Hibernate 3.2.x,现在我们计划将我们的hybernate映射文件改为hybernate + jpa支持映射文件.为此,我们创建了这样的文件

daoConfig.xml

 <beans:bean id="dataSource" 
  class="org.springframework.jndi.JndiObjectFactoryBean">
      <beans: property name="jndiName">
    <beans: value>java:/comp/jdbc/Paymentsdb</beans:value>  
   </beans: property>
  </beans: bean> 
 <beans: bean id="sessionFactory"
  class="org.springframework.orm.hibernate3.LocalSessionFactoryBean">
  <beans: property name="useTransactionAwareDataSource"
   value="true" />
  <beans: property name="dataSource">
   <beans:ref bean="dataSource"/>
  </beans: property>
  <beans: property name="hibernateProperties">
   <beans: props>
    <beans:prop key="hibernate.dialect">${database.target}</beans:prop>
    <beans:prop key="hibernate.connection.isolation">3</beans:prop>
    <beans:prop key="hibernate.current_session_context_class">jta</beans:prop>
    <beans:prop key="hibernate.transaction.factory_class">com.atomikos.icatch.jta.hibernate3.AtomikosJTATransactionFactory
    </beans:prop>
    <beans: prop key="hibernate.transaction.manager_lookup_class">com.atomikos.icatch.jta.hibernate3.TransactionManagerLookup
    </beans: prop>
    <beans: prop key="hibernate.connection.release_mode">on_close</beans: prop>
    <beans: prop key="hibernate.show_sql">false</beans: prop>
   </beans: props>
  </beans: property>
 </beans: bean>

 <beans: bean id="jpaTemplate"
  class="org.springframework.orm.jpa.JpaTemplate">
  <beans: property name="entityManagerFactory">
   <beans:ref bean="entityManagerFactory" />
  </beans: property> …

我坚持使用JPA 2.0中的CriteriaBuilder构建动态查询.我的应用程序是Spring 3.0,基于Hibernate 3.6.0 + JPA 2.0.实际上我有两个实体taUser,另一个是taContact,在我的taUser班级有一个属性,与taContact我的pojo类有多对一的关系(示例)

public class TaUser implements java.io.Serializable {
    private int userId;
    private TaContact taContact;
    public int getUserId() {
        return this.userId;
    }

    public void setUserId(int userId) {
        this.userId = userId;
    }
    public TaContact getTaContact() {
        return taContact;
    }

    public void setTaContact(TaContact taContact) {
        this.taContact = taContact;
    }

    }


   public class TaContact implements java.io.Serializable {

    private int contactId;

    public int getContactId() {
        return this.contactId;
    }

    public void setContactId(int contactId) …