小编use*_*611的帖子

我正在尝试重定向,而不将参数添加到我的URL.

@Controller
...
public class SomeController
{
  ...
  @RequestMapping("save/")
  public String doSave(...)
  {
    ...
    return "redirect:/success/";
  }

  @RequestMapping("success/")
  public String doSuccess(...)
  {
    ...
    return "success";
  }

重定向后,我的网址看起来总是这样:.../success/?param1=xxx&param2=xxx.由于我希望我的URL有点RESTful,并且在重定向后我从不需要params,我不希望它们被添加到重定向上.

任何想法如何摆脱它们？

由于WS客户端服务和端口的初始化需要很长时间,所以我喜欢在启动时初始化它们并重用相同的端口实例.初始化看起来像这样:

private static RequestContext requestContext = null;

static
{
    MyService service = new MyService(); 
    MyPort myPort = service.getMyServicePort(); 

    Map<String, Object> requestContextMap = ((BindingProvider) myPort).getRequestContext();
    requestContextMap = ((BindingProvider)myPort).getRequestContext(); 
    requestContextMap.put(BindingProvider.USERNAME_PROPERTY, uName); 
    requestContextMap.put(BindingProvider.PASSWORD_PROPERTY, pWord); 

    rc = new RequestContext();
    rc.setApplication("test");
    rc.setUserId("test");
}

我班上某个地方的电话:

myPort.someFunctionCall(requestContext, "someValue");

我的问题:这个调用是否是线程安全的？

我的目标是过滤最佳匹配.在我的例子中,我有一个人员列表,我想按姓氏和名字过滤.

匹配的优势将是:

1. 姓氏和名字匹配,返回第一场比赛
2. 只有姓氏匹配,返回第一场比赛
3. 没有比赛,抛出一些例外

我的代码到目前为止:

final List<Person> persons = Arrays.asList(
  new Person("Doe", "John"),
  new Person("Doe", "Jane"),
  new Person("Munster", "Herman");

Person person = persons.stream().filter(p -> p.getSurname().equals("Doe")).???

目前我有一个Web应用程序,我们使用web.xml来配置应用程序.web.xml有welcome-file-list.

<web-app>  
   ...
   <welcome-file-list>  
     <welcome-file>home.html</welcome-file>  
   </welcome-file-list>  
</web-app>  

我们计划使用spring框架并使用java类进行应用程序配置.

class MyApplication extends WebApplicationInitializer {
    public void onStartUp(ServletContext context){
        ...
    }
}

如何在此java类中指定welcome-file-list？

我在这里有一个非常天真的概念问题需要澄清.在我的应用程序中,我目前正在使用JSR 303 Hibernate Validator来验证带有@NotBlank和@NotNull注释的域模型,比如检查用户名,密码等.

public class Admin {

  private Long id;
  @NotBlank(message = "Username should not be null")
  private String username;
  @NotBlank(message = "Username should not be null")
  private String password;
  ...

但是为了验证像现有用户名这样的域逻辑,我仍然使用Spring的Validator接口

@Component
public class AdminValidator implements Validator {
  ...
  @Override
  public void validate(Object target, Errors errors) {
    Admin admin = (Admin) target;
    if (usernameAlradyExist())){
        errors.rejectValue("username", null, "This user already exist's in the system.");
    }

}

在控制器中,我正在使用它们

@RequestMapping(value = "/register", method = RequestMethod.POST)
public String register(@Valid @ModelAttribute("admin") Admin admin, …

我想知道我是否可以将这段xml配置映射到Spring JavaConfig:

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans" 
  xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
  xmlns:aop="http://www.springframework.org/schema/aop" 
  xmlns:tx="http://www.springframework.org/schema/tx" 
  xsi:schemaLocation="http://www.springframework.org/schema/beans   http://www.springframework.org/schema/beans/spring-beans.xsd
                      http://www.springframework.org/schema/aop     http://www.springframework.org/schema/aop/spring-aop.xsd
                      http://www.springframework.org/schema/tx      http://www.springframework.org/schema/tx/spring-tx.xsd"
  default-autowire="byName">

  <aop:config>
     <aop:pointcut id="serviceAnnotatedClass" expression="@within(org.springframework.stereotype.Service)" />
     <aop:advisor id="managerTx" advice-ref="txAdvice" pointcut-ref="serviceAnnotatedClass" order="20" />
  </aop:config>

  <tx:advice id="txAdvice" transaction-manager="transactionManager">
    <tx:attributes>
      <tx:method name="get*" read-only="true" />
      <tx:method name="find*" read-only="true" />
      <tx:method name="load*" read-only="true" />
      <tx:method name="is*" read-only="true" />
      <tx:method name="ownTransaction*" propagation="REQUIRES_NEW" rollback-for="Exception" />
      <tx:method name="*" rollback-for="Exception" />
    </tx:attributes>
  </tx:advice>

</beans>

到目前为止,我想出了如何替换aop:pointcut with

<aop:advisor id="managerTx" advice-ref="txAdvice" 
pointcut="com.myapp.configuration.AspectConfig.serviceAnnotatedClass()" order="20"/>

和

import org.aspectj.lang.annotation.Aspect;
import org.aspectj.lang.annotation.Pointcut;

@Aspect
public class AspectConfig
{

  @Pointcut("@within(org.springframework.stereotype.Service)") …

有没有办法用特定的jdk启动SonarQube服务器(v.3.7.4)？

我的情况:我的java-home设置为jdk 1.8,但SonarQube服务器有1.8的已知问题.所以我想用jdk 1.7启动服务器(不将我的java-home设置为1.7).我在bat文件中找不到任何东西.

操作系统:Windows 7; SonarQube服务器版本:3.7.4

我很难从OneToMany协会中删除孩子.我的实体:

@Entity
@Table(name = "PERSON")
public class PersonEntity extends BaseVersionEntity<Long> implements Comparable<PersonEntity>
{
  ...
  // bi-directional many-to-one association to Project
  @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "person", orphanRemoval = true)
  private final Set<ProjectEntity> projects = new HashSet<ProjectEntity>();
  ...

@Entity
@Table(name = "PROJECT")
public class ProjectEntity extends BaseVersionEntity<ProjectPK>
{
  @EmbeddedId
  private ProjectPK id;
  ...
  // bi-directional many-to-one association to UdbPerson
  @ManyToOne(fetch = FetchType.LAZY)
  @JoinColumn(name = "PERSON_ID", nullable = false, insertable = false, updatable = false)
  private PersonEntity person;
  ... …

此查询用于以一对多关系检索最后的记录(请参阅SQL连接:选择一对多关系中 的最后记录)

SELECT  p.*
FROM    customer c 
        INNER JOIN (
                      SELECT customer_id, MAX(date) MaxDate
                      FROM purchase
                      GROUP BY customer_id
                    ) MaxDates ON c.id = MaxDates.customer_id 
        INNER JOIN purchase p ON MaxDates.customer_id = p.customer_id
                    AND MaxDates.MaxDate = p.date;

我的问题:如何使用jpa criteria-api的subselect构建此连接？可能吗？如果没有,可以使用jpql吗？

我的代码到目前为止:

final CriteriaBuilder cb = entityManager.getCriteriaBuilder();
final CriteriaQuery<Purchase> query = cb.createQuery(Purchase.class);
final Root<CustomerEntity> root = query.from(Customer.class);

// here should come the join with the sub-select

final Path<Purchase> path = root.join(Customer_.purchases);
query.select(path);

final TypedQuery<Purchase> typedQuery = entityManager.createQuery(query);
return typedQuery.getResultList();

我正忙着自己的资源实施.该的getInputStream -方法不会被调用.

我的经纪人:

public class ResourceHandlerWrapperImpl extends
        ResourceHandlerWrapper {

  private final ResourceHandler wrapped;

  public ResourceHandlerWrapper(final ResourceHandler wrapped)
  {
    this.wrapped = wrapped;
  }

  @Override
  public ResourceHandler getWrapped()
  {
    return wrapped;
  }

  @Override
  public Resource createResource(final String resourceName, final String libraryName)
  {
    if (AppConstants.RESOURCE_MEDIA_LIB.equals(libraryName))
    {
      return new MediaResource(resourceName);
    }
    else
    {
      return super.createResource(resourceName, libraryName);
    }
  }

  /**
   * @see javax.faces.application.ResourceHandlerWrapper#libraryExists(java.lang.String)
   */
  @Override
  public boolean libraryExists(final String libraryName)
  {
    if (AppConstants.RESOURCE_MEDIA_LIB.equals(libraryName))
    {
      return true;
    }
    else
    {
      return super.libraryExists(libraryName);
    }
  }

  /** …