小编kub*_*a44的帖子

为什么JavaMail连接超时太长

在我的应用程序中,我连接到服务器以验证用户.这是代码:

try {
        Properties prop = new Properties();
        prop.put("mail.smtp.starttls.enable","true");
        prop.put("mail.smtp.auth", "true");
        prop.put("mail.smtp.connectiontimeout", 1000);


        Session session = Session.getInstance(prop, null);
        Transport transport = session.getTransport("smtp");
        transport.connect("mion.elka.pw.edu.pl", 587, registerLog, registerPass);
        transport.close();
        return true;
    } catch (NoSuchProviderException ex) {
        Logger.getLogger(RegisterController.class.getName()).log(Level.SEVERE, null, ex);
        return false;
    } catch(AuthenticationFailedException ex) {
        Logger.getLogger(RegisterController.class.getName()).log(Level.SEVERE, null, ex);
        return false;
    } catch (MessagingException ex) {
        Logger.getLogger(RegisterController.class.getName()).log(Level.SEVERE, null, ex);
        return false;
    }

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我将连接超时设置为1000毫秒= 1秒,但它被忽略.当我调试并设置错误的用户名和密码时,我抓住了

javax.mail.MessagingException: java.net.SocketTimeoutException: Read timed out

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不是在1000毫秒之后,而是在5000*60毫秒= 5分钟之后

怎么了？我怎样才能减少时间？

java jakarta-mail connection-timeout

kub*_*a44

2015 06-17

17
推荐指数

6
解决办法

4万
查看次数

hibernate查询列表:空或null？

如果我有使用Hibernate的方法,像这样:

public <T> T typedQuery(Query q, Class<T> type) {
    List<T> results = q.list();

    //result will be null or empty List ?
}

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如果查询不会从表中获取任何记录,那么result会是null空列表吗？

java null hibernate list jpql

kub*_*a44

2018 07-11

13
推荐指数

1
解决办法

1万
查看次数

为什么页面在JSF中更改,而不是URL？

我有简单的测试网络应用程序:第一页 - index.xhtml:

<?xml version='1.0' encoding='UTF-8' ?>
<html>
 <h:head>
    <title>Facelet Title</title>
 </h:head>
 <h:body>
    First page
    <h:form>
        <h:commandButton value="Go to Next page" action="next"/>
    </h:form>
 </h:body>
</html>

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和下一页 - next.xhtml:

<?xml version='1.0' encoding='UTF-8' ?>
<html>
<h:head>
    <title>Facelet Title</title>
 </h:head>
 <h:body>
    Next page
    <h:form>
        <h:commandButton value="Go to First page" action="index"/>
    </h:form>
 </h:body>
</html>

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当我运行我的应用程序时,我在浏览器上有该URL:

http://localhost:8080/Test/

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和index.xhtml作为第一页.

然后当我点击"转到下一页"我有网址

http://localhost:8080/Test/index.xhtml

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和next.xhtml页面.当我点击"转到第一页"页面更改(到index.xhtml)但是url它

http://localhost:8080/Test/next.xhtml

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web.xml中:

<?xml version="1.0" encoding="UTF-8"?>
<web-app>
 <context-param>
     <param-name>javax.faces.PROJECT_STAGE</param-name>
     <param-value>Development</param-value>
 </context-param>
 <listener>
     <listener-class>org.apache.myfaces.webapp.StartupServletContextListener</listener-class>
 </listener>
 <servlet>
     <servlet-name>Faces Servlet</servlet-name>
     <servlet-class>javax.faces.webapp.FacesServlet</servlet-class>
     <load-on-startup>1</load-on-startup>
 </servlet>
 <servlet-mapping>
     <servlet-name>Faces Servlet</servlet-name>
     <url-pattern>*.xhtml</url-pattern>
 </servlet-mapping> …

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navigation url jsf jsf-2

kub*_*a44

2017 05-13

7
推荐指数

2
解决办法

1万
查看次数

为什么EntityManager为空？

在我的Web应用程序中,我在Apache Tomcat(TomEE)/7.0.37服务器上使用OpenJPA.我使用Netbeans自动生成类("实体类来自数据库......"和"会话Bean来自实体类......").在SessionBean(例如UserFacade)我想获得EntityManager:

@Stateless
public class UserFacade extends AbstractFacade<User> {
  @PersistenceContext(unitName = "CollDocPU")
  private EntityManager em;

  @Override
  protected EntityManager getEntityManager() {
    return em;
  }
}

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但是当我通过上面的方式得到它时,我得到了null.当我通过:

@Override
protected EntityManager getEntityManager() {
    EntityManagerFactory emf = Persistence.createEntityManagerFactory("CollDocPU");
    EntityManager ecm = emf.createEntityManager(); 
    return ecm;
}

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ecm不是null,没关系

我的persistence.xml:

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">
<persistence-unit name="CollDocPU" transaction-type="RESOURCE_LOCAL">
<provider>org.apache.openjpa.persistence.PersistenceProviderImpl</provider>
<class>model.entity.StudentAddSolution</class>
<class>model.entity.Lecturer</class>
<class>model.entity.Solution</class>
<class>model.entity.Student</class>
<class>model.entity.Course</class>
<class>model.entity.File</class>
<class>model.entity.CourseHasLecturer</class> 
<class>model.entity.Mail</class>
<class>model.entity.StudentAtCourse</class>
<class>model.entity.Roles</class>
<class>model.entity.Task</class>
<class>model.entity.User</class>
<exclude-unlisted-classes>false</exclude-unlisted-classes>
<properties>
   <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:11080/myBase?zeroDateTimeBehavior=convertToNull"/>
   <property name="javax.persistence.jdbc.password" value="pass,"/>
   <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
   <property …

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java entitymanager openjpa persistence.xml tomee

kub*_*a44

2015 12-06

7
推荐指数

1
解决办法

3万
查看次数

我找到了一些答案:https://stackoverflow.com/a/21218921/2754014关于依赖注入.没有任何注释@Autowired,@Inject或,或@Resource.让我们假设这个示例TwoInjectionStylesbean 没有任何XML配置(简单除外)<context:component-scan base-package="com.example" />.

在没有指定注释的情况下注入是否正确？

java spring annotations dependency-injection autowired

kub*_*a44

lucky-day

7
推荐指数

2
解决办法

8572
查看次数

c:选择多个c:when - fall-through开关

如果我在我的jsf页面上有这个代码:

<c:choose>
    <c:when test="${empty example1}">
    </c:when>

    <c:when test="${empty example2}">
    </c:when>

    <c:otherwise>
    </c:otherwise>     
</c:choose>

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它将像java switch语句一样工作case和break- 如果第一次是真的,第二次不会测试,对吧？

我应该写什么来获得" switch声明case但没有break":

当第一个C:when是真的东西添加到页面,当第二个是真时,某些东西也会添加到页面

jsf loops jstl break

kub*_*a44

2013 11-10

4
推荐指数

1
解决办法

9567
查看次数

如何通过Java 8 Predicate仅过滤特定元素？

我收集List<Foo>的Foo要素:

class Foo {
  private TypeEnum type;
  private int amount;

  //getters, setters ...
}

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Foo类型可以TypeEnum.A和TypeEnum.B.

我想只Foo从列表中获取那些元素type == TypeEnum.B然后amount大于零(amount > 0)的元素.

我怎么能用Java 8 Streams filter()方法做到这一点？

如果我使用:

List<Foo> l = list.stream()
    .filter(i -> i.getType().equals(TypeEnum.B) && i.getAmount() > 0)
    .collect(Collectors.<Foo>toList());

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我得到了Foo元素,TypeEnum.B但没有TypeEnum.A.

java java-8 java-stream

kub*_*a44

2017 10-16

4
推荐指数

1
解决办法

415
查看次数

使用@ViewScoped时出现ViewExpiredException

我的h有问题:commandButton"登录":当我使用@ViewScoped并按下此按钮时会出现ViewExpiredException,但是当我使用@SessionScoped时,没有任何错误.

堆栈跟踪:

javax.faces.application.ViewExpiredException: /pages/register.xhtmlNo saved view state could be found for the view identifier: /pages/register.xhtml
at org.apache.myfaces.lifecycle.RestoreViewExecutor.execute(RestoreViewExecutor.java:132)
at org.apache.myfaces.lifecycle.LifecycleImpl.executePhase(LifecycleImpl.java:170)
at org.apache.myfaces.lifecycle.LifecycleImpl.execute(LifecycleImpl.java:117)
at javax.faces.webapp.FacesServlet.service(FacesServlet.java:197)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:305)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:210)
at org.primefaces.webapp.filter.FileUploadFilter.doFilter(FileUploadFilter.java:79)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:243)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:210)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:222)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:123)
at org.apache.tomee.catalina.OpenEJBValve.invoke(OpenEJBValve.java:45)
at org.apache.tomee.catalina.OpenEJBValve.invoke(OpenEJBValve.java:45)
at org.apache.tomee.catalina.OpenEJBValve.invoke(OpenEJBValve.java:45)
at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:472)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:171)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:99)
at org.apache.catalina.valves.AccessLogValve.invoke(AccessLogValve.java:936)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:118)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:407)
at org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1004)
at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:589)
at org.apache.tomcat.util.net.JIoEndpoint$SocketProcessor.run(JIoEndpoint.java:312)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615)
at java.lang.Thread.run(Thread.java:724)

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我的页面:

            <h:form>
                <h:panelGrid columns="2" >
                    <h:outputLabel value="Login:"/>
                    <h:inputText value="#{registerController.registerLog}"/>
                    <h:outputLabel value="#{msg.password}"/>
                    <h:inputSecret id="pass" …

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jsf myfaces jsf-2 viewexpiredexception view-scope

kub*_*a44

2013 09-14

3
推荐指数

1
解决办法

9948
查看次数

Java - 以相同的方法抛出和捕获

我在理解现有代码时遇到问题。我想知道 Java 如何管理抛出异常并以相同的方法捕获它。我在其他问题中找不到它，所以我准备了一个例子。在 Java 中运行下面的代码会输出什么？

public static void main(String [ ] args) {
     try{
         System.out.println("1");
         method();
     }
     catch(IOException e) {
         System.out.println("4");
     }
}

public static void method() throws IOException {
     try {
         System.out.println("2");
         throw new IOException();
     }
     catch(IOException e) {
         System.out.println("3");
     }
 }

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它会是1 2 3还是1 2 4？

java exception try-catch throw

kub*_*a44

2017 03-22

3
推荐指数

1
解决办法

5998
查看次数

为什么我得到javax.ejb.EJBTransactionRolledbackException？ - setRollbackOnly

在我的Web应用程序中,我在Apache Tomcat(TomEE)/7.0.37服务器上使用OpenJPA.

我的实体User.class:

@Entity
@Table(name = "USER")
public class User implements Serializable {
  private static final long serialVersionUID = 1L;
  @Id
  @GeneratedValue
  @Basic(optional = false)
  @Column(name = "id_user")
  private Integer idUser;
  @Size(max = 8)
  @Column(name = "login")
  private String login;
  @Size(max = 128)
  @Column(name = "password")
  private String password;
  @OneToOne(cascade = CascadeType.ALL, mappedBy = "user")
  private Lecturer lecturer;
  @OneToOne(cascade = CascadeType.ALL, mappedBy = "user")
  private Student student;
  @OneToMany(cascade = CascadeType.ALL, mappedBy = "user")
  private List<UserHasRoles> userHasRolesList;

  //constructors, getters, setters
} …

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java jpa openjpa tomee

kub*_*a44

2015 12-06

0
推荐指数

1
解决办法

3万
查看次数

如何使用jpa将utf-8字符正确插入MySQL表

我想通过我的Web应用程序插入MySQL数据库信息.我是在TomEE服务器上通过OpenJPA来做的,例如:

public void addLecturer(String name, String surname) {
    Lecturer lect = new Lecturer(name,surname);

    em.persist(lect);
}

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Lecturer是实体类.问题是应用程序以latin1编码插入数据.我想以UTF-8编码插入它.

当我在命令行终端中通过简单的MySQL命令插入数据时,我获得UTF-8编码的信息.

我看了这个,我看到character_set_server被设置为latin1,但我不知道这是问题还是如何改变它:

mysql> show variables like 'char%';
+--------------------------+----------------------------+
| Variable_name            | Value                      |
+--------------------------+----------------------------+
| character_set_client     | utf8                       |
| character_set_connection | utf8                       |
| character_set_database   | utf8                       |
| character_set_filesystem | binary                     |
| character_set_results    | utf8                       |
| character_set_server     | latin1                     |
| character_set_system     | utf8                       |
| character_sets_dir       | /usr/share/mysql/charsets/ |
+--------------------------+----------------------------+

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编辑

我有同样的问题:

JPA utf-8字符没有保留

和

UTF - 8与JPA和Glassfish 4.0 …

mysql encoding persistence jpa

kub*_*a44

2017 05-23

0
推荐指数

2
解决办法

8722
查看次数