小编bun*_*112的帖子

我有一个文档实体多对一地映射到项目实体.

当我调用document.getProject,在调试器中,在文档对象的项目字段中,我看到了一些东西Project$$EnhancerByCGLIB$$67a694bd.

如何检索实际项目对象？

我有时会得到这个例外.我真的不知道哪里有问题,我使用了abstracttablemodel的实现.我无法找到发生异常的精确位置,因为当我使用调试器并逐步运行异常时不会发生.

Exception occurred during event dispatching:
    java.lang.ArrayIndexOutOfBoundsException: 2 >= 2
    at java.util.Vector.elementAt(Vector.java:427)
    at javax.swing.table.DefaultTableColumnModel.getColumn(DefaultTableColumnModel.java:277)
    at javax.swing.plaf.basic.BasicTableUI.paintGrid(BasicTableUI.java:1931)
    at javax.swing.plaf.basic.BasicTableUI.paint(BasicTableUI.java:1767)
    at com.alee.laf.table.WebTableUI.paint(WebTableUI.java:52)
    at javax.swing.plaf.ComponentUI.update(ComponentUI.java:143)
    at javax.swing.JComponent.paintComponent(JComponent.java:752)
    at javax.swing.JComponent.paint(JComponent.java:1029)
    at javax.swing.JComponent.paintChildren(JComponent.java:862)
    at javax.swing.JComponent.paint(JComponent.java:1038)
    at javax.swing.JViewport.paint(JViewport.java:747)
    at javax.swing.JComponent.paintToOffscreen(JComponent.java:5124)
    at javax.swing.BufferStrategyPaintManager.paint(BufferStrategyPaintManager.java:278)
    at javax.swing.RepaintManager.paint(RepaintManager.java:1225)
    at javax.swing.JComponent._paintImmediately(JComponent.java:5072)
    at javax.swing.JComponent.paintImmediately(JComponent.java:4882)
    at javax.swing.RepaintManager.paintDirtyRegions(RepaintManager.java:786)
    at javax.swing.RepaintManager.paintDirtyRegions(RepaintManager.java:714)
    at javax.swing.RepaintManager.prePaintDirtyRegions(RepaintManager.java:694)
    at javax.swing.RepaintManager.access$700(RepaintManager.java:41)
    at javax.swing.RepaintManager$ProcessingRunnable.run(RepaintManager.java:1636)
    at java.awt.event.InvocationEvent.dispatch(InvocationEvent.java:209)
    at java.awt.EventQueue.dispatchEventImpl(EventQueue.java:641)
    at java.awt.EventQueue.access$000(EventQueue.java:84)
    at java.awt.EventQueue$1.run(EventQueue.java:602)
    at java.awt.EventQueue$1.run(EventQueue.java:600)
    at java.security.AccessController.doPrivileged(Native Method)
    at java.security.AccessControlContext$1.doIntersectionPrivilege(AccessControlContext.java:87)
    at java.awt.EventQueue.dispatchEvent(EventQueue.java:611)
    at java.awt.EventDispatchThread.pumpOneEventForFilters(EventDispatchThread.java:269)
    at java.awt.EventDispatchThread.pumpEventsForFilter(EventDispatchThread.java:184)
    at java.awt.EventDispatchThread.pumpEventsForFilter(EventDispatchThread.java:178)
    at java.awt.Dialog$1.run(Dialog.java:1046)
    at …

在纯Hibernate我可以做:

Blob blob= Hibernate.getLobCreator(sessionFactory.getCurrentSession()).createBlob(inputStream, len);

如何在jpa中执行此操作(以hibernate作为提供程序)？

在纯hibernate中,我为blob创建了用户类型,它使用了setBinaryStream预处理语句.这个解决方案对我很有用,我正在寻找一种方法将它移植到JPA.

我有一个Picture：

public class Picture implements java.io.Serializable {

    private byte[] picEncoded;
    private String Name;
    //etc

是否可以移动byte[]到另一个类而不在 db 中创建物理分隔的表？我需要使用一些继承策略吗？

编辑

单独实体中的 Blob：

pojo :

 public class PictureBlob implements java.io.Serializable {
        private Integer pictureBlobId;
        private byte[] blob;

hbm: :

<class name="PictureBlob" table="PICTURE">

<id name="pictureBlobId" type="int">
  <column length="200" name="PictureID"/>      
</id>

<property name="blob" type="byte[]" insert="false" update="false">
  <column name="PicEncoded" not-null="false"/>
</property>
</class>

图片：

hbm: :

  <one-to-one class="PictureBlob" constrained="true" name="pictureBlob" fetch="select"/>

如何插入新图片？

PictureBlob pictureBlob= new PictureBlob();
        pictureBlob.setBlob(new byte[]{84,32,22});
        Picture p = new Picture(); …

我需要使用有限的内存使用来验证大xml .到目前为止我找到的每个代码都会出现内存错误.

方法我试过:

 //method 1
        SAXParserFactory factory = SAXParserFactory.newInstance();
        factory.setValidating(false);
        factory.setNamespaceAware(true);

        SchemaFactory schemaFactory = SchemaFactory.newInstance("http://www.w3.org/2001/XMLSchema");
        factory.setSchema(schemaFactory.newSchema(new Source[] {new StreamSource(Thread.currentThread().getContextClassLoader().getResource("xmlresource/XSD_final2.xsd").getFile())}));
        SAXParser parser = factory.newSAXParser();
        XMLReader reader = parser.getXMLReader();
        reader.setErrorHandler(new SimpleErrorHandler());
        reader.parse(new InputSource(inputXml));
//method2 

XMLValidationSchemaFactory sf = XMLValidationSchemaFactory.newInstance(XMLValidationSchema.SCHEMA_ID_W3C_SCHEMA);
            XMLValidationSchema vs = sf.createSchema(Thread.currentThread().getContextClassLoader().getResource("xmlresource/XSD_final2.xsd"));
            XMLStreamReader2 sr = (XMLStreamReader2) XMLInputFactory2.newInstance().createXMLStreamReader(new FileInputStream(inputXml));
            sr.validateAgainst(vs);
            try {
              while (sr.hasNext()) {
                sr.next();
              }
              System.out.println("Validated ok!");
            } catch (XMLValidationException ve) {
              System.err.println("Validation problem: "+ve);
              isValid = false;
            }
            sr.close();

//方法3

      SchemaFactory factory = SchemaFactory.newInstance("http://www.w3.org/2001/XMLSchema");
          String fileName = Thread.currentThread().getContextClassLoader().getResource("xmlresource/XSD_final2.xsd").getFile(); …

我有2类:Msg和Task是1对1.

如果我尝试保存Msg实例而不设置任务实例,我得到.

org.hibernate.PropertyValueException: not-null property references a null or transient value: entity3.Msg.task

如何在没有任务的情况下启用保存Msg？我在Msg的映射文件中有这个,但它没有帮助

<many-to-one class="entity3.Task" fetch="select" name="task" not-null="false">
  <column name="TaskID" not-null="true" unique="true"/>
</many-to-one>

提前致谢!

我有一种方法可以在玻璃窗格上绘制动画图像,以模拟图标的脉冲.它需要一个图像(我使用组件的图标)和边界(按钮fe).我需要这个区域:

我知道按钮有getBounds(),标签有类似的东西吗？或者也许是Icon的坐标.两者都不错.

我使用xjc创建了一些类.

    public class MyType {

    @XmlElementRefs({
        @XmlElementRef(name = "MyInnerType", type = JAXBElement.class, required = false),

    })
    @XmlMixed
    protected List<Serializable> content;

    public List<Serializable> getContent() {
        if (content == null) {
            content = new ArrayList<Serializable>();
        }
        return this.content;
    }
}

但我不能使用添加内部元素

getContent().add(newItem);

因为MyInnerType不是Serializable.为什么它不是对象列表？我如何添加内部元素？

我在为新实体生成id时遇到了麻烦,我尝试过:

@Id
@GeneratedValue
private Long myId;

和

@Id
@GeneratedValue(generator="increment")
@GenericGenerator(name="increment", strategy = "increment")
private Long myId;

但是entityManager.persist我得到了Table "SEQUENCE" not found 纯粹的冬眠generator class="increment"对我来说很好.

我在Table模型中有这个代码:

public class DocumentProjectTableModel extends AbstractTableModel{

    private List<MyDocument> myDocuments;
    public String getValueAt(int row, int column) {
            String toReturn = null;
            MyDocument myDocument = myDocuments.get(row);
            SimpleDateFormat format = new SimpleDateFormat("dd.MM.yyyy");

            switch (column) {
                case 0:
                    if(myDocument.getProject().getRegDate()!=null) toReturn = format.format(myDocument.getProject().getRegDate());
                    break;
                case 1:
                    toReturn = myDocument.getProject().getRegNum();
                    break;
                case 2:
                    toReturn = myDocument.getProject().getDescription();
                    break;
                case 3:
                    toReturn = myDocument.getProject().getShortName();
                    break;
                case 4:
                    toReturn = myDocument.getProject().getSecondName()+myDocument.getProject().getFirstName()+myDocument.getProject().getMiddleName();
                    break;

            }
            return toReturn;
        }
//  some other stuff is not shown

我想更改每一行的背景颜色,例如,如果myDocument.getIsRegistered() == true,我希望此行有黄色背景,如果myDocument.getIsValid …