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将记录与postgresql中的先前记录进行比较

我在PostgreSQL DB中有一个表,如下所示:

 Client | Rate | StartDate|EndDate     
 A      | 1000 | 2005-1-1 |2005-12-31
 A      | 2000 | 2006-1-1 |2006-12-31
 A      | 3000 | 2007-1-1 |2007-12-31  
 B      | 5000 | 2006-1-1 |2006-12-31  
 B      | 8000 | 2008-1-1 |2008-12-31  
 C      | 2000 | 2006-1-1 |2006-12-31  
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如何得到这个结果?

 Client | Rate | StartDate|EndDate    |Pre Rate | Pre StartDate |Pre EndDate    
 A      | 1000 | 2005-1-1 |2005-12-31 |         |               |             
 A      | 2000 | 2006-1-1 |2006-12-31 | 1000    | 2005-1-1      |2005-12-31            
 A      | 3000 | …
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postgresql comparison max next seconds

9
推荐指数
1
解决办法
6675
查看次数

Eclipse Juno加载描述符错误

我刚刚将我的Eclipse从Galileo更新为Juno.我创建了一个新的工作区并导入了一个基于Java Spring MVC的项目.当我打开它时,它开始"加载描述符",然后弹出一条错误消息:在"为mta_pfm加载描述符"期间发生内部错误.如果我使用Galileo打开它,则没有错误.我试图在Juno中禁用XML文件的验证,但它不起作用.我现在不知道如何修复它.请帮忙.谢谢

 org.eclipse.emf.ecore.resource.impl.ResourceSetImpl$1DiagnosticWrappedException: org.eclipse.emf.ecore.xmi.IllegalValueException: Value '
    Container
  ' is not legal. (platform:/resource/mta_pfm/WebContent/WEB-INF/web.xml, 138, 14)
    at org.eclipse.emf.ecore.resource.impl.ResourceSetImpl.handleDemandLoadException(ResourceSetImpl.java:319)
    at org.eclipse.emf.ecore.resource.impl.ResourceSetImpl.demandLoadHelper(ResourceSetImpl.java:278)
    at org.eclipse.jem.internal.util.emf.workbench.ProjectResourceSetImpl.getResource(ProjectResourceSetImpl.java:1064)
    at org.eclipse.jst.jee.model.internal.JEE5ModelProvider.getModelResource(JEE5ModelProvider.java:169)
    at org.eclipse.jst.jee.model.internal.Web25ModelProvider.getModelObject(Web25ModelProvider.java:44)
    at org.eclipse.jst.jee.model.internal.JEE5ModelProvider.getModelObject(JEE5ModelProvider.java:215)
    at org.eclipse.jst.jee.model.internal.common.AbstractMergedModelProvider.loadProviders(AbstractMergedModelProvider.java:261)
    at org.eclipse.jst.jee.model.internal.common.AbstractMergedModelProvider.access$2(AbstractMergedModelProvider.java:255)
    at org.eclipse.jst.jee.model.internal.common.AbstractMergedModelProvider$LoadModelsWorkspaceRunnable.run(AbstractMergedModelProvider.java:278)
    at org.eclipse.core.internal.resources.Workspace.run(Workspace.java:2344)
    at org.eclipse.jst.jee.model.internal.common.AbstractMergedModelProvider.loadModel(AbstractMergedModelProvider.java:249)
    at org.eclipse.jst.jee.model.internal.common.AbstractMergedModelProvider.getMergedModel(AbstractMergedModelProvider.java:219)
    at org.eclipse.jst.jee.model.internal.common.AbstractMergedModelProvider.getModelObject(AbstractMergedModelProvider.java:139)
    at org.eclipse.jst.jee.ui.internal.navigator.Web25ContentProvider.getNewContentProviderInstance(Web25ContentProvider.java:69)
    at org.eclipse.jst.jee.ui.internal.navigator.LoadingJeeDDJob.run(LoadingJeeDDJob.java:50)
    at org.eclipse.core.internal.jobs.Worker.run(Worker.java:53)
Caused by: org.eclipse.emf.ecore.xmi.IllegalValueException: Value '
    Container
  ' is not legal. (platform:/resource/mta_pfm/WebContent/WEB-INF/web.xml, 138, 14)
    at org.eclipse.emf.ecore.xmi.impl.XMLHandler.setFeatureValue(XMLHandler.java:2657)
    at org.eclipse.emf.ecore.xmi.impl.XMLHandler.setFeatureValue(XMLHandler.java:2642)
    at org.eclipse.emf.ecore.xmi.impl.XMLHandler.endElement(XMLHandler.java:1564)
    at com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.endElement(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl.scanEndElement(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.XMLDocumentFragmentScannerImpl$FragmentContentDriver.next(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.XMLDocumentScannerImpl.next(Unknown Source)
    at com.sun.org.apache.xerces.internal.impl.XMLNSDocumentScannerImpl.next(Unknown …
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eclipse descriptor

9
推荐指数
1
解决办法
7160
查看次数

Spring @Async生成LazyInitializationExceptions

Spring MVC应用程序需要执行一个需要几分钟的密集计算工作.客户端希望以异步方式运行此作业.但是在我对方法启用@Async注释后(参见代码1)并获得错误(参见代码2).我的web.xml和web-appliaiton-context.xml也在下面给出.我试图找到解决方法来解决这个问题,但失败了.请帮忙.

代码1:

@Service
public class AsyncServiceImpl implements AsyncServiceInt{

    protected int pertArrSize = 1000;
    @Autowired
    protected TblvDao tblvDao1 = null;

    @Override
    @Async
    public void startSlowProcess(Integer scenarioId) throws SecurityException, IllegalArgumentException, IOException, ClassNotFoundException, NoSuchFieldException, IllegalAccessException {
        batchUpdateSummaryPertSim(scenarioId);

    }

    public void batchUpdateSummaryPertSim(Integer scenarioId) throws SecurityException, IllegalArgumentException, IOException, ClassNotFoundException, NoSuchFieldException, IllegalAccessException
    {
        double[] summaryArry = new double[pertArrSize];
        summaryArry = summaryPertSim_env(scenarioId, summaryArry);
    }

    @Transactional
    private double[] summaryPertSim_env(Integer  scenarioId,
    double[] summaryArry) throws IOException, ClassNotFoundException, SecurityException, IllegalArgumentException, NoSuchFieldException, IllegalAccessException {
        ScenarioTblv scenario = tblvDao1.getScenarioTblv(scenarioId);
        TblvResultSaved savedResult = …
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spring asynchronous lazy-initialization

1
推荐指数
2
解决办法
4832
查看次数