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SQL Server データベースからデータを取得したいので、jdbc ドライバーと eclipselink orm フレームワークを使用します。問題があります。これが私の例です。

public static void main(String[] args)
{
        EntityManager entityManager =  
                Persistence.createEntityManagerFactory("HRPSPU").createEntityManager();

        List<Personnel> personnelList = entityManager.createQuery("Select p FROM Personnel p",Personnel.class).getResultList();

        for (Personnel personnel : personnelList)
        {
                System.out.println(personnel.getName());
        }
    }

とエラー、

SLF4J: Class path contains multiple SLF4J bindings.
SLF4J: Found binding in [jar:file:/E:/tutorial/glassfish3/glassfish/modules/bean-validator.jar!/org/slf4j/impl/StaticLoggerBinder.class]
SLF4J: Found binding in [jar:file:/E:/tutorial/glassfish3/glassfish/modules/weld-osgi-bundle.jar!/org/slf4j/impl/StaticLoggerBinder.class]
SLF4J: See http://www.slf4j.org/codes.html#multiple_bindings for an explanation.
Tem 31, 2012 11:44:51 PM org.hibernate.validator.util.Version <clinit>
INFO: Hibernate Validator 4.2.0.Final
[EL Info]: 2012-07-31 23:44:51.929--ServerSession(451764492)--EclipseLink, version: Eclipse Persistence Services - 2.4.0.v20120608-r11652
[EL Info]: connection: 2012-07-31 23:44:52.951--ServerSession(451764492)--file:/E:/tutorial/workspace/HRPS/build/classes/_HRPSPU login successful
[EL Warning]: 2012-07-31 23:44:53.927--UnitOfWork(1400947054)--Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.4.0.v20120608-r11652): org.eclipse.persistence.exceptions.DatabaseException
**Internal Exception: com.microsoft.sqlserver.jdbc.SQLServerException: Invalid object name 'Personnel'.**
Error Code: 208
Call: SELECT IdentificationNumber, Age, Name, Surname FROM Personnel
Query: ReadAllQuery(referenceClass=Personnel sql="SELECT IdentificationNumber, Age, Name, Surname FROM Personnel")
Exception in thread "main" Local Exception Stack: 
Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.4.0.v20120608-r11652): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: com.microsoft.sqlserver.jdbc.SQLServerException: Invalid object name 'Personnel'.
Error Code: 208
Call: SELECT IdentificationNumber, Age, Name, Surname FROM Personnel
Query: ReadAllQuery(referenceClass=Personnel sql="SELECT IdentificationNumber, Age, Name, Surname FROM Personnel")
    at org.eclipse.persistence.exceptions.DatabaseException.sqlException(DatabaseException.java:333)
    at org.eclipse.persistence.internal.databaseaccess.DatabaseAccessor.basicExecuteCall(DatabaseAccessor.java:646)
    at org.eclipse.persistence.internal.databaseaccess.DatabaseAccessor.executeCall(DatabaseAccessor.java:537)
    at org.eclipse.persistence.internal.sessions.AbstractSession.basicExecuteCall(AbstractSession.java:1800)
    at org.eclipse.persistence.sessions.server.ServerSession.executeCall(ServerSession.java:566)
    at org.eclipse.persistence.internal.queries.DatasourceCallQueryMechanism.executeCall(DatasourceCallQueryMechanism.java:207)
    at org.eclipse.persistence.internal.queries.DatasourceCallQueryMechanism.executeCall(DatasourceCallQueryMechanism.java:193)
    at org.eclipse.persistence.internal.queries.DatasourceCallQueryMechanism.executeSelectCall(DatasourceCallQueryMechanism.java:264)
    at org.eclipse.persistence.internal.queries.DatasourceCallQueryMechanism.selectAllRows(DatasourceCallQueryMechanism.java:648)
    at org.eclipse.persistence.internal.queries.ExpressionQueryMechanism.selectAllRowsFromTable(ExpressionQueryMechanism.java:2681)
    at org.eclipse.persistence.internal.queries.ExpressionQueryMechanism.selectAllRows(ExpressionQueryMechanism.java:2634)
    at org.eclipse.persistence.queries.ReadAllQuery.executeObjectLevelReadQuery(ReadAllQuery.java:420)
    at org.eclipse.persistence.queries.ObjectLevelReadQuery.executeDatabaseQuery(ObjectLevelReadQuery.java:1149)
    at org.eclipse.persistence.queries.DatabaseQuery.execute(DatabaseQuery.java:852)
    at org.eclipse.persistence.queries.ObjectLevelReadQuery.execute(ObjectLevelReadQuery.java:1108)
    at org.eclipse.persistence.queries.ReadAllQuery.execute(ReadAllQuery.java:392)
    at org.eclipse.persistence.queries.ObjectLevelReadQuery.executeInUnitOfWork(ObjectLevelReadQuery.java:1196)
    at org.eclipse.persistence.internal.sessions.UnitOfWorkImpl.internalExecuteQuery(UnitOfWorkImpl.java:2875)
    at org.eclipse.persistence.internal.sessions.AbstractSession.executeQuery(AbstractSession.java:1602)
    at org.eclipse.persistence.internal.sessions.AbstractSession.executeQuery(AbstractSession.java:1584)
    at org.eclipse.persistence.internal.sessions.AbstractSession.executeQuery(AbstractSession.java:1549)
    at org.eclipse.persistence.internal.jpa.QueryImpl.executeReadQuery(QueryImpl.java:231)
    at org.eclipse.persistence.internal.jpa.QueryImpl.getResultList(QueryImpl.java:411)
    at entity.Main.main(Main.java:18)
Caused by: com.microsoft.sqlserver.jdbc.SQLServerException: Invalid object name 'Personnel'.
    at com.microsoft.sqlserver.jdbc.SQLServerException.makeFromDatabaseError(SQLServerException.java:216)
    at com.microsoft.sqlserver.jdbc.SQLServerStatement.getNextResult(SQLServerStatement.java:1515)
    at com.microsoft.sqlserver.jdbc.SQLServerPreparedStatement.doExecutePreparedStatement(SQLServerPreparedStatement.java:404)
    at com.microsoft.sqlserver.jdbc.SQLServerPreparedStatement$PrepStmtExecCmd.doExecute(SQLServerPreparedStatement.java:350)
    at com.microsoft.sqlserver.jdbc.TDSCommand.execute(IOBuffer.java:5696)
    at com.microsoft.sqlserver.jdbc.SQLServerConnection.executeCommand(SQLServerConnection.java:1715)
    at com.microsoft.sqlserver.jdbc.SQLServerStatement.executeCommand(SQLServerStatement.java:180)
    at com.microsoft.sqlserver.jdbc.SQLServerStatement.executeStatement(SQLServerStatement.java:155)
    at com.microsoft.sqlserver.jdbc.SQLServerPreparedStatement.executeQuery(SQLServerPreparedStatement.java:285)
    at org.eclipse.persistence.internal.databaseaccess.DatabaseAccessor.executeSelect(DatabaseAccessor.java:938)
    at org.eclipse.persistence.internal.databaseaccess.DatabaseAccessor.basicExecuteCall(DatabaseAccessor.java:609)
    ... 22 more

これは私のpersistance.xmlです。integratedsecurity=trueに接続したので、ユーザー名とパスワードがありませんでした...

    <?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="HRPSPU" transaction-type="RESOURCE_LOCAL">
             <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
            <class>entity.Personnel</class>
         <properties>
          <property name="javax.persistence.jdbc.url" value="jdbc:sqlserver://localhost:1433;databaseName=Arge;integratedSecurity=true;"/>
          <property name="javax.persistence.jdbc.driver" value="com.microsoft.sqlserver.jdbc.SQLServerDriver"/>
          <property name="javax.persistence.jdbc.password" value=""/>
          <property name="javax.persistence.jdbc.user" value=""/>
         </properties>
        </persistence-unit>
    </persistence>

私は Eclipse Juno IDE と Glassfish アプリケーション サーバーを使用しています。

4

3 に答える 3

1

問題を解決しました。データベースに独自のスキーマを作成すると、テーブル アノテーションにエンティティ クラスがインポートされます。例えば：

 @Table(name="Example",schema="Example",catalog="Example")
于 2012-08-01T18:39:40.320 に答える
0

現在のように見えるのは、データベースにそのようなテーブルがないということです。Eclipseリンクはマッピングを確認し、SQLを発行しますが、DBでは失敗します。SQL Serverで別の方法で呼び出されている場合は、適切なスキーマに接続していないか、@Table注釈が必要です。Personnel

@Entity
@Table(name="...")
public class Personnel{   }

エンティティとテーブルの定義を表示すると役立ちます。

于 2012-08-01T08:06:51.047 に答える
0

私は1つのデータベースとスキーマを持っています.私のテーブルはこのスキーマの中にあります.私はlocalhostに接続します.

パッケージ エンティティ。

import java.io.Serializable;
import javax.persistence.*;


/**
 * The persistent class for the Personnel database table.
 * 
 */
@Entity
@Table(name="Personnel")
public class Personnel implements Serializable {
    private static final long serialVersionUID = 1L;

    @Column(name="Age")
    private short age;

    @Column(name="IdentificationNumber")
    @GeneratedValue
    @Id
    private long identificationNumber;

    @Column(name="Name")
    private String name;

    @Column(name="Surname")
    private String surname;

    public Personnel() {
    }

    public short getAge() {
        return this.age;
    }

    public void setAge(short age) {
        this.age = age;
    }

    public long getIdentificationNumber() {
        return this.identificationNumber;
    }

    public void setIdentificationNumber(long identificationNumber) {
        this.identificationNumber = identificationNumber;
    }

    public String getName() {
        return this.name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getSurname() {
        return this.surname;
    }

    public void setSurname(String surname) {
        this.surname = surname;
    }
}
于 2012-08-01T18:31:07.037 に答える