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我在 MySQL 中创建了一个数据库，并在我的 servlet 中使用 JPA 访问数据库。这些是细节

  1. 实体名称 -> 注册用户

  2. 字段 id ，类型 -> 整数

所以根据我的查询，我试图找到 id 为 1001 的记录。 

EntityManager em = HibernateUtil.getInstance().getEntityManager();
        Query q = em
                .createQuery("SELECT record FROM RegisteredUser record WHERE record.id = 1001");
        RegisteredUser r = (RegisteredUser) q.getSingleResult();

但是在这样做时，我收到以下错误！

 javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '2 registered0_.id as id0_, registered0_.current_status as current2_0_, registere' at line 1
        at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1360)
        at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1288)
        at org.hibernate.ejb.QueryImpl.getSingleResult(QueryImpl.java:313)
        at com.aces.servlets.UserStatusServlet.getStatus(UserStatusServlet.java:193)
        at com.aces.servlets.UserStatusServlet.access$0(UserStatusServlet.java:188)
        at com.aces.servlets.UserStatusServlet$1.onComplete(UserStatusServlet.java:50)
        at org.apache.catalina.core.AsyncListenerWrapper.fireOnComplete(AsyncListenerWrapper.java:40)
        at org.apache.catalina.core.AsyncContextImpl.fireOnComplete(AsyncContextImpl.java:119)
        at org.apache.coyote.AsyncStateMachine.asyncPostProcess(AsyncStateMachine.java:190)
        at org.apache.coyote.AbstractProcessor.asyncPostProcess(AbstractProcessor.java:116)
        at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:593)
        at org.apache.tomcat.util.net.JIoEndpoint$SocketProcessor.run(JIoEndpoint.java:312)
        at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
        at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
        at java.lang.Thread.run(Unknown Source)
    Caused by: org.hibernate.exception.SQLGrammarException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '2 registered0_.id as id0_, registered0_.current_status as current2_0_, registere' at line 1
        at org.hibernate.exception.internal.SQLExceptionTypeDelegate.convert(SQLExceptionTypeDelegate.java:83)
        at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:49)
        at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:125)
        at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:110)
        at org.hibernate.engine.jdbc.internal.proxy.AbstractStatementProxyHandler.continueInvocation(AbstractStatementProxyHandler.java:129)
        at org.hibernate.engine.jdbc.internal.proxy.AbstractProxyHandler.invoke(AbstractProxyHandler.java:81)
        at com.sun.proxy.$Proxy54.executeQuery(Unknown Source)
        at org.hibernate.loader.Loader.getResultSet(Loader.java:1962)
        at org.hibernate.loader.Loader.doQuery(Loader.java:829)
        at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:289)
        at org.hibernate.loader.Loader.doList(Loader.java:2447)
        at org.hibernate.loader.Loader.doList(Loader.java:2433)
        at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2263)
        at org.hibernate.loader.Loader.list(Loader.java:2258)
        at org.hibernate.loader.hql.QueryLoader.list(QueryLoader.java:470)
        at org.hibernate.hql.internal.ast.QueryTranslatorImpl.list(QueryTranslatorImpl.java:355)
        at org.hibernate.engine.query.spi.HQLQueryPlan.performList(HQLQueryPlan.java:196)
        at org.hibernate.internal.SessionImpl.list(SessionImpl.java:1161)
        at org.hibernate.internal.QueryImpl.list(QueryImpl.java:101)
        at org.hibernate.ejb.QueryImpl.getSingleResult(QueryImpl.java:280)
        ... 12 more
    Caused by: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '2 registered0_.id as id0_, registered0_.current_status as current2_0_, registere' at line 1
        at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
        at sun.reflect.NativeConstructorAccessorImpl.newInstance(Unknown Source)
        at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(Unknown Source)
        at java.lang.reflect.Constructor.newInstance(Unknown Source)
        at com.mysql.jdbc.Util.handleNewInstance(Util.java:411)
        at com.mysql.jdbc.Util.getInstance(Util.java:386)
        at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:1052)
        at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:3609)
        at com.mysql.jdbc.MysqlIO.checkErrorPacket(MysqlIO.java:3541)
        at com.mysql.jdbc.MysqlIO.sendCommand(MysqlIO.java:2002)
        at com.mysql.jdbc.MysqlIO.sqlQueryDirect(MysqlIO.java:2163)
        at com.mysql.jdbc.ConnectionImpl.execSQL(ConnectionImpl.java:2624)
        at com.mysql.jdbc.PreparedStatement.executeInternal(PreparedStatement.java:2127)
        at com.mysql.jdbc.PreparedStatement.executeQuery(PreparedStatement.java:2293)
        at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
        at sun.reflect.NativeMethodAccessorImpl.invoke(Unknown Source)
        at sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source)
        at java.lang.reflect.Method.invoke(Unknown Source)
        at org.hibernate.engine.jdbc.internal.proxy.AbstractStatementProxyHandler.continueInvocation(AbstractStatementProxyHandler.java:122)
        ... 27 more

我的实体类（它是由 eclipse 生成的）

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

@Entity
@Table(name="registered_users")
@NamedQuery(name="RegisteredUser.findAll", query="SELECT r FROM RegisteredUser r")
public class RegisteredUser implements Serializable {
    private static final long serialVersionUID = 1L;

    @Id
    private int id;

    @Column(name="current_status")
    private byte currentStatus;

    private String password;

    private String username;

    public RegisteredUser() {
    }

    public int getId() {
        return this.id;
    }

    public void setId(int id) {
        this.id = id;
    }

    public byte getCurrentStatus() {
        return this.currentStatus;
    }

    public void setCurrentStatus(byte currentStatus) {
        this.currentStatus = currentStatus;
    }

    public String getPassword() {
        return this.password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    public String getUsername() {
        return this.username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

}

提前致谢 ：）

4

2 回答 2

1

错误在您的查询中：

SELECT record FROM RegisteredUser record WHERE record.id LIKE 1001;

您混淆了record用于两种不同事物的 sql 解释器。第一个用于列，第二个用于获取的行。

尝试这个：

SELECT * FROM RegisteredUser record WHERE record.id LIKE 1001;

我也相信LIKE带有字符串的关键字词，我不确定您的 record.id 是整数还是 varchar。

如果您分享有关您的表格的更多详细信息以及您需要获取的确切内容，我可以提供更好的输入。

于 2013-08-18T06:46:17.603 回答
0

为了获得一个对象，您应该使用 JPA object(...)语法。尝试这个：

SELECT object(record) FROM RegisteredUser as record WHERE record.id = 1001
于 2013-08-18T07:38:41.247 回答