Question

我正在开发通过Hibernate框架访问运行SQL Server 2012的数据库的应用程序。但是，我不知道如何使用序列来生成要添加的新记录的ID。每当我尝试将新的对象实例保存到数据库表时，都会出现异常。要保存的类如下：

package com.xantrix.webapp.entities;

import java.io.Serializable;
import java.util.Date;

import javax.persistence.Basic;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.SequenceGenerator;
import javax.persistence.Table;
import javax.persistence.TableGenerator;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;

@Entity
@Table(name = "Coupons")
public class Coupons implements Serializable
{
    private static final long serialVersionUID = -2788720560904709897L;

    @Id
    //@GeneratedValue(strategy=GenerationType.TABLE)
    /*
    @TableGenerator(
        name="Coupons",
        table="Progressivi",
        pkColumnName="Tipo",
        valueColumnName="Progressivo",
        allocationSize=200
        )
        */
    @SequenceGenerator(name="Coup_Gen", sequenceName="Test_Seq")
    @GeneratedValue(generator="Coup_Gen")
    //@GeneratedValue(generator="Coupons")
    @Column(name = "Id")
    private long id;

    @Temporal(TemporalType.TIME)
    @Column(name = "Data")
    private Date dataCreaz;

    @Basic
    private int idDeposito;

    @Basic
    private int qta;

    @Basic
    private double valore;

    @ManyToOne
    @JoinColumn(name = "IdCliente", referencedColumnName = "CODFIDELITY")
    private Clienti cliente;

    public Coupons()
    {

    }

    public long getId() {
        return id;
    }

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

    public Date getDataCreaz() {
        return dataCreaz;
    }

    public void setDataCreaz(Date dataCreaz) {
        this.dataCreaz = dataCreaz;
    }

    public int getIdDeposito() {
        return idDeposito;
    }

    public void setIdDeposito(int idDeposito) {
        this.idDeposito = idDeposito;
    }

    public int getQta() {
        return qta;
    }

    public void setQta(int qta) {
        this.qta = qta;
    }

    public double getValore() {
        return valore;
    }

    public void setValore(double valore) {
        this.valore = valore;
    }

    public Clienti getCliente() {
        return cliente;
    }

    public void setCliente(Clienti cliente) {
        this.cliente = cliente;
    }
}

如何更改Spring用来通过Spring MVC 5 Web App查询Microsoft SQL Server序列的SQL查询？

由于错误，我无法将所有Microsoft SQL Server序列与Spring MVC 5 Web App的Hibernate一起使用

Invalid object name <sequence>.

该问题影响Microsoft SQL Server 2012（但也影响Microsoft SQL Server 2018）。

我在Spring Tool Suite 4中将Spring MVC 5框架与Hibernate一起使用。该服务器是Apache Tomcat 9.0.19。

我创建了一个名为

的用户

WebClient1

在SQL Server Management Studio（Microsoft SQL Server 2012）中。

我创建了以下序列：

USE [AlphaShop]
GO

USE [AlphaShop]
GO

/****** Object:  Sequence [dbo].[Test_Seq]    Script Date: 09/07/2019 10:56:28 ******/
CREATE SEQUENCE [dbo].[Test_Seq] 
 AS [bigint]
 START WITH 500
 INCREMENT BY 50
 MINVALUE 500
 MAXVALUE 9223372036854775807
 CACHE 
GO

以下SQL查询在Microsoft SQL Server Management Studio中成功运行：

SELECT NEXT VALUE FOR dbo.Test_Seq;

但是，如果我尝试访问以下Web应用程序URL，则会出现HTTP Status 500错误： http://localhost:8080/alphashop/coupons/aggiungi/67000023 其中文本“ 67000023”是现有用户的ID，如以下SQL查询的结果所示：

select * from CLIENTI
where CODFIDELITY = 67000023

在Spring Tool Suite 4的控制台中，我可以看到最新调用的SQL查询：

Hibernate: 
    select
        next_val as id_val 
    from
        Test_Seq with (updlock,
        rowlock)
lug 10, 2019 12:26:38 PM org.hibernate.id.enhanced.TableStructure$1$1 execute
ERROR: could not read a hi value
com.microsoft.sqlserver.jdbc.SQLServerException: Il nome di oggetto 'Test_Seq' non è valido.
    at com.microsoft.sqlserver.jdbc.SQLServerException.makeFromDatabaseError(SQLServerException.java:256)
    at com.microsoft.sqlserver.jdbc.SQLServerStatement.getNextResult(SQLServerStatement.java:1621)

lug 10, 2019 12:26:38 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
WARN: SQL Error: 208, SQLState: S0002
lug 10, 2019 12:26:38 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
ERROR: Il nome di oggetto 'Test_Seq' non è valido.
lug 10, 2019 12:26:38 PM org.apache.catalina.core.StandardWrapperValve invoke
GRAVE: Servlet.service() for servlet [dispatcher] in context with path [/alphashop] threw exception [Request processing failed; nested exception is javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: error performing isolated work] with root cause
com.microsoft.sqlserver.jdbc.SQLServerException: Il nome di oggetto 'Test_Seq' non è valido.
    at com.microsoft.sqlserver.jdbc.SQLServerException.makeFromDatabaseError(SQLServerException.java:256)
    at com.microsoft.sqlserver.jdbc.SQLServerStatement.getNextResult(SQLServerStatement.java:1621)

如您所见，Hibernate生成错误的SQL查询

    select
        next_val as id_val 
    from
        Test_Seq with (updlock,
        rowlock)

代替：

SELECT NEXT VALUE FOR dbo.Test_Seq;

成功。

通过Web浏览器Mozilla Firefox 68.0在Apache Tomcat 9的HTML页面中呈现的HTTP Status 500错误为：

Type Exception Report
Message Request processing failed; nested exception is javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: error performing isolated work
Description The server encountered an unexpected condition that prevented it from fulfilling the request.
Exception
org.springframework.web.util.NestedServletException: Request processing failed; nested exception is javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: error performing isolated work
    org.springframework.web.servlet.FrameworkServlet.processRequest(FrameworkServlet.java:1013)
Root Cause
javax.persistence.PersistenceException: org.hibernate.exception.SQLGrammarException: error performing isolated work
    org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:154)
    org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:181)
Root Cause
org.hibernate.exception.SQLGrammarException: error performing isolated work
    org.hibernate.exception.internal.SQLStateConversionDelegate.convert(SQLStateConversionDelegate.java:106)
    org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:42)
Root Cause
com.microsoft.sqlserver.jdbc.SQLServerException: Il nome di oggetto 'Test_Seq' non è valido.
    com.microsoft.sqlserver.jdbc.SQLServerException.makeFromDatabaseError(SQLServerException.java:256)
    com.microsoft.sqlserver.jdbc.SQLServerStatement.getNextResult(SQLServerStatement.java:1621)

在对象浏览器中，我可以看到序列“ dbo.Test_Seq”。该数据库的名称是AlphaShop。

我尝试执行以下SQL命令，但不能解决问题：

ALTER LOGIN WebClient1 WITH DEFAULT_DATABASE = [AlphaShop];

用户的身份验证类型为“ SQL Server身份验证”。用户具有“公共”角色。

我无法弄清楚如何使用序列来生成要添加的新记录的ID；每当我尝试将新的对象实例保存到数据库表时，都会出现异常。

请帮助。提前谢谢了。操作系统：Microsoft Windows 10 64bit版本2019.3

Answer 1

我找到了一个解决方案：您需要在“ application.properties”文件中更改Hibernate属性以指定正确的方言（在我的情况下，正确的方言是“ SQLServer2012Dialect”）：

#Hibernate properties
hibernate.dialect = org.hibernate.dialect.SQLServer2012Dialect
hibernate.show_sql = true
hibernate.format_sql = true
hibernate.hbm2ddl.auto = validate

代替错误

#Hibernate properties
hibernate.dialect =  org.hibernate.dialect.SQLServerDialect
hibernate.show_sql = true
hibernate.format_sql = true
hibernate.hbm2ddl.auto = validate

SQL Server 2012/2018-无效的对象名称

1 个答案: