InvalidDataAccessResourceUsageException：org.hibernate.exception.SQLGrammarException：

Question

我有以下問題我已經卡了一段時間：

我收到此錯誤： org.springframework.dao.InvalidDataAccessResourceUsageException：無法准備語句； SQL [調用休眠序列的下一個值]； 嵌套異常是 org.hibernate.exception.SQLGrammarException：無法准備語句

我發現有這個錯誤的人會得到它，因為他們使用保留字作為表名，但我認為這不是我的問題。

我的兩個 model 類如下。 我正在跳過 getter/setter 和 constructors

@Entity
@Table(name = "GATEWAY_MODEL")
public class GetewayModel implements Serializable {

    private static final long serialVersionUID = 1L;
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;
    @Column(name = "serial_number", nullable = false, length = 12, updatable = true)
    private String serialNumber;
    @Column(name = "name", nullable = false, length = 12, updatable = true)
    private String name;
    @Column(name = "ipFour", nullable = false, length = 12, updatable = true)
    private String ipFour;
    @Column(name = "peripheral_devices", updatable = true)
    @OneToMany(cascade = CascadeType.ALL, mappedBy = "gateway")
    private Set<PeripheralDevicesModel> peripheralDevices = new HashSet<PeripheralDevicesModel>();


@Entity
@Table(name = "PERIPHERAL_DIVICES_MODEL")
public class PeripheralDevicesModel {
    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;
    @Column(name = "uID", nullable = false)
    private String uID;
    @Column(name = "vendor", nullable = false)
    private String vendor;
    @Column(name = "date_created", nullable = false)
    private Date dateCreated;
    @Enumerated(EnumType.STRING)
    @Column(name = "status")
    private Status status;
    @JsonIgnore
    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "gateway")
    private GetewayModel gateway;

然后在主要的 class 中，我嘗試輸入一些數據，如下所示：

@Bean
CommandLineRunner initDatabase(GatewayRepository gatewayRepo, PeripheralDevicesRepository devicesRepo) {

    Set<PeripheralDevicesModel> devicesSet = new HashSet<>();

    GetewayModel gateway = new GetewayModel();
    gateway.setId(123l);
    gateway.setSerialNumber("1123");
    gateway.setName("gateway");
    gateway.setIpFour("1.160.10.240");
    
    PeripheralDevicesModel devices = new PeripheralDevicesModel();
    devices.setId(1234l);
    devices.setuID("2");
    devices.setDateCreated(new Date());
    devices.setGateway(gateway);
    devices.setStatus(Status.OFFLINE);
    devices.setVendor("vendor");
    devicesSet.add(devices);
    
    gateway.setPeripheralDevices(devicesSet);
    return args -> {
        gatewayRepo.save(gateway);
        devicesRepo.save(devices);
    };

我猜這是一些問題，因為我的 model 數據中的 OneToMany 關系。

我從堆棧跟蹤中獲得了更多信息

調用 hibernate_sequence 的下一個值 2020-06-26 08:34:53 - SQL 錯誤：90036，SQLState：90036 2020-06-26 08:34:53 - 找不到序列“HIBERNATE_SEQUENCE”； SQL 語句：調用 hibernate_sequence [90036-200] 的下一個值 2020-06-26 08:34:53 -

您是否知道如何解決此問題或為什么它不起作用。 謝謝

配置.屬性：

spring.datasource.url=jdbc:h2:mem:testdb;DB_CLOSE_DELAY=-1
spring.datasource.driverClassName=org.h2.Driver
spring.datasource.username=sa
spring.datasource.password=
spring.jpa.database-platform=org.hibernate.dialect.H2Dialect

# Enabling H2 Console
spring.h2.console.enabled=true

# Custom H2 Console URL
spring.h2.console.path=/h2

spring.jpa.hibernate.ddl-auto=none
 
#Turn Statistics on and log SQL stmts
spring.jpa.show-sql=true
spring.jpa.properties.hibernate.format_sql=true
spring.jpa.properties.hibernate.generate_statistics=false
#logging.level.org.hibernate.type=trace
#logging.level.org.hibernate.stat=debug
 
logging.pattern.console=%d{yyyy-MM-dd HH:mm:ss} - %msg%n

更新：在生成 GenerationType.IDENTITY 之后，我現在收到以下錯誤：

Hibernate: 
    select
        getewaymod0_.id as id1_0_1_,
        getewaymod0_.ip_four as ip_four2_0_1_,
        getewaymod0_.name as name3_0_1_,
        getewaymod0_.serial_number as serial_n4_0_1_,
        peripheral1_.gateway as gateway5_1_3_,
        peripheral1_.id as id1_1_3_,
        peripheral1_.id as id1_1_0_,
        peripheral1_.date_created as date_cre2_1_0_,
        peripheral1_.gateway as gateway5_1_0_,
        peripheral1_.uid as uid3_1_0_,
        peripheral1_.vendor as vendor4_1_0_ 
    from
        gateway_model getewaymod0_ 
    left outer join
        peripheral_divices_model peripheral1_ 
            on getewaymod0_.id=peripheral1_.gateway 
    where
        getewaymod0_.id=?
2020-06-26 16:42:04 - SQL Error: 42102, SQLState: 42S02
2020-06-26 16:42:04 - Table "GATEWAY_MODEL" not found; SQL statement:
select getewaymod0_.id as id1_0_1_, getewaymod0_.ip_four as ip_four2_0_1_, getewaymod0_.name as name3_0_1_, getewaymod0_.serial_number as serial_n4_0_1_, peripheral1_.gateway as gateway5_1_3_, peripheral1_.id as id1_1_3_, peripheral1_.id as id1_1_0_, peripheral1_.date_created as date_cre2_1_0_, peripheral1_.gateway as gateway5_1_0_, peripheral1_.uid as uid3_1_0_, peripheral1_.vendor as vendor4_1_0_ from gateway_model getewaymod0_ left outer join peripheral_divices_model peripheral1_ on getewaymod0_.id=peripheral1_.gateway where getewaymod0_.id=? [42102-200]
2020-06-26 16:42:04 - HHH000327: Error performing load command
org.hibernate.exception.SQLGrammarException: could not prepare statement

Answer 1

我更改了 2 件事（除了代碼中的一些語法/錯字錯誤）：

1. 添加cascade=CascadeType.ALL如下：

    @JsonIgnore @ManyToOne(cascade=CascadeType.ALL, fetch = FetchType.EAGER) @JoinColumn(name = "gateway") private GatewayModel gateway;

2. 您不能添加具有nullable = false列的實體：
   devices.setGateway(新 GetewayModel());

   devices.setGateway(gateway);

否則，它在H2上工作正常。

更新：
要查找的語法/錯字錯誤：

• @Table(name = "PERIPHERAL_DIVICES_MODEL")需要是@Table(name = "PERIPHERAL_DEVICES_MODEL")
• public class GetewayModel需要是public class GatewayModel
• private GetewayModel gateway; 需要是private GatewayModel gateway;