PSQLException：列索引超出范圍：7，列數：6 - Postgres初學者

Question

這是我正在構建用於示例網站的數據庫。

我正在使用游戲1.3

這是結構：

@Entity
public class BlogUser extends BlogModel {

    @Id
    @Column(nullable = false)
    @GeneratedValue(strategy = GenerationType.TABLE)
    public Long id;

    public String firstName;

    public String lastName;

    @Column(nullable = false)
    public String username;

    @Column(nullable = false)
    public String password;

    @Column(nullable = false)
    public String email;

    public BlogUser(String fName, String lName, String uName, String pWord, String email1) {
        firstName = fName;
        lastName = lName;
        username = uName;
        password = pWord;
        email = email1;
    }
}

這是來自控制器的命令

public static void registerSubmit(String firstName, String lastName, String username, String password, String email) throws FileNotFoundException {

    BlogUser user = new BlogUser(firstName, lastName, username, password, email);

    //persists the entity in the database
    try {
        user.save();
    }
    catch (PersistenceException e) {
        File file = new File("testsalotofthem.txt");
        PrintStream ps = new PrintStream(file);
        e.printStackTrace(ps);
    }
}

有些東西有手工完成的屬性，有些則沒有，只是我試圖了解所有可以設置的方法（允許的是什么，你需要手動做什么，可以用哪種方式處理）。

我試圖在數據庫中保留用戶。 它包含所有數據，但由於某種原因，它說我試圖使用7列而不是6列。

數據來自網站，該命令是信息進入用戶的地方。

任何解決這個問題的技巧都會有所幫助，因為我很難過。

這是堆棧跟蹤：

javax.persistence.PersistenceException: org.hibernate.exception.DataException: could not insert: [models.BlogUser]
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1387)
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1310)
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1316)
at org.hibernate.ejb.AbstractEntityManagerImpl.flush(AbstractEntityManagerImpl.java:999)
at play.db.jpa.JPABase._save(JPABase.java:45)
at play.db.jpa.GenericModel.save(GenericModel.java:232)
at controllers.Application.registerSubmit(Application.java:50)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:497)
at play.mvc.ActionInvoker.invokeWithContinuation(ActionInvoker.java:524)
at play.mvc.ActionInvoker.invoke(ActionInvoker.java:475)
at play.mvc.ActionInvoker.invokeControllerMethod(ActionInvoker.java:451)
at play.mvc.ActionInvoker.invokeControllerMethod(ActionInvoker.java:446)
at play.mvc.ActionInvoker.invoke(ActionInvoker.java:160)
at play.server.PlayHandler$NettyInvocation.execute(PlayHandler.java:267)
at play.Invoker$Invocation$1.apply(Invoker.java:290)
at play.Invoker$Invocation$1.apply(Invoker.java:287)
at play.db.jpa.JPA.withTransaction(JPA.java:258)
at play.db.jpa.JPA.withinFilter(JPA.java:217)
at play.db.jpa.JPAPlugin$TransactionalFilter.withinFilter(JPAPlugin.java:299)
at play.Invoker$Invocation.withinFilter(Invoker.java:271)
at play.Invoker$Invocation.run(Invoker.java:287)
at play.server.PlayHandler$NettyInvocation.run(PlayHandler.java:245)
at java.util.concurrent.Executors$RunnableAdapter.call(Executors.java:511)
at java.util.concurrent.FutureTask.run(FutureTask.java:266)
at java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask.access$201(ScheduledThreadPoolExecutor.java:180)
at java.util.concurrent.ScheduledThreadPoolExecutor$ScheduledFutureTask.run(ScheduledThreadPoolExecutor.java:293)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1142)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617)
at java.lang.Thread.run(Thread.java:745)

Caused by: org.hibernate.exception.DataException: could not insert: [models.BlogUser]
at org.hibernate.exception.internal.SQLStateConversionDelegate.convert(SQLStateConversionDelegate.java:135)
at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:49)
at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:125)
at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:3087)
at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:3509)
at org.hibernate.action.internal.EntityInsertAction.execute(EntityInsertAction.java:88)
at org.hibernate.engine.spi.ActionQueue.execute(ActionQueue.java:377)
at org.hibernate.engine.spi.ActionQueue.executeActions(ActionQueue.java:369)
at org.hibernate.engine.spi.ActionQueue.executeActions(ActionQueue.java:286)
at org.hibernate.event.internal.AbstractFlushingEventListener.performExecutions(AbstractFlushingEventListener.java:339)
at org.hibernate.event.internal.DefaultFlushEventListener.onFlush(DefaultFlushEventListener.java:52)
at org.hibernate.internal.SessionImpl.flush(SessionImpl.java:1234)
at org.hibernate.ejb.AbstractEntityManagerImpl.flush(AbstractEntityManagerImpl.java:996)
... 28 more

Caused by: org.postgresql.util.PSQLException: The column index is out of range: 7, number of columns: 6.
at org.postgresql.core.v3.SimpleParameterList.bind(SimpleParameterList.java:53)
at org.postgresql.core.v3.SimpleParameterList.setLiteralParameter(SimpleParameterList.java:114)
at org.postgresql.jdbc2.AbstractJdbc2Statement.bindLiteral(AbstractJdbc2Statement.java:2172)
at org.postgresql.jdbc2.AbstractJdbc2Statement.setLong(AbstractJdbc2Statement.java:1227)
at com.mchange.v2.c3p0.impl.NewProxyPreparedStatement.setLong(NewProxyPreparedStatement.java:170)
at org.hibernate.type.descriptor.sql.BigIntTypeDescriptor$1.doBind(BigIntTypeDescriptor.java:57)
at org.hibernate.type.descriptor.sql.BasicBinder.bind(BasicBinder.java:93)
at org.hibernate.type.AbstractStandardBasicType.nullSafeSet(AbstractStandardBasicType.java:280)
at org.hibernate.type.AbstractStandardBasicType.nullSafeSet(AbstractStandardBasicType.java:275)
at org.hibernate.type.ComponentType.nullSafeSet(ComponentType.java:358)
at org.hibernate.persister.entity.AbstractEntityPersister.dehydrateId(AbstractEntityPersister.java:2823)
at org.hibernate.persister.entity.AbstractEntityPersister.dehydrate(AbstractEntityPersister.java:2792)
at org.hibernate.persister.entity.AbstractEntityPersister.insert(AbstractEntityPersister.java:3064)
... 37 more

編輯：從play jpa GenericModel.java中添加了save（）方法

public <T extends JPABase> T save() {
    _save();
    return (T) this;
}

Edit2：添加了BlogUser表

            Table "public.bloguser"
  Column   |          Type          | Modifiers 
-----------+------------------------+-----------
 id        | bigint                 | not null
 email     | character varying(255) | not null
 firstname | character varying(255) | 
 lastname  | character varying(255) | 
 password  | character varying(255) | not null
 username  | character varying(255) | not null
Indexes:
    "bloguser_pkey" PRIMARY KEY, btree (id)

Answer 1

這有點舊，但我剛剛在Play 1.x上構建的舊應用程序遇到了同樣的錯誤，我升級了一些庫，所以我認為這可能會幫助其他人。 以下之一可能會修復錯誤：

1. 缺少列 - 顯而易見的一個，但可能是錯誤的最可能原因。 在開發模式下將jpa.default.ddl = update添加到application.conf文件通常會為您解決此問題。
2. 判別器列 - （這是我的問題所在）如果你升級了hibernate，那么你可能需要將@Column（insertable = false，updatable = false）注釋添加到你的超類中的discriminator字段。 較新版本的JPA沒有與舊版本相同的@Column默認值（相信它的2.1更改它）。