Playframework 2.5 SQL错误或缺少数据库（没有这样的表：information_schema.sequences）

时间：2017-01-27 23:04:45

标签： java database sqlite playframework playframework-2.5

我已经从https://www.playframework.com/download

下载了Play Java Starter示例

我需要使用sqlite数据库而不是内存中的h2数据库。我更改了application.conf内的必要行：

default.driver = org.sqlite.JDBC
default.url = "jdbc:sqlite:database.sqlite"

还将libraryDependencies += "org.xerial" % "sqlite-jdbc" % "3.8.11.2"添加到build.sbt

当我尝试运行应用程序时，出现以下错误：

[info] application - Creating Pool for datasource 'default'
[info] p.a.d.HikariCPConnectionPool - datasource [default] bound to JNDI as DefaultDS
[info] p.a.d.DefaultDBApi - Database [default] connected at jdbc:sqlite:database.sqlite
[warn] o.h.e.j.s.SqlExceptionHelper - SQL Error: 1, SQLState: null
[error] o.h.e.j.s.SqlExceptionHelper - [SQLITE_ERROR] SQL error or missing database (no such table: information_schema.sequences)
[error] application -

其次是：

! @72oobm82b - Internal server error, for (GET) [/] ->

play.api.UnexpectedException: Unexpected exception[ProvisionException: Unable to provision, see the following errors:

1) Error injecting constructor, javax.persistence.PersistenceException: [PersistenceUnit: defaultPersistenceUnit] Unable to build Hibernate SessionFactory
  at play.db.jpa.DefaultJPAApi$JPAApiProvider.<init>(DefaultJPAApi.java:39)
  at play.db.jpa.DefaultJPAApi$JPAApiProvider.class(DefaultJPAApi.java:34)
  while locating play.db.jpa.DefaultJPAApi$JPAApiProvider
  while locating play.db.jpa.JPAApi
    for parameter 1 at controllers.PersonController.<init>(PersonController.java:21)
  while locating controllers.PersonController
    for parameter 1 at router.Routes.<init>(Routes.scala:32)
  while locating router.Routes
  while locating play.api.inject.RoutesProvider
  while locating play.api.routing.Router
    for parameter 0 at play.api.http.JavaCompatibleHttpRequestHandler.<init>(HttpRequestHandler.scala:200)
  while locating play.api.http.JavaCompatibleHttpRequestHandler
  while locating play.api.http.HttpRequestHandler
    for parameter 4 at play.api.DefaultApplication.<init>(Application.scala:221)
  at play.api.DefaultApplication.class(Application.scala:221)
  while locating play.api.DefaultApplication
  while locating play.api.Application

1 error]
    at play.core.server.DevServerStart$$anonfun$mainDev$1$$anon$1$$anonfun$get$1$$anonfun$apply$1$$anonfun$1.apply(DevServerStart.scala:180)
    at play.core.server.DevServerStart$$anonfun$mainDev$1$$anon$1$$anonfun$get$1$$anonfun$apply$1$$anonfun$1.apply(DevServerStart.scala:131)
    at scala.Option.map(Option.scala:146)
    at play.core.server.DevServerStart$$anonfun$mainDev$1$$anon$1$$anonfun$get$1$$anonfun$apply$1.apply(DevServerStart.scala:131)
    at play.core.server.DevServerStart$$anonfun$mainDev$1$$anon$1$$anonfun$get$1$$anonfun$apply$1.apply(DevServerStart.scala:129)
    at scala.util.Success.flatMap(Try.scala:231)
    at play.core.server.DevServerStart$$anonfun$mainDev$1$$anon$1$$anonfun$get$1.apply(DevServerStart.scala:129)
    at play.core.server.DevServerStart$$anonfun$mainDev$1$$anon$1$$anonfun$get$1.apply(DevServerStart.scala:121)
    at scala.concurrent.impl.Future$PromiseCompletingRunnable.liftedTree1$1(Future.scala:24)
    at scala.concurrent.impl.Future$PromiseCompletingRunnable.run(Future.scala:24)
Caused by: com.google.inject.ProvisionException: Unable to provision, see the following errors:

1) Error injecting constructor, javax.persistence.PersistenceException: [PersistenceUnit: defaultPersistenceUnit] Unable to build Hibernate SessionFactory
  at play.db.jpa.DefaultJPAApi$JPAApiProvider.<init>(DefaultJPAApi.java:39)
  at play.db.jpa.DefaultJPAApi$JPAApiProvider.class(DefaultJPAApi.java:34)
  while locating play.db.jpa.DefaultJPAApi$JPAApiProvider
  while locating play.db.jpa.JPAApi
    for parameter 1 at controllers.PersonController.<init>(PersonController.java:21)
  while locating controllers.PersonController
    for parameter 1 at router.Routes.<init>(Routes.scala:32)
  while locating router.Routes
  while locating play.api.inject.RoutesProvider
  while locating play.api.routing.Router
    for parameter 0 at play.api.http.JavaCompatibleHttpRequestHandler.<init>(HttpRequestHandler.scala:200)
  while locating play.api.http.JavaCompatibleHttpRequestHandler
  while locating play.api.http.HttpRequestHandler
    for parameter 4 at play.api.DefaultApplication.<init>(Application.scala:221)
  at play.api.DefaultApplication.class(Application.scala:221)
  while locating play.api.DefaultApplication
  while locating play.api.Application

1 error
    at com.google.inject.internal.InjectorImpl$2.get(InjectorImpl.java:1025)
    at com.google.inject.internal.InjectorImpl.getInstance(InjectorImpl.java:1051)
    at play.api.inject.guice.GuiceInjector.instanceOf(GuiceInjectorBuilder.scala:405)
    at play.api.inject.guice.GuiceInjector.instanceOf(GuiceInjectorBuilder.scala:400)
    at play.api.inject.guice.GuiceApplicationBuilder.build(GuiceApplicationBuilder.scala:123)
    at play.api.inject.guice.GuiceApplicationLoader.load(GuiceApplicationLoader.scala:21)
    at play.core.server.DevServerStart$$anonfun$mainDev$1$$anon$1$$anonfun$get$1$$anonfun$apply$1$$anonfun$1$$anonfun$2.apply(DevServerStart.scala:168)
    at play.core.server.DevServerStart$$anonfun$mainDev$1$$anon$1$$anonfun$get$1$$anonfun$apply$1$$anonfun$1$$anonfun$2.apply(DevServerStart.scala:164)
    at play.utils.Threads$.withContextClassLoader(Threads.scala:21)
    at play.core.server.DevServerStart$$anonfun$mainDev$1$$anon$1$$anonfun$get$1$$anonfun$apply$1$$anonfun$1.apply(DevServerStart.scala:164)
Caused by: javax.persistence.PersistenceException: [PersistenceUnit: defaultPersistenceUnit] Unable to build Hibernate SessionFactory
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.persistenceException(EntityManagerFactoryBuilderImpl.java:967)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:892)
    at org.hibernate.jpa.HibernatePersistenceProvider.createEntityManagerFactory(HibernatePersistenceProvider.java:58)
    at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:55)
    at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:39)
    at play.db.jpa.DefaultJPAApi.lambda$start$1(DefaultJPAApi.java:60)
    at java.lang.Iterable.forEach(Iterable.java:75)
    at play.db.jpa.DefaultJPAApi.start(DefaultJPAApi.java:59)
    at play.db.jpa.DefaultJPAApi$JPAApiProvider.<init>(DefaultJPAApi.java:46)
    at play.db.jpa.DefaultJPAApi$JPAApiProvider$$FastClassByGuice$$dcd4cdbd.newInstance(<generated>)
Caused by: org.hibernate.exception.GenericJDBCException: Unable to build DatabaseInformation
    at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:47)
    at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:111)
    at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:97)
    at org.hibernate.tool.schema.internal.Helper.buildDatabaseInformation(Helper.java:140)
    at org.hibernate.tool.schema.internal.AbstractSchemaMigrator.doMigration(AbstractSchemaMigrator.java:92)
    at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.performDatabaseAction(SchemaManagementToolCoordinator.java:183)
    at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.process(SchemaManagementToolCoordinator.java:72)
    at org.hibernate.internal.SessionFactoryImpl.<init>(SessionFactoryImpl.java:309)
    at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:445)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:889)
Caused by: java.sql.SQLException: [SQLITE_ERROR] SQL error or missing database (no such table: information_schema.sequences)
    at org.sqlite.core.DB.newSQLException(DB.java:890)
    at org.sqlite.core.DB.newSQLException(DB.java:901)
    at org.sqlite.core.DB.throwex(DB.java:868)
    at org.sqlite.core.NativeDB.prepare(Native Method)
    at org.sqlite.core.DB.prepare(DB.java:211)
    at org.sqlite.jdbc3.JDBC3Statement.executeQuery(JDBC3Statement.java:81)
    at com.zaxxer.hikari.pool.ProxyStatement.executeQuery(ProxyStatement.java:111)
    at com.zaxxer.hikari.pool.HikariProxyStatement.executeQuery(HikariProxyStatement.java)
    at org.hibernate.tool.schema.extract.internal.SequenceInformationExtractorH2DatabaseImpl.extractMetadata(SequenceInformationExtractorH2DatabaseImpl.java:37)
    at org.hibernate.tool.schema.extract.internal.DatabaseInformationImpl.initializeSequences(DatabaseInformationImpl.java:65)

数据库文件database.sqlite将自动在根文件夹内创建，其中包含0个字节。

我错过了什么吗？

1 个答案:

答案 0 :(得分：0)

在SQLite func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: [NSObject: AnyObject]?) -> Bool { // Override point for customization after application launch. var soundID: SystemSoundID = 0 let soundFile: String = NSBundle.mainBundle().pathForResource("symphony", ofType: "wav")! let soundURL: NSURL = NSURL(fileURLWithPath: soundFile) AudioServicesCreateSystemSoundID(soundURL, &soundID) AudioServicesPlayAlertSound(soundID) Fabric.with([Twitter.self]) //Firebase configuration FIRApp.configure() //Resource code from stackoverflow to create UNUserNotificationCenter let center = UNUserNotificationCenter.current() center.requestAuthorization(options:[.badge, .alert, .sound]) { (granted, error) in // Enable or disable features based on authorization. } application.registerForRemoteNotifications() return true } 中不存在。等效的是它的系统表information_schema。但是，默认情况下，表sqlite_master不可用。 sequences由字段构成：

sqlite_master

不幸的是，你似乎还需要做一些额外的工作才能从H2过渡到SQLite。

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