玩2.5，hibernate：table没有映射 [英] play 2.5, hibernate : table is not mapped

问题描述

我一直在

以外发生异常（例外情况是：org.hibernate.hql.internal.ast.QuerySyntaxException：用户未被映射）

调用此代码时发生。

  TypedQuery< User> query = jpaApi.em（）。createQuery（从用户u中选择u，其中u.email =：email和u.secretHash =：secretHash，User.class）
 .setParameter（email，parameter.getEmail （））
 .setParameter（secretHash，hashAlgorithm.hash（parameter.getPassword（）））; 
  

但是如果这个项目是由actirvator start（开发env）运行的，这个异常没有发生。

它的意思是......只有在生产环境中，我得到了这个异常。

我该如何解决它。

请帮助我。

感谢您的帮助，我的项目信息在

• 系统环境：
  
  $ b <1> Play：2.5.4
  
  
  

  2）hibernate：5.2.1.final
  
  

  
  

  • 异常消息：

    
    
    

    引起者：org.hibernate.hql.internal.ast.QuerySyntaxException：用户未映射[从用户u中选择u u.email =：email和u .secretHash =：secretHash]
    at org.hibernate.hql.internal.ast.QuerySyntaxException.generateQueryException（QuerySyntaxException.java:79）
    at org.hibernate.QueryException.wrapWithQueryString（QueryException.java:103）在org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile（QueryTranslatorImpl.java:218）
    
    在org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile（QueryTranslatorImpl.java:142）
    在org.hibernate.engine.query.spi.HQLQueryPlan。（HQLQueryPlan.java:115）
    在org.hibernate.engin e.query.spi.HQLQueryPlan。（HQLQueryPlan.java:77）
    at org.hibernate.engine.query.spi.QueryPlanCache.getHQLQueryPlan（QueryPlanCache.java:152）
    org.hibernate.internal组织：在org.hibernate.internal.AbstractSharedSessionContract.createQuery（AbstractSharedSessionContract.java:623）
    .AbstractSharedSessionContract.getQueryPlan（AbstractSharedSessionContract.java:521）
    ... 52个
    所致。 hibernate.hql.internal.ast.QuerySyntaxException：用户未映射
    在org.hibernate.hql.internal.ast.util.SessionFactoryHelper.requireClassPersister（SessionFactoryHelper.java:171）
    在org.hibernate。 hql.internal.ast.tree.FromElementFactory.addFromElement（FromElementFactory.java:91）
    at org.hibernate.hql.internal.ast.tree.FromClause.addFromElement（FromClause.java:79）
    at org.hibernate.hql.internal.ast.HqlSqlWalker.createFromElement（HqlSqlWalker.java:321）
    在org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromElement（HqlSqlBaseWa lker.java:3690）美元，org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromElementList（HqlSqlBaseWalker.java:3579 B $ B）
    在org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromClause（ HqlSqlBaseWalker.java:718）美元，org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.query（HqlSqlBaseWalker.java:574 b $ b）
    在org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectStatement（ HqlSqlBaseWalker.java:311）美元，org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.statement（HqlSqlBaseWalker.java:259 b $ b）
    在org.hibernate.hql.internal.ast.QueryTranslatorImpl.analyze（ QueryTranslatorImpl.java:262）
    在org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile（QueryTranslatorImpl.java:190）
    ... 58 more

  
  

  • persistence.xml

     < persistence xmlns =http://java.sun.com / xml / ns / persistence
     xmlns：xsi =http://www.w3.org/2001/XMLSchema-instance
     xsi：schemaLocation =http://java.sun.com / xml / ns /持久性http ：//java.sun.com/xml/ns/persistence/persistence_2_0.xsd
     version =2.0> 
    < persistence-unit name =Hoth-PersistenceUnittransaction-type =RESOURCE_LOCAL> 
    < provider> org.hibernate.jpa.HibernatePersistenceProvider< / provider> 
    <非-jta-data-source> DefaultDS< /非-jta-data-source> 
    <属性> 
    < property name =hibernate.dialectvalue =org.hibernate.dialect.PostgreSQL94Dialect/> 
     
    < property name =hibernate.show_sqlvalue =true/> 
    < property name =hibernate.format_sqlvalue =true/> 
    < property name =hibernate.use_sql_commentsvalue =true/> 
     
    < property name =hibernate.max_fetch_depthvalue =5/> 
     
    < property name =hibernate.hbm2ddl.autovalue =update/> 
     
    < property name =hibernate.jdbc.batch_sizevalue =50/> 
    < property name =hibernate.jdbc.batch_versioned_datavalue =true/> 
    < property name =hibernate.order_insertsvalue =true/> 
    < / properties> 
    < / persistence-unit> 
      

  
  

  • 实体类代码
    
    

      @Entity 
     @Table（name =users）
     @Getter @Setter @ToString @ EqualsAndHashCode（of =email）
     public class User {
     @Id 
     private String email; 
     
     @Column（name =secret_hash）
     private String secretHash; 
     
     @Column（name =accessible_at）
     private LocalDateTime accessibleAt; 
     
    私人字符串性别; 
     
     @Column（name =birth_year）
     private Integer birthYear; 
     
     @Column（name =picture_name）
     private String pictureName; 
     
     @Column（name =picture_url）
     private String pictureUrl; 
     
    私人字符串昵称; 
     
      

  
  
  
  

  解决方案

  Persistence.xml中缺少用户映射。我无法从片段中看到用户定义的包，但假设它位于包persistence.models中。 persistence.xml应该是这样的：

   < persistence-unit name =Hoth-PersistenceUnittransaction-type = RESOURCE_LOCAL> 
  < provider> org.hibernate.jpa.HibernatePersistenceProvider< / provider> 
  <非-jta-data-source> DefaultDS< /非-jta-data-source> 
  < class> persistence.models.User< / class> 
  <属性> 
   ... 
  < / properties> 
    

  
  

  I have been getting a exception below

  (Exception is : org.hibernate.hql.internal.ast.QuerySyntaxException: User is not mapped )

  It occured when this code is called.

  TypedQuery<User> query = jpaApi.em().createQuery("select u from User u where u.email = :email and u.secretHash = :secretHash", User.class)
              .setParameter("email", parameter.getEmail())
              .setParameter("secretHash", hashAlgorithm.hash(parameter.getPassword()));
  

  but if this project is running by "actirvator start" (develop env), this exception did not occur.

  it mean ... only on Production environment i got this exception.

  how can i fix it.

  please help me.

  thanks for your help and my project information is below

  • System environment :

  1) Play : 2.5.4

  2) hibernate : 5.2.1.final

  • Exception Message :

    Caused by: org.hibernate.hql.internal.ast.QuerySyntaxException: User is not mapped [select u from User u where u.email = :email and u.secretHash = :secretHash] at org.hibernate.hql.internal.ast.QuerySyntaxException.generateQueryException(QuerySyntaxException.java:79) at org.hibernate.QueryException.wrapWithQueryString(QueryException.java:103) at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:218) at org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:142) at org.hibernate.engine.query.spi.HQLQueryPlan.(HQLQueryPlan.java:115) at org.hibernate.engine.query.spi.HQLQueryPlan.(HQLQueryPlan.java:77) at org.hibernate.engine.query.spi.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:152) at org.hibernate.internal.AbstractSharedSessionContract.getQueryPlan(AbstractSharedSessionContract.java:521) at org.hibernate.internal.AbstractSharedSessionContract.createQuery(AbstractSharedSessionContract.java:623) ... 52 more Caused by: org.hibernate.hql.internal.ast.QuerySyntaxException: User is not mapped at org.hibernate.hql.internal.ast.util.SessionFactoryHelper.requireClassPersister(SessionFactoryHelper.java:171) at org.hibernate.hql.internal.ast.tree.FromElementFactory.addFromElement(FromElementFactory.java:91) at org.hibernate.hql.internal.ast.tree.FromClause.addFromElement(FromClause.java:79) at org.hibernate.hql.internal.ast.HqlSqlWalker.createFromElement(HqlSqlWalker.java:321) at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromElement(HqlSqlBaseWalker.java:3690) at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromElementList(HqlSqlBaseWalker.java:3579) at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.fromClause(HqlSqlBaseWalker.java:718) at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.query(HqlSqlBaseWalker.java:574) at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectStatement(HqlSqlBaseWalker.java:311) at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.statement(HqlSqlBaseWalker.java:259) at org.hibernate.hql.internal.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:262) at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:190) ... 58 more

  • persistence.xml

    <persistence xmlns="http://java.sun.com/xml/ns/persistence"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd"
    version="2.0">
    <persistence-unit name="Hoth-PersistenceUnit" transaction-type="RESOURCE_LOCAL">
        <provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
        <non-jta-data-source>DefaultDS</non-jta-data-source>
        <properties>
            <property name="hibernate.dialect" value="org.hibernate.dialect.PostgreSQL94Dialect" />
    
            <property name="hibernate.show_sql" value="true" />
            <property name="hibernate.format_sql" value="true" />
            <property name="hibernate.use_sql_comments" value="true" />
    
            <property name="hibernate.max_fetch_depth" value="5" />
    
            <property name="hibernate.hbm2ddl.auto" value="update" />
    
            <property name="hibernate.jdbc.batch_size" value="50" />
            <property name="hibernate.jdbc.batch_versioned_data" value="true" />
            <property name="hibernate.order_inserts" value="true" />
        </properties>
    </persistence-unit>
    

  • Entity Class Code

    @Entity
    @Table (name = "users")
    @Getter @Setter @ToString @EqualsAndHashCode (of = "email")
    public class User {
    @Id
    private String email;
    
    @Column(name = "secret_hash")
    private String secretHash;
    
    @Column(name = "accessed_at")
    private LocalDateTime accessedAt;
    
    private String gender;
    
    @Column(name = "birth_year")
    private Integer birthYear;
    
    @Column(name = "picture_name")
    private String pictureName;
    
    @Column(name = "picture_url")
    private String pictureUrl;
    
    private String nickname;
    }
    

  解决方案

  The User mapping is missing in the persistence.xml. I cannot see from the snippets in what package the User is defined, but let's suppose it is in the package persistence.models. The persistence.xml should be like this:

  <persistence-unit name="Hoth-PersistenceUnit" transaction-type="RESOURCE_LOCAL">
  <provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
  <non-jta-data-source>DefaultDS</non-jta-data-source>
  <class>persistence.models.User</class>
  <properties>
      ...
  </properties>
  

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