2017-08-30 87 views
0

我正在使用JPA在我的公司开发多租户应用程序。所以我使用TABLE_PER_TENANT功能将租户特定数据分成不同的模式。将数据推入数据库是没有问题的,可以正常工作。问题只是从数据库中读取数据,应用程序找不到我的命名查询。JPA:名称的查询名称:...未找到

我已经尝试过不同的东西:

  • 重新编译和部署应用程序
  • 移动的persistence.xml到不同的文件夹
  • 经过错误日志和调试的整个应用程序,无法找到其原因
  • 只好看看所有可能的解决方案,我可以找到关于谷歌和尝试他们...

我在这里错过了什么?有任何想法吗?

BaseObject.java

@MappedSuperclass 
@Multitenant(MultitenantType.TABLE_PER_TENANT) 
@TenantTableDiscriminator(type = TenantTableDiscriminatorType.SCHEMA, contextProperty="tenant.id") 
public abstract class BaseObject { 

    /** 
    * The (globally unique) ID of the object. 
    */ 
    @Id 
    @Column(name = "GUID", length = 36) 
    private String guid = UUID.randomUUID().toString(); 

    /** 
    * The {@link Date} the object was created at. 
    */ 
    @Temporal(TemporalType.TIMESTAMP) 
    @Column(name = "CREATION_DATE", updatable = false) 
    private Date createdAt = null; 

    /** 
    * The {@link Date} the object was last modified at. 
    */ 
    @Temporal(TemporalType.TIMESTAMP) 
    @Column(name = "MODIFICATION_DATE") 
    private Date lastModifiedAt = null; 

    /** 
    * ID of the user who created the object. 
    */ 
    @Column(name = "CREATED_BY", updatable = false, length = 20) 
    private String createdBy = null; 

    /** 
    * ID of the user who was the last to modify the object. 
    */ 
    @Column(name = "MODIFIED_BY", length = 20) 
    private String lastModifiedBy = null; 

    // Methods... 
} 

Plant.java

@Entity 
@Table(name = "PLANT_POLLUTION_DATA") 
@NamedQueries({ @NamedQuery(name = "Plant.getPlants", query = "SELECT c FROM Plant c"), 
       @NamedQuery(name = "Plant.getPlantById", query = "SELECT c FROM Plant c WHERE c.id = :id") }) 

@XmlRootElement(name = "plantlist") 
@XmlAccessorType(XmlAccessType.FIELD) 
public class Plant extends BaseObject implements Serializable { 

    /** 
    * The <code>serialVersionUID</code> of the {@link Plant} class. 
    */ 
    private static final long serialVersionUID = 1L; 

    @Column(name = "ID", length = 36, nullable = true) 
    String id = null; 

    @Column(name = "O3", length = 10, nullable = true) 
    String o3 = null; 

    @Column(name = "DATE_FIELD") 
    java.sql.Date dateField = null; 

    @Column(name = "location", length = 36, nullable = true) 
    String location = null; 

    // Methods... 
} 

NamedQuery调用

String tenantId = getTenantId(); 
Map<String, String> props = new HashMap<String, String>(); 
props.put("tenant.id", tenantId); 
EntityManager em = this.getEntityManagerFactory().createEntityManager(props); 
Query query = em.createNamedQuery("Plant.getPlantById"); 
query.setParameter("id", id); 
retVal = query.getResultList(); 

周的persistence.xml(位于 “file.war/WEB-INF /类/ META-INF/persistence.xml中”)

<?xml version="1.0" encoding="UTF-8"?> 
<persistence version="2.1" xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd"> 
    <persistence-unit name="pollutionmonitoring" transaction-type="RESOURCE_LOCAL"> 
     <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider> 
     <class>com.company.cloud.samples.pollutionmonitoring.model.BaseObject</class> 
     <class>com.company.cloud.samples.pollutionmonitoring.model.Plant</class> 
     <properties> 
      <property name="eclipselink.ddl-generation" value="create-or-extend-tables"/> 
     </properties> 
    </persistence-unit> 
</persistence> 
+0

做任何查询的工作,或者它只是你的命名查询无法找到?尝试设置启动时部署持久性属性以查看是否有帮助https://www.eclipse.org/eclipselink/documentation/2.4/jpa/extensions/p_deploy_on_startup.htm – Chris

+0

正如您在我的答案中所看到的,我可能已经解决了我的问题。在另一个班上有一个完美的查询,只是在这个班上。但我仍然不知道为什么我的解决方法确实有效...... – Burdel

+0

这就是我的评论 - 找出其他选项和潜在问题。如果你对你的解决方法很好,很好。 – Chris

回答

0

我可以只通过添加以下2行代码解决这个问题自己:

... 
EntityManager em = this.getEntityManagerFactory().createEntityManager(props); 

em.getTransaction().begin(); 
em.getTransaction().commit(); 

Query query = em.createNamedQuery("Plant.getPlantById"); 
... 

我不知道这个变化究竟,但它为我工作:)也许有人知道为什么这个解决方案确实工作?

PS:我只是从另一个类中复制它,我必须将数据添加到数据库,这是两个类之间的唯一区别。

+0

这意味着你的JPA提供者有一个错误,你应该向他们报告这个错误。没有兼容的JPA提供商会要求这样做。 –