Java spring boot jpa:从与表架构无关的jpa查询返回自定义对象
以下是我的配置: Java:1.8 Spring Boot::v2.3.1.0版本 我有以下型号/类别: 图书:图书课程如下。请注意,这不是一个与数据库中的表相关的实体,或者可以将其视为从多个表中获取的数据,这些表只具有必需的属性。简而言之,我不会为以下对象提供表模式Java spring boot jpa:从与表架构无关的jpa查询返回自定义对象,java,spring-boot,spring-data-jpa,Java,Spring Boot,Spring Data Jpa,以下是我的配置: Java:1.8 Spring Boot::v2.3.1.0版本 我有以下型号/类别: 图书:图书课程如下。请注意,这不是一个与数据库中的表相关的实体,或者可以将其视为从多个表中获取的数据,这些表只具有必需的属性。简而言之,我不会为以下对象提供表模式 package com.myservice.model; public class Book { private long bookId; private String bookName; private
package com.myservice.model;
public class Book {
private long bookId;
private String bookName;
private String bookType;
private String author;
public Book(long bookId, String bookName, String bookType, String author) {
this.bookId = bookId;
this.bookName = bookName;
this.bookType = bookType;
this.author = author;
}
public long getBookId() {
return bookId;
}
public void setBookId(long bookId) {
this.bookId = bookId;
}
public String getBookName() {
return bookName;
}
public void setBookName(String bookName) {
this.bookName = bookName;
}
public String getBookType() {
return bookType;
}
public void setBookType(String bookType) {
this.bookType = bookType;
}
public String getAuthor() {
return author;
}
public void setAuthor(String author) {
this.author = author;
}
}
和上面一样,我还有一节课
package com.myservice.model;
public class Business {
private String businessName;
private String businessType;
private String owner;
public Book(String businessName, String businessType, String owner) {
this.businessName = businessName;
this.bookType = bookType;
this.owner = owner;
}
public String getBusinessName() {
return businessName;
}
public void setBusinessName(String businessName) {
this.businessName = businessName;
}
public String getBusinessType() {
return businessType;
}
public void setBusinessType(String businessType) {
this.businessType = businessType;
}
public String getOwner() {
return owner;
}
public void setOwner(String owner) {
this.owner = owner;
}
}
我需要输出如下对象
import java.util.List;
public class Configuration {
private List<Book> books;
private List<Business> businesses;
private List<String> authors;
private List<String> owners;
// Constructor and getters/setters omitted for brevity
}
实施:
@Repository
@Transactional(readOnly = true)
public class ConfigurationRepositoryImpl implements ConfigurationRepositoryCustom {
@PersistenceContext
EntityManager entityManager;
private List<Book> getBooks() {
Query query = entityManager.createNativeQuery(
"SELECT book_id, book, bookType, author FROM TABLE1 JOIN TABLE2 <Other criterias>"
, Book.class
);
return query.getResultList();
}
private List<Business> getBusinesses() {
List<Business> businesses = new ArrayList<>();
businesses.add(new Business("ABC", "Constructions", "ABC PQR" ));
businesses.add(new Business("XYZ", "Constructions", "PQR CCC" ));
businesses.add(new Business("PQR", "Education", "ABC PQR" ));
return businesses;
}
@Override
public Configuration getConfigData() {
List<Book> books = getBooks();
List<Business> businesses = getBusinesses();
List<String> authors = books.stream().map(Book::getAuthor).collect(Collectors.toList());
List<String> owners = businesses.stream().map(Book::getOwner).collect(Collectors.toList());
Configuration config = new Configuration(
books,
businesses,
authors,
owners
);
return config;
}
}
现在,我的配置不是DB中的表,我仍然添加了属性“id”
接下来,我得到这个错误:
Caused by: java.lang.IllegalArgumentException: Not a managed type: class com.myservice.model.Configuration
at org.hibernate.metamodel.internal.MetamodelImpl.managedType(MetamodelImpl.java:582) ~[hibernate-core-5.4.17.Final.jar:5.4.17.Final]
at org.hibernate.metamodel.internal.MetamodelImpl.managedType(MetamodelImpl.java:85) ~[hibernate-core-5.4.17.Final.jar:5.4.17.Final]
at org.springframework.data.jpa.repository.support.JpaMetamodelEntityInformation.<init>(JpaMetamodelEntityInformation.java:75) ~[spring-data-jpa-2.3.1.RELEASE.jar:2.3.1.RELEASE]
at org.springframework.data.jpa.repository.support.JpaEntityInformationSupport.getEntityInformation(JpaEntityInformationSupport.java:66) ~[spring-data-jpa-2.3.1.RELEASE.jar:2.3.1.RELEASE]
at org.springframework.data.jpa.repository.support.JpaRepositoryFactory.getEntityInformation(JpaRepositoryFactory.java:229) ~[spring-data-jpa-2.3.1.RELEASE.jar:2.3.1.RELEASE]
at org.springframework.data.jpa.repository.support.JpaRepositoryFactory.getTargetRepository(JpaRepositoryFactory.java:179) ~[spring-data-jpa-2.3.1.RELEASE.jar:2.3.1.RELEASE]
at org.springframework.data.jpa.repository.support.JpaRepositoryFactory.getTargetRepository(JpaRepositoryFactory.java:162) ~[spring-data-jpa-2.3.1.RELEASE.jar:2.3.1.RELEASE]
at org.springframework.data.jpa.repository.support.JpaRepositoryFactory.getTargetRepository(JpaRepositoryFactory.java:72) ~[spring-data-jpa-2.3.1.RELEASE.jar:2.3.1.RELEASE]
Caused by: org.hibernate.MappingException: Could not determine type for: java.util.List, at table: configuration, for columns: [org.hibernate.mapping.Column(books)]
at org.hibernate.mapping.SimpleValue.getType(SimpleValue.java:499) ~[hibernate-core-5.4.17.Final.jar:5.4.17.Final]
at org.hibernate.mapping.SimpleValue.isValid(SimpleValue.java:466) ~[hibernate-core-5.4.17.Final.jar:5.4.17.Final]
at org.hibernate.mapping.Property.isValid(Property.java:227) ~[hibernate-core-5.4.17.Final.jar:5.4.17.Final]
我的数据库中当然没有这样的表
因此,基本上我正在寻找一种基于其他对象(少数来自DB,少数来自其他调用等)创建自定义(此处为配置)对象的方法,而该自定义对象不在DB中。那么如何实现同样的功能呢?您不需要这个接口
import com.myservice.model.Configuration;
public interface ConfigurationRepositoryCustom {
Configuration getConfigData();
}
public interface ConfigurationRepository extends JpaRepository<Configuration, Long>, ConfigurationRepositoryCustom {
}
public interface ConfigurationRepository扩展了JpaRepository、ConfigurationRepositoryCustom{
}
相反,可以直接在代码中使用
ConfigurationRepositoryCustom
,也可以使用名称ConfigurationRepository
(无需添加自定义后缀)。您不需要此接口
import com.myservice.model.Configuration;
public interface ConfigurationRepositoryCustom {
Configuration getConfigData();
}
public interface ConfigurationRepository extends JpaRepository<Configuration, Long>, ConfigurationRepositoryCustom {
}
public interface ConfigurationRepository扩展了JpaRepository、ConfigurationRepositoryCustom{
}
相反,可以直接在代码中使用
ConfigurationRepositoryCustom
,也可以使用名称ConfigurationRepository
(无需使用自定义后缀)。您只能为@Entity class创建一个JpaRepository接口,这是原始“非托管类型”IllegalArgumentException的来源。由于Configuration
不是数据库中的表,因此不需要ConfigurationRepository
。只需删除它(以及配置中的@Entity注释和id字段),您只能为@Entity类创建一个JpaRepository接口,这就是原始“非托管类型”IllegalArgumentException的来源。由于Configuration
不是数据库中的表,因此不需要ConfigurationRepository
。只需删除它(以及配置中的@Entity注释和id字段)
public interface ConfigurationRepository extends JpaRepository<Configuration, Long>, ConfigurationRepositoryCustom {
}