spring boot jpa：從與表架構無關的jpa查詢返回自定義對象

Question

這是我的配置：Java: 1.8 Spring Boot :: v2.3.1.RELEASE

我有以下模型/類：

書籍：書籍類如下。 請注意，這不是與 DB 中的表相關的實體，也可以將其視為從多個表中提取的數據，僅具有必需的屬性。 簡而言之，我不會有以下對象的表模式。

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;
    }

}

與上面類似，我還有一個班級，

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;
    }

}

我需要作為以下對象的輸出

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

}

所以，我試圖擁有一個存儲庫

import com.myservice.model.Configuration;
import org.springframework.data.jpa.repository.JpaRepository;

public interface ConfigurationRepository extends JpaRepository<Configuration, Long>, ConfigurationRepositoryCustom {

}

自定義界面

import com.myservice.model.Configuration;

public interface ConfigurationRepositoryCustom {

    Configuration getConfigData();
}

執行：

@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;
    }
}

基本上，如果您看到 Configuration 對象不在 DB 中。

當我啟動應用程序時，它首先拋出了這個錯誤：

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]

對於上述，我添加了@Entity？ 對我的 Configuration 類的注釋

接下來，它拋出了錯誤：

Caused by: org.hibernate.AnnotationException: No identifier specified for entity: com.myservice.model.Configuration
    at org.hibernate.cfg.InheritanceState.determineDefaultAccessType(InheritanceState.java:266) ~[hibernate-core-5.4.17.Final.jar:5.4.17.Final]
    at org.hibernate.cfg.InheritanceState.getElementsToProcess(InheritanceState.java:211) ~[hibernate-core-5.4.17.Final.jar:5.4.17.Final]
    at org.hibernate.cfg.AnnotationBinder.bindClass(AnnotationBinder.java:781) ~[hibernate-core-5.4.17.Final.jar:5.4.17.Final]
    

現在，我的配置不是數據庫中的表，我仍然添加了屬性“id”

@Id
@GeneratedValue
private long id;

接下來，我收到此錯誤：

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]

我的數據庫中當然沒有這樣的表。

因此，基本上我正在尋找一種基於其他對象（很少來自 DB，很少來自其他調用等）創建自定義（此處為配置）對象的方法，並且此自定義對象不在 DB 中。 那么如何實現相同的呢？

Answer 1

你不需要這個接口

public interface ConfigurationRepository extends JpaRepository<Configuration, Long>, ConfigurationRepositoryCustom {

}

相反，直接在您的代碼中使用ConfigurationRepositoryCustom ，您也可以使用名稱ConfigurationRepository （無需以 custom 為后綴）。