Spring Data JPA 存储过程

7
我正在尝试调用一个Oracle存储过程,以下是我的实体和仓库类。
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.NamedStoredProcedureQuery;
import javax.persistence.ParameterMode;
import javax.persistence.StoredProcedureParameter;

@NamedStoredProcedureQuery(name = "casedetails", procedureName = "APPS.GET_SR_DATA", resultClasses = Case.class, parameters = {
        @StoredProcedureParameter(name = "p_sr_number", mode = ParameterMode.IN, type = String.class),
        @StoredProcedureParameter(name = "p_results", mode = ParameterMode.REF_CURSOR, type = void.class),
        @StoredProcedureParameter(name = "p_retVal", mode = ParameterMode.OUT, type = Long.class),
        @StoredProcedureParameter(name = "p_errMsg", mode = ParameterMode.OUT, type = String.class) })
/* }) */
@Entity
public class Case implements Serializable {

    @Id
    @Column(name = "sr_number")
    private String caseNumber;
    @Column(name = "problem_summary")
    private String caseTitle;
    @Column(name = "sr_type")
    private String caseCategory;
}

我的仓库类,

import org.springframework.data.repository.query.Param;
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.data.jpa.repository.query.Procedure;

public interface CaseRepository extends JpaRepository<Case,String>{

    @Procedure(name = "casedetails", procedureName = "APPS.GET_SR_DATA")
    List<Case> findByCaseNumber(@Param("caseNumber") String caseNumber);
}

以下是persistence.xml文件。
<?xml version="1.0"?>
<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 persistence_1_0.xsd" version="1.0">

    <persistence-unit name="case-rest-service" transaction-type="RESOURCE_LOCAL">
        <provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
        <exclude-unlisted-classes>false</exclude-unlisted-classes>
    </persistence-unit>

</persistence>

项目pom.xml条目
<dependency>
            <groupId>org.springframework.hateoas</groupId>
            <artifactId>spring-hateoas</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.data</groupId>
            <artifactId>spring-data-jpa</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.data</groupId>
            <artifactId>spring-data-commons</artifactId>
        </dependency>
<dependency>
            <groupId>org.eclipse.persistence</groupId>
            <artifactId>eclipselink</artifactId>
            <version>2.6.0-M3</version>
        </dependency>

        <dependency>
            <groupId>org.hibernate</groupId>
            <artifactId>hibernate-validator</artifactId>
            <scope>runtime</scope>
        </dependency>
        <dependency>
            <groupId>org.hibernate</groupId>
            <artifactId>hibernate-entitymanager</artifactId>
        </dependency>

我尝试运行代码库测试时,遇到了以下错误:

Object of class [org.springframework.data.jpa.repository.query.PartTreeJpaQuery] must be an instance of class org.springframework.data.jpa.repository.query.StoredProcedureJpaQuery; nested exception is java.lang.IllegalArgumentException: Object of class [org.springframework.data.jpa.repository.query.PartTreeJpaQuery] must be an instance of class org.springframework.data.jpa.repository.query.StoredProcedureJpaQuery

我以前从未使用过Spring Data JPA与存储过程,但我很感激这里的任何反馈/输入。


你解决了吗? - Opal
有人知道如何解决这个问题吗? - Alter Hu
我在@Procedure注释上也遇到了同样的问题。 - alltej
1个回答

0

我从未找到解决方案,所以采用了不同的方法

@Repository 
public interface MyObjectRepository extends CrudRepository<MyObject, String>  {
    @Query(value = "EXECUTE [dbo].[myProc] :fieldName, :pages", nativeQuery = true)
    Set<MyObject> findAllByFieldName(@Param("fieldName") String fieldName, @Param("pages") int pages);
}

这是一个类似的问题,我建议相同的答案question

这个链接可能会有所帮助SPRING DATA JPA中的数据仓库


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