java - Spring JPA 多对多，额外列未更新

Question

提案和公司之间存在多对多关系。该关系连同与提案相关的公司职能 ID 一起存储在表 proposal_companies 中。

以下是(我认为的)实体的相关部分:

提案

@Entity
@Table(name="sales.proposals")
public class Proposal implements Serializable {
    private static final long serialVersionUID = 1L;
    
    private int proposalID;
    private List<ProposalCompany> companies = new ArrayList<>();
    
    public Proposal() {}
    
    @Id
    @Column(name="pk_proposalid")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @JsonView({View.ProposalView.class,View.ProposalRevisionView.class})
    public int getProposalID() {
        return proposalID;
    }

    public void setProposalID(int proposalID) {
        this.proposalID = proposalID;
    }

    @OneToMany(mappedBy="proposal")
    @JsonView(View.ProposalView.class)
    public List<ProposalCompany> getCompanies() {
        return companies;
    }
    
    public void setCompanies(List<ProposalCompany> companies) {
        this.companies = companies;
    }
    
    public void addCompany(Company company) {
        ProposalCompany proposalCompany = new ProposalCompany(this, company);
        this.companies.add(proposalCompany);
        company.getProposals().add(proposalCompany);
    }
    
    public void removeCompany(Company company) {
        for (Iterator<ProposalCompany> iterator = companies.iterator(); iterator.hasNext(); ) {
            ProposalCompany proposalCompany = iterator.next();
 
            if (proposalCompany.getProposal().equals(this) &&
                    proposalCompany.getCompany().equals(company)) {
                iterator.remove();
                proposalCompany.getCompany().getProposals().remove(proposalCompany);
                proposalCompany.setProposal(null);
                proposalCompany.setCompany(null);
            }
        }
    }
}

公司

@Entity
@Table(name="companies.companies")
public class Company implements Serializable {
    private static final long serialVersionUID = 1L;
    
    private int companyID;
    private List<ProposalCompany> proposals = new ArrayList<>();
    
    public Company() {}
    
    @Id
    @Column(name="pk_companyid")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @JsonView(View.AllCompaniesView.class)
    public int getCompanyID() {
        return companyID;
    }

    public void setCompanyID(int companyID) {
        this.companyID = companyID;
    }

    @OneToMany(mappedBy="company")
    @JsonView(View.SingleCompanyView.class)
    public List<ProposalCompany> getProposals() {
        return proposals;
    }

    public void setProposals(List<ProposalCompany> proposals) {
        this.proposals = proposals;
    }
}

提案公司

@Entity
@Table(name="sales.proposal_companies")
public class ProposalCompany implements Serializable {
    private static final long serialVersionUID = 1L;
    
    private Proposal proposal;
    private Company company;
    private JobFunction jobFunction;

    public ProposalCompany() {}
    
    public ProposalCompany(Proposal proposal, Company company) {
        this.proposal = proposal;
        this.company = company;
    }

    @Id
    @ManyToOne(cascade=CascadeType.ALL)
    @JoinColumn(name="fk_proposalid")
    @JsonView(View.SingleCompanyView.class)
    public Proposal getProposal() {
        return proposal;
    }

    public void setProposal(Proposal proposal) {
        this.proposal = proposal;
    }

    @Id
    @ManyToOne(cascade=CascadeType.ALL)
    @JoinColumn(name="fk_companyid")
    @JsonView(View.ProposalView.class)
    public Company getCompany() {
        return company;
    }

    public void setCompany(Company company) {
        this.company = company;
    }

    @ManyToOne
    @JoinColumn(name="fk_jobfunctionid")
    @JsonView(View.ProposalView.class)
    public JobFunction getJobFunction() {
        return jobFunction;
    }

    public void setJobFunction(JobFunction jobFunction) {
        this.jobFunction = jobFunction;
    }

    @Override
    public String toString() {
        return "Proposal: "+(proposal==null ? "NULL" : proposal.toString())+" / Company: "+(company==null ? "NULL" : company.toString())+" / Function: "+(jobFunction==null ? "NULL": jobFunction.toString());
    }
}

这对于数据库中的现有条目效果很好。当我从 UI 添加公司，或者甚至尝试更改附加到提案的现有公司的功能时，就会出现问题。 (我使用的是 Angular 5，它将 JSON 格式的 Proposal 对象传递给 Spring。)

当我添加新公司时，出现以下错误:

2018-01-30 09:02:34,187 https-jsse-nio-8493-exec-8 ERROR [/].[dispatcherServlet] - Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Request processing failed; 
nested exception is org.springframework.orm.jpa.JpaObjectRetrievalFailureException: Unable to find com.couplingcorp.sales.domain.proposals.ProposalCompany with id Proposal: Proposal ID: 2923 / Company: 10 - Company Name / Function: NULL; 
nested exception is javax.persistence.EntityNotFoundException: Unable to find com.couplingcorp.sales.domain.proposals.ProposalCompany with id Proposal: Proposal ID: 2923 / Company: 10 - Company Name / Function: NULL] with root cause

即使这是传入的 JSON:

companies   […]
    0   {…}
        company {…}
            companyID:      10
            companyName:    Company Name
            companyAbbr:
        jobFunction {…}
            jobFunctionID:  4
            jobFunctionName:    Sole Handler
            jobFunctionDefinition:  Company who purchases and uses the product

以下是 Spring 在服务层处理 JSON 的方式:

@Override
public Proposal update(Proposal proposal) throws EntityNotFoundException {
    Optional<Proposal> prop = repository.findById(proposal.getProposalID());
    if (!prop.isPresent()) {
        throw new EntityNotFoundException();
    }

    // in cases where a ProposalCompany has been added or updated, we need to set up the objects appropriately
    proposal.getCompanies().stream()
        .filter((pc) -> pc.getProposal()==null)
        .forEachOrdered((pc) -> {
            LOGGER.debug("The proposal is not set for "+pc.toString());
            pc.setProposal(proposal);
            LOGGER.debug("The proposal is now set for "+pc.toString());
        });
    
    return repository.save(proposal);
}

我可以在日志中看到，在保存提案时，ProposalCompany 对象看起来没问题:

2018-01-30 09:02:34,112 https-jsse-nio-8493-exec-8 DEBUG proposals.ProposalServiceImpl - The proposal is not set for Proposal: NULL / Company: 10 - Company Name / Function: 4: Sole Handler
2018-01-30 09:02:34,112 https-jsse-nio-8493-exec-8 DEBUG proposals.ProposalServiceImpl - The proposal is now set for Proposal: Proposal ID: 2923 / Company: 10 - Company Name / Function: 4: Sole Handler

我错过了什么会导致它找不到该实体，即使它显然拥有所需的所有信息？

我正在使用 Spring Boot 2.0.0.M7 和 Hibernate 5.2.12。

这里是从调用 save 处开始的完整日志的链接(我不能把它放在这里，因为它太长了): https://pastebin.com/2AjVA2zp

编辑:我还尝试了其他两种方法。

第二个如所述here 。这看起来很有希望，但我收到错误“字段列表”中的未知列“companies0_.proposal_pk_proposalid”。不知何故，使用嵌入的 ID 时，它无法正确形成 SQL。如果我忽略 JSON 的公司部分，那么一切都会正常。但我需要了解这些公司，所以这不是一个有效的选择。

第三种方法与第二种类似，只是我在 @Embeddable PropoasalCompanyID 类中使用了实际的 Proposal 和 Company 对象，但我无法弄清楚如何映射 ManyToMany 方面.

Answer 1

这很容易。您忘记了正确级联。

首先，您不应该从子实体级联到父实体。所以代替:

@ManyToOne(cascade=CascadeType.ALL)

你应该有:

@ManyToOne

其次，应该在父端使用级联:

@OneToMany(mappedBy="company", cascade = CascadeType.ALL, orphanRemoval = true)

和

@OneToMany(mappedBy="proposal", cascade = CascadeType.ALL, orphanRemoval = true)

第三，您应该对 @ManyToOne 关联使用 FetchType.LAZY，因为默认情况下它们是 EAGER，这对性能不利。

所以，用这个代替:

@ManyToOne(fetch = FetchType.LAZY)