Spring JPA多对多列没有更新

Question

我和提案公司之间有很多关系.该关系proposal_companies与该提案相关的公司功能的ID一起存储在表中.

以下是实体的相关部分(我认为):

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;
    }
}

ProposalCompany

@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());
    }
}

这适用于数据库中的现有条目.当我从UI添加公司,甚至尝试更改附加到提案的现有公司的功能时,问题就出现了.(我使用的是Angular 5,它将ProposalJSON格式的对象传递给Spring.)

当我添加新公司时,这是错误:

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JSON进来:

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在服务层上处理JSON的方法:

@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,看起来ProposalCompany对象是正常的:

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实体,即使它显然拥有它需要的所有信息？

我使用Spring Boot 2.0.0.M7和Hibernate 5.2.12.

这是从save调用点到完整日志的链接(我不能把它放在这里,因为它太长了):https: //pastebin.com/2AjVA2zp

编辑: 我还尝试了另外两种方法.

第二个是这里描述的.这看起来很有希望,但我得到了错误Unknown column 'companies0_.proposal_pk_proposalid' in 'field list'.不知何故,嵌入式ID无法正确形成SQL.如果我忽略了JSON的公司部分,那么一切都有效.但我需要了解这些公司,因此这不是一个有效的选择.

第三种方法类似于第二种方法,除了我在@Embeddable PropoasalCompanyID类中使用了实际的Proposal和Company对象,但在那里我无法弄清楚如何映射ManyToMany方面.

Answer 1

这很简单.你忘记了正确的级联.

首先,您不应该从子实体级联到父级实体.所以代替:

@ManyToOne(cascade=CascadeType.ALL)

你应该有:

@ManyToOne

其次,你应该在父方使用casacade:

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

和

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

有关更多详细信息,请查看此文章.

第三,你应该使用FetchType.LAZY的@ManyToOne,因为它们在默认情况下EAGER协会,这就是不好的表现.

所以,请改用:

@ManyToOne(fetch = FetchType.LAZY)