主页 / coding / 问题 / 79280605
Accepted
androidDeweleper
Asked: 2024-12-14 20:13:46 +0800 CST

用于删除关系的 JPA 查询

  • 772
@Entity
data class Election(
    @Id @UuidGenerator val electionId: String = "",
    var name: String = "",
    var description: String = "",
    var startDate: LocalDateTime? = null,
    var endDate: LocalDateTime? = null,
    var electionCode: String = "", // auto generated to be unique

    @ManyToOne(cascade = [CascadeType.ALL]) @Column(nullable = false) var electoralCommission: ElectoralCommission = ElectoralCommission(),

    @OneToMany(mappedBy = "elections") var candidates: MutableList<Candidate> = mutableListOf(),

    @OneToMany var eligibleVoters: MutableList<Voter> = mutableListOf(),

    @Enumerated(EnumType.STRING) var electionStatus: ElectionStatus = ElectionStatus.UPCOMING
)

@Entity
data class Voter(
    @Id
    @UuidGenerator
    @Column(name = "voter_id")
    private val _voterId: String = "",

    val name: String = "",
    val email: String = "",
    val phoneNumber: String = "",
    val dateOfBirth: LocalDateTime = LocalDateTime.now(),

    @Column(name = "voter_code", unique = true, length = 8, updatable = false) var voterCode: String = "",

    @Enumerated(EnumType.STRING) val voterStatus: VoterStatus = VoterStatus.REGISTERED,

    @ManyToOne
    @JoinColumn(name = "electoral_commission_id") val electoralCommission: ElectoralCommission? = null
) {
    fun toCandidate(election: Election): Candidate {
        return Candidate(
            name = name,
            email = email,
            phoneNumber = phoneNumber,
            voterCode = voterCode,
            candidateStatus = CandidateStatus.REGISTERED,
            election = election
        )
    }
}

选举实体有一个选民名单（多对一，单向）。当有人投票时，该选民与选举的关系将结束。我无法弄清楚如何使用 jpa 实现这一点。我尝试使用election_voter表进行查询，但 jpa 给出无法解析查询的错误。查询是

从election_voter e中删除，其中e.voter_voter_code = :voterCode

sql

1 个回答

  1. Best Answer

    您在问题中提到：

    我曾尝试使用election_voter表进行查询......

    所以让我们从选民的表定义开始。

    CREATE TABLE voter
(
    voter_id    UUID PRIMARY KEY,
    election_id UUID NULL,
    FOREIGN KEY (election_id) REFERENCES election (election_id)
);

    这应该允许选民在没有选举的情况下存在。

    当仅使用此代码运行代码时，Hibernate将尝试将记录插入中间表，因此我们需要对 Election 类进行修改。

    @OneToMany 
@JoinColumn(name="election_id") // let Hibernate know about the FK
var eligibleVoters: MutableList<Voter> = mutableListOf()

    为了使用上述方法将选民从选举中分离出来，我们需要一个本机查询：

    @NativeQuery("update voter set election_id = null where voter_id = :voterId")

    这是不可取的，所以让我们尝试使用这样的查询：

    @Query("UPDATE Voter v SET v.electionId = NULL WHERE v.voterId = :voterId")

    查询表明我们需要添加electionId到选民：

    @Column(name = "election_id")
val electionId: String = "" // or var if updating it here

    并再次更改选举

    @OneToMany(mappedBy = "electionId")
var eligibleVoters: MutableList<Voter> = mutableListOf()
    • 2

相关问题