异常：参数值元素与预期类型不匹配

Question

我正在尝试使用Spring数据执行一个IN查询。我的模型如下所示：

@Entity
@Table(name = "customer", schema = "public", catalog = "postgres")
public class CustomerEntity {
    private int id;
    private String name;
    private int balance;
    private String bankId;

    @Id
    @Column(name = "id")
    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

    @Basic
    @Column(name = "name")
    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    @Basic
    @Column(name = "balance")
    public int getBalance() {
        return balance;
    }

    public void setBalance(int balance) {
        this.balance = balance;
    }

    @Basic
    @Column(name = "bank_id")
    public String getBankId() {
        return bankId;
    }

    public void setBankId(String bankId) {
        this.bankId = bankId;
    }

我的仓库界面是这样的：

@Repository
public interface TransactionsRepository extends JpaRepository<TransactionsEntity, Long> {

    List<TransactionsEntity> findByCustomerIdIn(List<CustomerEntity> customerEntities);

}

问题是当我尝试执行此代码List时

我得到这个例外：

原因：org.springframework.dao.InvalidDataAccessApiUsageException:Parameter value元素[org.example.domain.admin。CustomerEntity@6a1a2a4]与预期类型[java.lang.String（n/a）]不匹配；嵌套的异常是java.lang.IllegalArgumentException:Parameter-value元素[org.example.domain.admin]。CustomerEntity@6a1a2a4]与预期类型[java.lang.String（n/a）]不匹配。

更新：TransactionsEntity.class:

@Entity
@Table(name = "transactions", schema = "public", catalog = "postgres")
public class TransactionsEntity {

    private String id;
    private String amount;
    private String customerId;

    @Id
    @Column(name = "id")
    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    @Basic
    @Column(name = "amount")
    public String getAmount() {
        return amount;
    }

    public void setAmount(String amount) {
        this.amount = amount;
    }

    @Basic
    @Column(name = "customer_id")
    public String getCustomerId() {
        return customerId;
    }

    public void setCustomerId(String customerId) {
        this.customerId = customerId;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;

        TransactionsEntity that = (TransactionsEntity) o;

        if (id != null ? !id.equals(that.id) : that.id != null) return false;
        if (amount != null ? !amount.equals(that.amount) : that.amount != null) return false;
        if (customerId != null ? !customerId.equals(that.customerId) : that.customerId != null) return false;

        return true;
    }

    @Override
    public int hashCode() {
        int result = id != null ? id.hashCode() : 0;
        result = 31 * result + (amount != null ? amount.hashCode() : 0);
        result = 31 * result + (customerId != null ? customerId.hashCode() : 0);
        return result;
    }
}




Answer 1

正如在异常中所说的，Spring需要一个字符串，因为您的事务实体中的客户id是一个字符串，但您正在输入一个客户属性。相反，您应该输入一个列表

顺便说一句，假设您将客户id设置为客户属性的id，您的客户id不应该是int？

然后你可以做类似的事情

List<Integer> customerIds = customerEntitiesList.stream().map(CustomerEntity::getId).collect(Collectors.toList());