java - JPA 无法使用 START_OBJECT TOKEN 作为搜索键

Question

当我尝试获取之前插入数据库的更新数据时，我的代码中遇到了一些问题。 postman 的错误代码:

"JSON parse error: Can not deserialize instance of java.lang.Integer out of START_OBJECT token; nested exception is com.fasterxml.jackson.databind.JsonMappingException: Can not deserialize instance of java.lang.Integer out of START_OBJECT token\n at [Source: java.io.PushbackInputStream@647f9992; line: 1, column: 1]"

我有 2 个名为 Payment 和 Order 的实体，其中 Payment 将 Orders 作为 @OnetoOne 实体保存。付款类别:

@Entity
@Table(name = "tbl_payment")
public class Payment {

    @Id
    @GeneratedValue
    @Column(name = "payment_id")
    Integer id;
    @OneToOne(cascade=CascadeType.ALL)
    private Order order;
    @Column(name = "payment_receivable")
    private double receivable;
    @Column(name = "payment_outstanding")
    private double outstanding;
    @Column(name = "payment_status")
    private String status;

    public Payment(Order order, double receivable, double outstanding, String status) {
        this.order = order;
        this.receivable = receivable;
        this.outstanding = outstanding;
        this.status = status;
    }

    public Payment() {
    }

    @Override
    public String toString() {
        return "Payment{" +
                "id=" + id +
                ", order=" + order +
                ", receivable=" + receivable +
                ", outstanding=" + outstanding +
                ", status='" + status + '\'' +
                '}';
    }

    public Integer getId() {
        return id;
    }

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

    public Order getOrder() {
        return order;
    }

    public void setOrder(Order order) {
        this.order = order;
    }

    public double getReceivable() {
        return receivable;
    }

    public void setReceivable(double receivable) {
        this.receivable = receivable;
    }

    public double getOutstanding() {
        return outstanding;
    }

    public void setOutstanding(double outstanding) {
        this.outstanding = outstanding;
    }

    public String getStatus() {
        return status;
    }

    public void setStatus(String status) {
        this.status = status;
    }
}

订单类别:

@Entity
@Table(name = "tbl_order")
public class Order {

    @Id
    @GeneratedValue
    @Column(name = "order_id")
    Integer id;
    @Column(name = "order_amount")
    int amount;
    @Column(name = "sale_id")
    String sale_id;

    public Order(int amount, String sale_id) {
        this.amount = amount;
        this.sale_id = sale_id;
    }

    public Order() {
    }

    @Override
    public String toString() {
        return "Order{" +
                "id=" + id +
                ", amount=" + amount +
                ", sale_id='" + sale_id + '\'' +
                '}';
    }

    public Integer getId() {
        return id;
    }

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

    public int getAmount() {
        return amount;
    }

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

    public String getSale_id() {
        return sale_id;
    }

    public void setSale_id(String sale_id) {
        this.sale_id = sale_id;
    }
}

在我的 Controller 中，我将传入 payment_id 并使用它来将付款条目从未付状态更新为已付状态。这就是我收到上述错误的地方。

服务片段:

@Transactional
    @Modifying
    public Payment clearPaymentStatus(Integer p){
        TypedQuery query = em.createQuery("update Payment p set p.payment_status = ?1 where p.id = ?5", Payment.class);
        query.setParameter(1, "PAID");
        query.setParameter(5, new Integer(p));
        return ((Payment) query.getSingleResult());
    }

更新: Controller 片段

@RequestMapping("/clearpayment")
public @ResponseBody ResponseEntity<Payment> ClearPayment(@RequestBody Integer payment_id){
    return new ResponseEntity<Payment>( paymentService.clearPaymentStatus(payment_id), HttpStatus.ACCEPTED);
}

postman 的添加工作正常，只有更新有问题。任何建议都会很棒。谢谢!

Answer 1

JPA 与 JSON 无关。 JSON 只是一种输入格式。您可能正在尝试通过某种 Controller 提供一些 json 对象。该 Controller 隐式尝试为您转换对象，但失败了。通常，发生这种情况是因为 JSON 格式无效，或者至少无法映射到对象层次结构。因此，请仔细检查您的 JSON 并尝试了解应如何映射它。

现在，很难从您发布的数据中得知更多信息，但很明显您在引擎期望看到整数的位置提供了一个对象(方括号内的对象)。它内部有一些关系映射。

因此，正如我们的同事已经说过的，恰好序列化期间失败的对象，稍后在流程中，应该传递给 JPA 引擎进行数据库操作，但这个特定问题与 JPA 无关.