Spring JPA 双向无法评估 toString [英] Spring JPA bi-directional cannot evaluate toString

问题描述

我已经通过 @JsonIdentityInfo 解决了 JSON 递归循环到 Baeldung 的博客

请问有什么问题，为什么?

解决方案

好吧，我的猜测是 Registration.toString() 打印列表中每个付款的字符串表示，并且由于 Payment.toString() 包含Registration 的字符串表示，Registration.toString() 被再次调用，进而调用Payment.toString() 再次，依此类推.

尝试在 Payment.toString() 中返回一个空字符串以查看问题是否消失.

I have resolved JSON recursive loop with @JsonIdentityInfothrough to Baeldung's blog1 (Thanks)

But now, another error occurs :

Method threw 'java.lang.StackOverflowError' exception. Cannot evaluate com.mezoo.tdc.model.Payment.toString()

Here my Registration object :

    @Entity
    @Table(name="Registration")
    @JsonIdentityInfo( generator = ObjectIdGenerators.PropertyGenerator.class, property = "uuid")
    public class Registration implements Serializable {

       /*some private variables..*/

      // Bidirectional relationship
      @OneToMany(mappedBy="registration", cascade = {CascadeType.PERSIST, CascadeType.REMOVE, CascadeType.MERGE}, fetch = FetchType.LAZY)
      private List<Payment> payment;

                @Override
      public String toString() {
         return MoreObjects.toStringHelper(this)
            .add("payment", payment)
            .toString();
         }
    }

Now, Payment object :

    @Entity
    @Table(name="Payment")
    @JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "uuid")
    public class Payment implements Serializable {
       @ManyToOne
       @JoinColumn(name = "registration")
       private Registration registration;

       @Override
       public String toString() {
       return MoreObjects.toStringHelper(this)
            .add("registration", registration)
            .toString();
       }
    }

This is, what I see in debugger :

Please, what is wrong and why ?

解决方案

Well, my guess is that Registration.toString() prints the string representation of each payment in the list, and since Payment.toString() includes the string representation of Registration, Registration.toString() is called again, which in turn calls Payment.toString() again, and so on.

Try to return an empty string in Payment.toString() to see if the problem goes away.

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