杰克逊将缺少的属性反序列化为空Optional< T> [英] Jackson. Deserialize missing properties as empty Optional<T>

问题描述

假设我有一个这样的课程:

Let's say I have a class like this:

public static class Test {

    private Optional<String> something;

    public Optional<String> getSomething() {
        return something;
    }

    public void setSomething(Optional<String> something) {
        this.something = something;
    }
    
}

如果我反序列化此JSON，则会得到一个空的Optional:

If I deserialize this JSON, I get an empty Optional:

{"something":null}

但是如果缺少属性(在这种情况下只是空JSON)，我将得到null而不是 Optional< T> .我当然可以自己初始化字段，但是我认为最好有一种用于 null 和缺少属性的机制.那么有没有办法使杰克逊将缺少的属性反序列化为空的 Optional< T> ?

But if property is missing (in this case just empty JSON), I get null instead of Optional<T>. I could initialize fields by myself of course, but I think it would be better to have one mechanism for null and missing properties. So is there a way to make jackson deserialize missing properties as empty Optional<T>?

推荐答案

可选实际上并不是要用作字段，而应更多地用作返回值.为什么没有:

Optional is not really meant to be used as a field but more as a return value. Why not have:

public static class Test {
  private String something;
  public Optional<String> getSomething() {
    return Optional.ofNullable(something);
  }
  public void setSomething(String something) {
    this.something = something;
  }
}

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