Java 如何使用 Jackson 定义可选的 json 字段
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How to define optional json field using Hymanson
提问by mkorszun
I have an object with one optional field and can not find proper annotation to model it. Any ideas what is the proper way to do it with Hymanson?
我有一个带有一个可选字段的对象，但找不到合适的注释来对其进行建模。任何想法与Hyman逊一起做的正确方法是什么？
采纳答案by mwhs
In Hymanson you cannot make the difference between optional and non-optional fields. Just declare any field in your POJO. If a field is not present in your JSON structure then Hymanson will not call the setter. You may keep track of wether a setter has been called with a flag in the POJO.
在 Hymanson 中，您无法区分可选字段和非可选字段。只需在您的 POJO 中声明任何字段。如果您的 JSON 结构中不存在字段，则 Hymanson 将不会调用 setter。您可以跟踪是否在 POJO 中使用标志调用了 setter。
回答by pyb
Coming late to the party...
聚会迟到了...
Using Hymanson 2.8.6 via Spring HttpMessageConverter 4.3.6, I had to change my setterparameter to the unwrapped type, like so:
通过 Spring HttpMessageConverter 4.3.6 使用 Hymanson 2.8.6，我不得不将我的 setter参数更改为解包类型，如下所示：
class Foo {
private Optional<Bar> bar;
public void setBar(Bar bar) { // NOT Optional<Bar>, this gives me Optional.empty()
this.bar = Optional.of(bar);
}
// getter doesn't need to be changed
}
