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如何仅序列化变量的名称并忽略结构中枚举字段的值(serde)

[英]How to serialise only name of variant and ignore value for enum field in struct (serde)

Given the definitions:给定定义:

#[derive(Serialize, Deserialize)]
enum Bar {
  A(i64),
  B(u64),
}

#[derive(Serialize, Deserialize)]
struct Foo {
  bar: Bar,
}

the JSON serialisation for JSON 序列化为

Foo {
  bar: Bar::A(123),
}

would be:将会:

{
  "bar": "A"
}

It would be ideal to add an attribute to the field in the struct instead of inside the enum definition (the enum will be reused in a struct field where the value needs to be serialised too)将属性添加到结构中的字段而不是枚举定义中是理想的(枚举将在值也需要序列化的结构字段中重用)

The attribute #[serde(skip)] can be used on tuple variant fields:属性#[serde(skip)]可用于元组变量字段:

use serde::{Serialize, Deserialize}; // 1.0.126
use serde_json; // 1.0.64

#[derive(Serialize, Deserialize)]
enum Bar {
    A(#[serde(skip)] i64),
    B(#[serde(skip)] u64),
}

#[derive(Serialize, Deserialize)]
struct Foo {
    bar: Bar,
}

fn main() {
    let foo = Foo { bar: Bar::A(123) };
    println!("{}", serde_json::to_string(&foo).unwrap());
}
{"bar":"A"}

If modifying Bar isn't an option, you'll have to do it a bit more manually via #[serde(with =...)] or #[serde(serialize_with =...)] :如果修改Bar不是一个选项,您必须通过#[serde(with =...)]#[serde(serialize_with =...)]手动进行一些操作:

use serde::{Serialize, Deserialize, ser::Serializer}; // 1.0.126
use serde_json; // 1.0.64

#[derive(Serialize, Deserialize)]
enum Bar {
    A(i64),
    B(u64),
}

#[derive(Serialize, Deserialize)]
struct Foo {
    #[serde(serialize_with = "bar_name_only")]
    bar: Bar,
}

fn bar_name_only<S>(bar: &Bar, serializer: S) -> Result<S::Ok, S::Error>
where
    S: Serializer,
{
    let name = match bar {
        Bar::A(_) => "A",
        Bar::B(_) => "B",
    };

    serializer.serialize_str(name)
}

fn main() {
    let foo = Foo { bar: Bar::A(123) };
    println!("{}", serde_json::to_string(&foo).unwrap());
}
{"bar":"A"}

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