如何在不包含枚举变体名称的情况下序列化枚举？

Question

我试图将枚举序列化为JSON字符串.我Serialize为我的枚举实现了特征,因为它在文档中有描述,但我总是得到{"offset":{"Int":0}}而不是想要的{"offset":0}.

extern crate serde;
extern crate serde_json;

use std::collections::HashMap;

use serde::ser::{Serialize, Serializer};

#[derive(Debug)]
enum TValue<'a> {
    String(&'a str),
    Int(&'a i32),
}

impl<'a> Serialize for TValue<'a> {
    fn serialize<S>(&self, serializer: S) -> Result<S::Ok, S::Error>
    where
        S: Serializer,
    {
        match *self {
            TValue::String(ref s) => serializer.serialize_newtype_variant("TValue", 0, "String", s),
            TValue::Int(i) => serializer.serialize_newtype_variant("TValue", 1, "Int", &i),
        }
    }
}

fn main() {
    let offset: i32 = 0;
    let mut request_body = HashMap::new();
    request_body.insert("offset", TValue::Int(&offset));
    let serialized = serde_json::to_string(&request_body).unwrap();
    println!("{}", serialized); // {"offset":{"Int":0}}
}

Answer 1

您可以使用untagged将生成所需输出的属性.你不需要Serialize用这个来实现自己:

#[derive(Debug, Serialize)]
#[serde(untagged)]
enum TValue<'a> {
    String(&'a str),
    Int(&'a i32),
}

如果你想Serialize自己实现,我相信你想跳过你的变体,所以你不应该使用serialize_newtype_variant()它,因为它暴露你的变体.您应该使用 serialize_str()和serialize_i32()直接:

impl<'a> Serialize for TValue<'a> {
    fn serialize<S>(&self, serializer: S) -> Result<S::Ok, S::Error>
    where
        S: Serializer,
    {
        match *self {
            TValue::String(s) => serializer.serialize_str(s),
            TValue::Int(i) => serializer.serialize_i32(*i),
        }
    }
}

它产生所需的输出:

{"offset":0}