如何在不包含枚举变体名称的情况下序列化枚举?

本文介绍了如何在不包含枚举变体名称的情况下序列化枚举?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在尝试将枚举序列化为 JSON 字符串.我按照文档中的描述为我的枚举实现了 Serialize 特性，但我总是得到 {"offset":{"Int":0}} 而不是所需的{"offset":0}.

I am trying to serialize an enum to a JSON string. I implemented Serialize trait for my enum as it is described in the docs, but I always get {"offset":{"Int":0}} instead of the desired {"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}} }

推荐答案

您可以使用 untagged 属性将产生所需的输出.你不需要自己实现 Serialize :

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

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

If you wanted to implement Serialize yourself, I believe you want to skip your variant so you should not use serialize_newtype_variant() as it exposes your variant. You should use serialize_str() and serialize_i32() directly:

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}