如何在不包含枚举变量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量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.正如文档中所述，我为我的枚举实现了Serialize特质，但是我总是得到{"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> c3> 属性，它将产生所需的输出.您无需自己实现Serialize:

You can use the untagged attribute which will produce the desired output. You won't need to implement Serialize yourself with this:

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

如果您想自己实现Serialize，我相信您想跳过您的变体，因此您不应使用serialize_newtype_variant()，因为它会暴露您的变体.您应该直接使用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), } } }

它将产生所需的输出:

{"offset":0}