问题
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}}
}
回答1:
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),
}
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),
}
}
}
It produces the desired output:
{"offset":0}
来源:https://stackoverflow.com/questions/52034764/how-do-i-serialize-an-enum-without-including-the-name-of-the-enum-variant