Use $value and $unflatten together #430
Description
I have some XML that looks like:
<root>
<state>B</state>
</root>which I would like to represent in rust as:
enum State {
A,
B,
C,
}
struct Root {
state: State,
}I have tried combining $unflatten and $value to make this happen, but it doesn't seem to work:
Cargo.toml:
[package]
name = "quick-xml-test"
version = "0.1.0"
edition = "2021"
# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html
[dependencies]
serde = { version = "1.0.139", features = ["derive"] }
quick-xml = { version = "0.23.0", features = ["serialize" ] }src/main.rs:
mod working {
use serde::{Deserialize, Serialize};
#[derive(Debug, Deserialize, Serialize)]
enum State {
A,
B,
C,
}
#[derive(Debug, Deserialize, Serialize)]
struct StateOuter {
#[serde(rename = "$value")]
state: State,
}
#[derive(Debug, Deserialize, Serialize)]
pub struct Root {
state: StateOuter,
}
}
mod not_working {
use serde::{Deserialize, Serialize};
#[derive(Debug, Deserialize, Serialize)]
#[serde(rename = "$value")]
enum State {
A,
B,
C,
}
#[derive(Debug, Deserialize, Serialize)]
pub struct Root {
#[serde(rename = "$unflatten=state")]
state: State,
}
}
fn main() {
let xml = "<root><state>B</state></root>";
println!("Deserialized (working): {:?}", quick_xml::de::from_str::<working::Root>(&xml));
println!("Deserialized (not working): {:?}", quick_xml::de::from_str::<not_working::Root>(&xml));
}Output:
Deserialized (working): Ok(Root { state: StateOuter { state: B } })
Deserialized (not working): Err(Custom("unknown variant `state`, expected one of `A`, `B`, `C`"))
Is it possible to do without introducing the extra StateOuter struct?