rust - 有没有办法用 Serde 定义标签字段？

Question

我想要这样的东西:

#[derive(Debug, Serialize, Deserialize)]
struct MyStruct {
    field1: String,
    field2: Option<u64>,
    #[serde(tag(value = "tag_value"))]
    tag: ()
}

#[serde(tag(value = "tag_value"))] 并不是 Serde 提供的实际属性，它只是表达一个想法。我知道我可以自己完成所有序列化，使用远程、(de)serialize_with 等，但这些需要大量样板代码。

思路是标签字段必须存在，序列化为JSON的MyStruct应该是:

{
    "field1": "foo",
    "field2": 42,
    "tag": "tag_value"
}

如果“tag”字段缺失或映射到与“tag_value”不同的值，反序列化一定会失败。

Answer 1

使用单个变体枚举:

use serde; // 1.0.104
use serde_json; // 1.0.48

#[derive(Debug, serde::Serialize, serde::Deserialize)]
#[serde(rename_all = "snake_case")]
enum Tag {
    TagValue,
}

#[derive(Debug, serde::Serialize, serde::Deserialize)]
struct MyStruct {
    field1: String,
    field2: Option<u64>,
    tag: Tag,
}

fn main() {
    let s = MyStruct {
        field1: "foo".to_string(),
        field2: Some(42),
        tag: Tag::TagValue,
    };

    // The tag is included when serializing
    println!("{:?}", serde_json::to_string(&s));

    // Tag is required when deserializing
    println!(
        "{:?}",
        serde_json::from_str::<MyStruct>(
            "{\"field1\":\"foo\",\"field2\":42,\"tag\":\"tag_value\"}"
        )
    );
    println!(
        "{:?}",
        serde_json::from_str::<MyStruct>("{\"field1\":\"foo\",\"field2\":42}")
    );

    // A bad tag fails
    println!(
        "{:?}",
        serde_json::from_str::<MyStruct>("{\"field1\":\"foo\",\"field2\":42,\"tag\":\"oops\"}")
    );
}

这打印

Ok("{\"field1\":\"foo\",\"field2\":42,\"tag\":\"tag_value\"}")
Ok(MyStruct { field1: "foo", field2: Some(42), tag: TagValue })
Err(Error("missing field `tag`", line: 1, column: 28))
Err(Error("unknown variant `oops`, expected `tag_value`", line: 1, column: 40))

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