可序列化结构字段的多种可能类型

Question

我正在编写一个简单的书签管理器，它将以这样的 JSON 格式保存数据；

{
    "children": [
        {
            "name": "Social",
            "children": [
                {
                    "name": "Facebook",
                    "url": "https://facebook.com",
                    "faviconUrl": "",
                    "tags": [],
                    "keyword": "",
                    "createdAt": 8902351,
                    "modifiedAt": 90981235
                }
            ],
            "createdAt": 235123534,
            "modifiedAt": 23531235
        }
    ]
}

我试图通过创建一个共同的 Entry 特征来编写 children 字段以允许两种可能的类型（Directory 和 Bookmark），但我正在打一堵墙，因为我无法从 serde 为 Entry 特征实现 Serialize 特征。

use serde::{Serialize, Deserialize, Serializer};

#[derive(Serialize, Deserialize)]
struct Root<'a> {
    children: Vec<&'a dyn Entry>,
}

#[derive(Serialize, Deserialize)]
struct Directory<'a> {
    name: String,
    created_at: u64,
    children: Vec<&'a dyn Entry>,
    modified_at: u64
}

#[derive(Serialize, Deserialize)]
struct Bookmark {
    name: String,
    url: String,
    favicon_url: String,
    tags: Vec<String>,
    keyword: String,
    created_at: u64,
    modified_at: u64,
}

trait Entry {
    fn is_directory(&self) -> bool;
}

impl Entry for Directory<'_> {
    fn is_directory(&self) -> bool {
        true
    }
}

impl Entry for Bookmark {
    fn is_directory(&self) -> bool {
        false
    }
}

// can't do this
impl Serialize for Entry {}

是否有可能完成这项工作，或者我应该创建一个不包含具有多个可能值的字段的不同结构？我正在考虑将 JSON 加载为 HashMap<String, serde_json::Value> 并循环遍历哈希映射，但我想知道是否有更优雅的方法来执行此操作。

Answer 1

如果您只是将 Entry 设置为枚举而不是特征，并将 &dyn Entry 更改为 Entry，那么一切都应该正常工作，除了您最终会在您的 JSON，以及一个附加标签条目，告诉您该条目的类型。正如 Masklinn 在评论中指出的那样，这种情况也不正确，但可以使用 #[serde(rename_all = "camelCase")].

修复

use serde::{Deserialize, Serialize};

#[derive(Serialize, Deserialize)]
#[serde(rename_all = "camelCase")]
struct Root {
    children: Vec<Entry>,
}

#[derive(Serialize, Deserialize)]
#[serde(rename_all = "camelCase")]
struct Directory {
    name: String,
    created_at: u64,
    children: Vec<Entry>,
    modified_at: u64,
}

#[derive(Serialize, Deserialize)]
#[serde(rename_all = "camelCase")]
struct Bookmark {
    name: String,
    url: String,
    favicon_url: String,
    tags: Vec<String>,
    keyword: String,
    created_at: u64,
    modified_at: u64,
}

#[derive(Serialize, Deserialize)]
#[serde(rename_all = "camelCase")]
enum Entry {
    Directory(Directory),
    Bookmark(Bookmark),
}

如果你真的不想要额外的关卡和标签，那么你可以使用serde(untagged)注释来Entry。

#[derive(Deserialize, Serialize, Debug)]
#[serde(untagged)]
enum Entry {
    Directory(Directory),
    Bookmark(Bookmark),
}

如果您需要更多的灵活性，您可以创建一个包含两者所有字段的中间结构 BookmarkOrDirectory，其中只出现在其中一个的字段为 Option，然后实现 TryFrom<BookmarkOrDirectory> 为 Entry 并使用 serde(try_from=...) and serde(into=...) 将 to/from 转换为适当的形式。下面是一个示例实现。它编译，但有一些 todo! 分散在其中，并使用 String 作为错误类型，这是 hacky - 当然未经测试。

use core::convert::TryFrom;
use serde::{Deserialize, Serialize};

#[derive(Serialize, Deserialize)]
struct Root {
    children: Vec<Entry>,
}

#[derive(Clone)]
struct Directory {
    name: String,
    created_at: u64,
    children: Vec<Entry>,
    modified_at: u64,
}

#[derive(Clone)]
struct Bookmark {
    name: String,
    url: String,
    favicon_url: String,
    tags: Vec<String>,
    keyword: String,
    created_at: u64,
    modified_at: u64,
}

#[derive(Serialize, Deserialize, Clone)]
#[serde(try_from = "BookmarkOrDirectory", into = "BookmarkOrDirectory")]
enum Entry {
    Directory(Directory),
    Bookmark(Bookmark),
}

#[derive(Serialize, Deserialize)]
struct BookmarkOrDirectory {
    name: String,
    url: Option<String>,
    favicon_url: Option<String>,
    tags: Option<Vec<String>>,
    keyword: Option<String>,
    created_at: u64,
    modified_at: u64,
    children: Option<Vec<Entry>>,
}

impl BookmarkOrDirectory {
    pub fn to_directory(self) -> Result<Directory, (Self, String)> {
        // Check all the fields are there
        if !self.children.is_some() {
            return Err((self, "children is not set".to_string()));
        }
        // TODO: Check extra fields are not there
        Ok(Directory {
            name: self.name,
            created_at: self.created_at,
            children: self.children.unwrap(),
            modified_at: self.modified_at,
        })
    }
    pub fn to_bookmark(self) -> Result<Bookmark, (Self, String)> {
        todo!()
    }
}

impl TryFrom<BookmarkOrDirectory> for Entry {
    type Error = String;
    fn try_from(v: BookmarkOrDirectory) -> Result<Self, String> {
        // Try to parse it as direcory
        match v.to_directory() {
            Ok(directory) => Ok(Entry::Directory(directory)),
            Err((v, mesg1)) => {
                // if that fails try to parse it as bookmark
                match v.to_bookmark() {
                    Ok(bookmark) => Ok(Entry::Bookmark(bookmark)),
                    Err((_v, mesg2)) => Err(format!("unable to convert to entry - not a bookmark since '{}', not a directory since '{}'", mesg2, mesg1))
                }
            }
        }
    }
}

impl Into<BookmarkOrDirectory> for Bookmark {
    fn into(self) -> BookmarkOrDirectory {
        todo!()
    }
}

impl Into<BookmarkOrDirectory> for Directory {
    fn into(self) -> BookmarkOrDirectory {
        todo!()
    }
}

impl Into<BookmarkOrDirectory> for Entry {
    fn into(self) -> BookmarkOrDirectory {
        match self {
            Entry::Bookmark(bookmark) => bookmark.into(),
            Entry::Directory(directory) => directory.into(),
        }
    }
}

可序列化结构字段的多种可能类型

Multiple possible types for a serializable struct's field

traits

rust

serde