我正在编写一个简单的书签管理器,它将以 JSON 格式保存数据,如下所示;
{
"children": [
{
"name": "Social",
"children": [
{
"name": "Facebook",
"url": "https://facebook.com",
"faviconUrl": "",
"tags": [],
"keyword": "",
"createdAt": 8902351,
"modifiedAt": 90981235
}
],
"createdAt": 235123534,
"modifiedAt": 23531235
}
]
}
我尝试写children
通过创建公共(public) Directory
字段允许两种可能的类型( Bookmark
和 Entry
)特质,但我遇到了困难,因为我无法实现 Serialize
来自serde
的特征对于Entry
特质。
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 {}
是否有可能实现这项工作,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 HashMap ,但我想知道是否有一些更优雅的方法来做到这一点。
最佳答案
如果您只是将 Entry 设为枚举而不是特征,并更改 &dyn Entry
到只是Entry
,那么一切都应该正常工作,除非您最终会在 JSON 中得到一个额外的级别,以及一个额外的标记条目,告诉您该条目是什么类型。正如 Masklinn 在评论中指出的那样,这种情况也是不正确的,但可以使用 #[serde(rename_all = "camelCase")]
进行修复。 .
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),
}
如果您确实不需要额外的级别和标签,那么您可以使用 serde(untagged)注释 Entry
.
#[derive(Deserialize, Serialize, Debug)]
#[serde(untagged)]
enum Entry {
Directory(Directory),
Bookmark(Bookmark),
}
如果您需要更多的灵活性,您可以创建一个中间结构 BookmarkOrDirectory
包含两者的所有字段,其中仅出现在其中一个的字段为 Option
然后执行TryFrom<BookmarkOrDirectory>
对于 Entry
并使用serde(try_from=...)和 serde(into=...)转换为适当的形式/从适当的形式转换。下面是一个示例实现。它可以编译,但有一些 todo!
分散在其中,并使用String
作为一种错误类型,这是很hacky的 - 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(),
}
}
}
关于rust - 可序列化结构体字段的多种可能类型,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/67594909/