使用 Serde 反序列化时如何忽略额外的元组项? ("trailing characters" 错误)
How can I ignore extra tuple items when deserializing with Serde? ("trailing characters" error)
Serde 在反序列化为常规结构时忽略未知的命名字段。在反序列化为元组结构(例如来自异构 JSON 数组)时,我如何类似地忽略额外的项目?
例如,此代码忽略额外的 "c"
字段就好了:
#[derive(Serialize, Deserialize, Debug)]
pub struct MyStruct { a: String, b: i32 }
fn test_deserialize() -> MyStruct {
::serde_json::from_str::<MyStruct>(r#"
{
"a": "foo",
"b": 123,
"c": "ignore me"
}
"#).unwrap()
}
// => MyStruct { a: "foo", b: 123 }
相比之下,这对元组中的额外项目失败:
#[derive(Serialize, Deserialize, Debug)]
pub struct MyTuple(String, i32);
fn test_deserialize_tuple() -> MyTuple {
::serde_json::from_str::<MyTuple>(r#"
[
"foo",
123,
"ignore me"
]
"#).unwrap()
}
// => Error("trailing characters", line: 5, column: 13)
我想在我的数据格式中允许额外的项目以实现向前兼容性。让 Serde 在反序列化时忽略额外的元组项的最简单方法是什么?
您可以实施忽略序列其余部分的自定义 Visitor
。请注意,必须消耗整个序列。这是一个重要的部分(尝试删除它,你会得到同样的错误):
// This is very important!
while let Some(IgnoredAny) = seq.next_element()? {
// Ignore rest
}
这是一个工作示例:
use std::fmt;
use serde::de::{self, Deserialize, Deserializer, IgnoredAny, SeqAccess, Visitor};
use serde::Serialize;
#[derive(Serialize, Debug)]
pub struct MyTuple(String, i32);
impl<'de> Deserialize<'de> for MyTuple {
fn deserialize<D>(deserializer: D) -> Result<Self, D::Error>
where
D: Deserializer<'de>,
{
struct MyTupleVisitor;
impl<'de> Visitor<'de> for MyTupleVisitor {
type Value = MyTuple;
fn expecting(&self, formatter: &mut fmt::Formatter) -> fmt::Result {
formatter.write_str("struct MyTuple")
}
fn visit_seq<V>(self, mut seq: V) -> Result<Self::Value, V::Error>
where
V: SeqAccess<'de>,
{
let s = seq
.next_element()?
.ok_or_else(|| de::Error::invalid_length(0, &self))?;
let n = seq
.next_element()?
.ok_or_else(|| de::Error::invalid_length(1, &self))?;
// This is very important!
while let Some(IgnoredAny) = seq.next_element()? {
// Ignore rest
}
Ok(MyTuple(s, n))
}
}
deserializer.deserialize_seq(MyTupleVisitor)
}
}
fn main() {
let two_elements = r#"["foo", 123]"#;
let three_elements = r#"["foo", 123, "bar"]"#;
let tuple: MyTuple = serde_json::from_str(two_elements).unwrap();
assert_eq!(tuple.0, "foo");
assert_eq!(tuple.1, 123);
let tuple: MyTuple = serde_json::from_str(three_elements).unwrap();
assert_eq!(tuple.0, "foo");
assert_eq!(tuple.1, 123);
}
对于 JSON,我将结合 RawValue
和自定义反序列化:
use serde::{Deserialize, Deserializer};
#[derive(Debug)]
struct MyTuple(String, i32);
#[derive(Deserialize, Debug)]
struct MyTupleFutureCompat<'a>(
String,
i32,
#[serde(default, borrow)] Option<&'a serde_json::value::RawValue>,
);
impl<'de> Deserialize<'de> for MyTuple {
fn deserialize<D>(deserializer: D) -> Result<Self, D::Error>
where
D: Deserializer<'de>,
{
let t: MyTupleFutureCompat = Deserialize::deserialize(deserializer)?;
Ok(MyTuple(t.0, t.1))
}
}
fn main() -> Result<(), Box<dyn std::error::Error>> {
let json = r#"[
"foo",
123,
"ignore me"
]"#;
let d: MyTuple = serde_json::from_str(json)?;
println!("{:?}", d);
Ok(())
}
另请参阅:
- How to transform fields during deserialization using Serde?
Serde 在反序列化为常规结构时忽略未知的命名字段。在反序列化为元组结构(例如来自异构 JSON 数组)时,我如何类似地忽略额外的项目?
例如,此代码忽略额外的 "c"
字段就好了:
#[derive(Serialize, Deserialize, Debug)]
pub struct MyStruct { a: String, b: i32 }
fn test_deserialize() -> MyStruct {
::serde_json::from_str::<MyStruct>(r#"
{
"a": "foo",
"b": 123,
"c": "ignore me"
}
"#).unwrap()
}
// => MyStruct { a: "foo", b: 123 }
相比之下,这对元组中的额外项目失败:
#[derive(Serialize, Deserialize, Debug)]
pub struct MyTuple(String, i32);
fn test_deserialize_tuple() -> MyTuple {
::serde_json::from_str::<MyTuple>(r#"
[
"foo",
123,
"ignore me"
]
"#).unwrap()
}
// => Error("trailing characters", line: 5, column: 13)
我想在我的数据格式中允许额外的项目以实现向前兼容性。让 Serde 在反序列化时忽略额外的元组项的最简单方法是什么?
您可以实施忽略序列其余部分的自定义 Visitor
。请注意,必须消耗整个序列。这是一个重要的部分(尝试删除它,你会得到同样的错误):
// This is very important!
while let Some(IgnoredAny) = seq.next_element()? {
// Ignore rest
}
这是一个工作示例:
use std::fmt;
use serde::de::{self, Deserialize, Deserializer, IgnoredAny, SeqAccess, Visitor};
use serde::Serialize;
#[derive(Serialize, Debug)]
pub struct MyTuple(String, i32);
impl<'de> Deserialize<'de> for MyTuple {
fn deserialize<D>(deserializer: D) -> Result<Self, D::Error>
where
D: Deserializer<'de>,
{
struct MyTupleVisitor;
impl<'de> Visitor<'de> for MyTupleVisitor {
type Value = MyTuple;
fn expecting(&self, formatter: &mut fmt::Formatter) -> fmt::Result {
formatter.write_str("struct MyTuple")
}
fn visit_seq<V>(self, mut seq: V) -> Result<Self::Value, V::Error>
where
V: SeqAccess<'de>,
{
let s = seq
.next_element()?
.ok_or_else(|| de::Error::invalid_length(0, &self))?;
let n = seq
.next_element()?
.ok_or_else(|| de::Error::invalid_length(1, &self))?;
// This is very important!
while let Some(IgnoredAny) = seq.next_element()? {
// Ignore rest
}
Ok(MyTuple(s, n))
}
}
deserializer.deserialize_seq(MyTupleVisitor)
}
}
fn main() {
let two_elements = r#"["foo", 123]"#;
let three_elements = r#"["foo", 123, "bar"]"#;
let tuple: MyTuple = serde_json::from_str(two_elements).unwrap();
assert_eq!(tuple.0, "foo");
assert_eq!(tuple.1, 123);
let tuple: MyTuple = serde_json::from_str(three_elements).unwrap();
assert_eq!(tuple.0, "foo");
assert_eq!(tuple.1, 123);
}
对于 JSON,我将结合 RawValue
和自定义反序列化:
use serde::{Deserialize, Deserializer};
#[derive(Debug)]
struct MyTuple(String, i32);
#[derive(Deserialize, Debug)]
struct MyTupleFutureCompat<'a>(
String,
i32,
#[serde(default, borrow)] Option<&'a serde_json::value::RawValue>,
);
impl<'de> Deserialize<'de> for MyTuple {
fn deserialize<D>(deserializer: D) -> Result<Self, D::Error>
where
D: Deserializer<'de>,
{
let t: MyTupleFutureCompat = Deserialize::deserialize(deserializer)?;
Ok(MyTuple(t.0, t.1))
}
}
fn main() -> Result<(), Box<dyn std::error::Error>> {
let json = r#"[
"foo",
123,
"ignore me"
]"#;
let d: MyTuple = serde_json::from_str(json)?;
println!("{:?}", d);
Ok(())
}
另请参阅:
- How to transform fields during deserialization using Serde?