Closed BenJeau closed 9 months ago
Use #[ts(optional)]
For individual fields labeled as #[serde(default)]
, ts_rs could label the fields as optional. Currently a #[ts(optional)]
marker only works on Option<>
types, though. This would help me specifically with boolean types which are often given as optional types and only serialized if true
.
For individual fields labeled as
#[serde(default)]
, ts_rs could label the fields as optional. Currently a#[ts(optional)]
marker only works onOption<>
types, though. This would help me specifically with boolean types which are often given as optional types and only serialized iftrue
.
You could do it like this:
use serde::Serialize;
#[derive(TS, Serialize)]
#[ts(export)]
struct Foo {
#[ts(optional)]
#[serde(skip_serializing_if = "should_skip", default = "Default::default")]
my_optional_bool: Option<bool>,
}
fn should_skip(field: &Option<bool>) -> bool {
matches!(field, None | Some(false))
}
Right, this is what we were doing before. However, because Rust doesn't have a True
object, our code was required to "know" that the Option was always Option(true) or None, but never Option(false), and sometimes we would unwittingly add Option(false) in there if, for example, option chaining wasn't correctly written. So my preference would be to allow ts_rs to generate the question mark property if there was a serde(default), as that should in theory allow for the property to be declared optional.
I think this is data modeling error. You're saying your data should only have two valid states, but it has three. Something more appropriate would be using Option<()>
and implementing (De)Serialize to convert it to ?: boolean
:
use ts_rs::TS;
use serde::{Deserialize, Serialize};
#[derive(TS, Serialize, Deserialize, PartialEq, Debug)]
#[ts(export)]
struct Foo {
#[ts(optional, as = "Option<bool>")]
#[serde(skip_serializing_if = "Option::is_none", default = "Default::default", with = "deser")]
my_optional_bool: Option<()>,
}
mod deser {
use serde::{Serializer, Serialize, Deserializer, Deserialize};
pub fn serialize<S: Serializer>(value: &Option<()>, serializer: S) -> Result<S::Ok, S::Error> {
value.map(|_| true).serialize(serializer)
}
pub fn deserialize<'de, D: Deserializer<'de>>(deserializer: D) -> Result<Option<()>, D::Error> {
Ok(Option::<bool>::deserialize(deserializer)?.filter(|&x| x).map(|_| ()))
}
}
#[test]
fn test() {
let none = Foo { my_optional_bool: None };
let some = Foo { my_optional_bool: Some(()) };
// Type definition
assert_eq!(Foo::inline(), "{ my_optional_bool?: boolean, }");
// Serializing
assert_eq!(serde_json::to_string(&none).unwrap(), "{}");
assert_eq!(serde_json::to_string(&some).unwrap(), r#"{"my_optional_bool":true}"#);
// Deserializing
assert_eq!(serde_json::from_str::<Foo>(r#"{"my_optional_bool":true}"#).unwrap(), some);
assert_eq!(serde_json::from_str::<Foo>(r#"{"my_optional_bool":false}"#).unwrap(), none); // `false` becomes `None`!
assert_eq!(serde_json::from_str::<Foo>("{}").unwrap(), none);
}
If you really want to use Option<bool>
, then your code should handle Some(false)
, because it is a valid value for that type, and therefore, sooner or later, it will show up in your codebase. You can't just go "Eh, it'll never happen", because as you said, it already does. Make your invalid states unrepresentable with Option<()>
or handle all representable states appropriately with something like
match optional_bool {
Some(true) => todo!("True case"),
Some(false) | None => todo!("False case")
}
or
if optional_bool.filter(|&x| x).is_some() {
todo!("True case"),
} else {
todo!("False case")
}
or something else that handles all three cases
My usecase is one where nonexistence of a value is equivalent to passing that value as false. Serde allows that using the default macro. In our code, we treat the value as boolean because there is either true or [false/undefined]. We don’t want to use Option
Unfortunately, ts_rs only allows for optional fields to be declared as Option
For me, it makes complete sense for a non-Option field with #[serde(default)] to be rendered as “field?: type” in ts. I would support a feature that made that conversion possible.
For me, it makes complete sense for a non-Option field with #[serde(default)] to be rendered as “field?: type” in ts. I would support a feature that made that conversion possible.
We've had something somewhat similar in a previous version, but ended up removing it.
The problem is that we don't know what you're going to do with your types - If you're sending them to a rust server, for example, then yeah, #[serde(default)]
should be field?: Type
, but if they represent a server response, then they should be field: Type
.
Like #[serde(default)]
, there are a lot of "asymmetric" serde attributes we don't support, like skip_serializing_if
, skip_serializing
and skip_deserializing
. Whichever way we handeled them, it'd be wrong in 50% of the cases.
So yeah, this all gets kinda tricky.
Whether we can make #[ts(optional)]
work on non-Option
fields might be worth investigating, but I'm afraid we'll hit lots of edge-cases doing that. Maybe just doing #[ts(type = "number | undefined")]
is good enough?
In my mind ts_rs is a code generator, but doesn’t explicitly guarantee that the generated data is consistent or logical. So for our case, generating an optional parameter with a question mark should be something the user can do at their own behest. In typescript, a parameter that is declared optional can be removed from the instantiation, but one declared type | optional
still needs to be instantiated.
I think a ts(optional) would be great that supports arbitrary objects. And yes, if we were to be fully bidirectionally consistent then we would need to use serde(default) and skip_serializing_if(x). But that’s our own responsibility IMO.
Fair enough! I've noted this in #294.
Some thought needs to be put into this to figure out if
#[ts(optional)]
for any type
#[ts(optional)]
are to add the ?
, but to remove the | null
from the type.#[ts(optional)]
should work.Given that ts optional is only available for Option types, this won’t be a breaking release if the functionality is kept as is for option types. In fact it should basically “just work” if you remove the requirement for Option in the macro. Can submit a PR if helpful
I think it's important that #[ts(optional)]
behaves predictably, so using it on Option<T>
and T
should do the same.
However, on Option<T>
, it generates field?: T
.
Only #[ts(optional = nullable)]
generates field?: T | null
. That's the behaviour we'd want when used on a non-Option
type - keep the type the same, and add a ?
to the field.
So right now, we could support #[ts(optional = nullable)]
on every type, while disallowing #[ts(optional)]
.
Or we accept that they behave differently, but I'm really not sure about that.
I prefer that our rust code use boolean rather than Option<()>.
Then this becomes even simpler, no custom (de)serialization required:
use ts_rs::TS;
use serde::{Deserialize, Serialize};
#[derive(TS, Serialize, Deserialize, PartialEq, Debug)]
#[ts(export)]
struct Foo {
#[ts(optional, as = "Option<bool>")]
#[serde(skip_serializing_if = "std::ops::Not::not", default = "Default::default")]
my_optional_bool: bool,
}
#[test]
fn test() {
let falsy = Foo { my_optional_bool: false };
let truthy = Foo { my_optional_bool: true };
// Type definition
assert_eq!(Foo::inline(), "{ my_optional_bool?: boolean, }");
// Serializing
assert_eq!(serde_json::to_string(&falsy).unwrap(), "{}"); // `false` is not serialized
assert_eq!(serde_json::to_string(&truthy).unwrap(), r#"{"my_optional_bool":true}"#);
// Deserializing
assert_eq!(serde_json::from_str::<Foo>(r#"{"my_optional_bool":true}"#).unwrap(), truthy);
assert_eq!(serde_json::from_str::<Foo>(r#"{"my_optional_bool":false}"#).unwrap(), falsy);
assert_eq!(serde_json::from_str::<Foo>("{}").unwrap(), falsy); // `undefined` is deserialized into `false`
}
With #[ts(as = "...")]
you can just make TS
interpret your type as if it were Option<T>
instead of T
, and then add #[ts(optional)]
to change the type to field?: T
instead of field: T | null
.
The feature you're asking for already exists through as
+ optional
That's a very nice solution! That didn't even cross my mind.
This is a good intermediate solution for sure! It would be nice to have a solution that can use the type as defined, so that there's no duplication between as = Option<X>
and field: X
But I appreciate that recommendation, I can implement that solution now.
As a real example, here's my (now updated) definition of a struct:
#[derive(Debug, Deserialize, Serialize, TS, Clone, PartialEq)]
#[serde(rename_all = "camelCase")]
#[ts(export, export_to = "ts/")]
pub struct FilePropertySelection {
#[ts(optional, as = "Option<bool>")]
#[serde(default, skip_serializing_if = "std::ops::Not::not")]
pub everything: bool,
/// Return all properties within the given groups (along with properties)
#[ts(optional, as = "Option<HashSet<EntityPropertyGroupId>>")]
#[serde(default, skip_serializing_if = "HashSet::is_empty")]
pub groups: HashSet<EntityPropertyGroupId>,
/// Return all given properties (along with groups)
/// WARN: If a property is specified, and it doesn't exist, and error will
/// occur
#[ts(optional, as = "Option<HashSet<FilePropertyId>>")]
#[serde(default, skip_serializing_if = "HashSet::is_empty")]
pub properties: HashSet<FilePropertyId>,
}
As you can see, the use of as = ...
does work, it just doesn't quite get me to the point where I feel like the problem is "solved". I would love to not have to use the duplicative syntax if possible. But again, this is a choice for your side, as I can technically achieve what I want.
I would love to not have to use the duplicative syntax if possible.
We could allow as
to parse Option<_>
and have _
be converted into the field's original type
use ts_rs::TS;
use serde::{Deserialize, Serialize};
#[derive(TS, Serialize, Deserialize, PartialEq, Debug)]
#[ts(export)]
struct Foo {
#[ts(optional, as = "Option<_>")]
#[serde(skip_serializing_if = "std::ops::Not::not", default)]
my_optional_bool: bool,
}
#[test]
fn test() {
let falsy = Foo { my_optional_bool: false };
let truthy = Foo { my_optional_bool: true };
// Type definition
assert_eq!(Foo::inline(), "{ my_optional_bool?: boolean, }");
// Serializing
assert_eq!(serde_json::to_string(&falsy).unwrap(), "{}"); // `false` is not serialized
assert_eq!(serde_json::to_string(&truthy).unwrap(), r#"{"my_optional_bool":true}"#);
// Deserializing
assert_eq!(serde_json::from_str::<Foo>(r#"{"my_optional_bool":true}"#).unwrap(), truthy);
assert_eq!(serde_json::from_str::<Foo>(r#"{"my_optional_bool":false}"#).unwrap(), falsy);
assert_eq!(serde_json::from_str::<Foo>("{}").unwrap(), falsy); // `undefined` is deserialized into `false`
}
Your example would then become
#[derive(Debug, Deserialize, Serialize, TS, Clone, PartialEq)]
#[serde(rename_all = "camelCase")]
#[ts(export, export_to = "ts/")]
pub struct FilePropertySelection {
#[ts(optional, as = "Option<_>")]
#[serde(default, skip_serializing_if = "std::ops::Not::not")]
pub everything: bool,
/// Return all properties within the given groups (along with properties)
#[ts(optional, as = "Option<_>")]
#[serde(default, skip_serializing_if = "HashSet::is_empty")]
pub groups: HashSet<EntityPropertyGroupId>,
/// Return all given properties (along with groups)
/// WARN: If a property is specified, and it doesn't exist, and error will
/// occur
#[ts(optional, as = "Option<_>")]
#[serde(default, skip_serializing_if = "HashSet::is_empty")]
pub properties: HashSet<FilePropertyId>,
}
Sounds great!
I still think the best/easiest solution is to just allow for the #[ts(optional)]
to be used on non-option types, but your solution is appropriate as well.
Great! I have already coded it on my work computer and will submit a PR on monday
Check out #299
Hi thanks for the library!
For a struct with serde default for some fields, the fields becomes optional when Deserializing and if not define will use the default value. I think the resulting TypeScript code should reflect that. The following Rust code:
Should result in the following TypeScript types if we only care about deserialization