英文:
Using Zod with Recursive types and infer them properly
问题
我有这样的模式：
export const barSchema = z.object({
id: z.string(),
foo: z.object(fooSchema),
});
export const fooSchema = z.object({
id: z.string(),
bar: z.object(barSchema),
});
export type BarType = z.infer<typeof barSchema>;
但是barSchema给我带来了TypeScript错误：
'barSchema'隐式具有类型'any'，因为它没有类型注释，并且在其自己的初始化程序中直接或间接引用。
我真的想要使用类型推断的功能。我该如何解决这个问题？
英文:
I have schemas like so:
export const barSchema = z.object({
id: z.string(),
foo: z.object(fooSchema),
});
export const fooSchema = z.object({
id: z.string(),
bar: z.object(barSchema),
});
export type BarType = z.infer<typeof barSchema>;
However barSchema gives me the typescript error:
'barSchema' implicitly has type 'any' because it does not have a type annotation and is referenced directly or indirectly in its own initializer.
I really want to use the ability to infer types. How can I do that and resolve the issue.
答案1
得分: 1
抱歉，根据 Zod 文档的说法，我认为你所要求的可能是不可能的。
你可以在 Zod 中定义递归模式，但由于 TypeScript 的限制，它们的类型不能被静态推断。相反，你需要手动定义类型定义，并将其提供给 Zod 作为“类型提示”。
英文:
Unfortunately, I don't think what you're asking for is possible if the Zod documentation is to be believed.
> You can define a recursive schema in Zod, but because of a limitation of TypeScript, their type can't be statically inferred. Instead you'll need to define the type definition manually, and provide it to Zod as a "type hint".
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