英文:
Create a schema that has some dynamic keys and some static keys with zod
问题
我想要实现类似这种类型的结果，但使用zod。
type a = {
[key: string]: string;
name: string;
surname: string;
}
因此，这是一个具有一些静态字段但接受其他动态字段的类型。
我知道 z.record(z.string(), z.string())，但似乎它只能重新创建类型只有动态键的情况，如：
type zod = {
[key: string]: string;
}
我还尝试使用 merge：
const dynamicPropertiesSchema = z.record(z.string());
export const PersonSchema = z.object({
anagraphics: z.object({
name: z.string(),
surname: z.string(),
}).merge(dynamicPropertiesSchema),
notes: z.array(z.string()),
});
但我收到以下错误，强调了 .merge() 参数：
“类型参数 'ZodRecord <ZodString, ZodString>' 无法分配给类型参数 'AnyZodObject'。 类型 'ZodRecord <ZodString, ZodString>' 缺少类型 'ZodObject <any, any, any, { [x: string]: any; }, { [x: string]: any; }>' 的以下属性：_cached，_getCached，shape，strict，以及其他14个.ts(2345)”
总之，我想要类似这样的东西：
const mixedSchema = z.object({
z.record(z.string(), z.string()),
name: z.string(),
surname: z.string(),
});
是否有办法实现这个？
感谢您的帮助！
英文:
I want to achieve a result similar to this type, but using zod.
type a = {
[key: string]: string;
name: string;
surname: string;
}
So a type that has some static fields but accepts other dynamic ones.
I know about z.record(z.string(), z.string()), but it seems like it can only recreate the case where the type only has dynamic keys, as in:
type zod = {
[key: string]: string;
}
I also tried using merge:
const dynamicPropertiesSchema = z.record(z.string());
export const PersonSchema = z.object({
anagraphics: z.object({
name: z.string(),
surname: z.string(),
}).merge(dynamicPropertiesSchema),
notes: z.array(z.string()),
});
But I get the following error highlighting the .merge() parameter:
Argument of type 'ZodRecord<ZodString, ZodString>' is not assignable to parameter of type 'AnyZodObject'. Type 'ZodRecord<ZodString, ZodString>' is missing the following properties from type 'ZodObject<any, any, any, { [x: string]: any; }, { [x: string]: any; }>': _cached, _getCached, shape, strict, and 14 more.ts(2345)
So, all in all, I'd like something like this:
const mixedSchema = z.object({
z.record(z.string(), z.string()),
name: z.string(),
surname: z.string(),
});
Would this be able to be achieved somehow?
Thanks for your help!
答案1
得分: 3
我认为实现这个的正确方式是使用交集：
```ts
const schema = z.intersection(z.object({
name: z.string(),
surname: z.string(),
}), z.record(z.string(), z.string()));
type T = z.infer<typeof schema>;
// ^? { name: string; surname: string } & Record<string, string>
类似于在普通的TS类型中使用的方式：
type T = { name: string; surname: string } & Record<string, string>;
<details>
<summary>英文:</summary>
I believe the correct way to accomplish this would be to use an intersection:
```ts
const schema = z.intersection(z.object({
name: z.string(),
surname: z.string(),
}), z.record(z.string(), z.string()));
type T = z.infer<typeof schema>;
// ^? { name: string; surname: string } & Record<string, string>
Similar to how you would do it with plain TS types:
type T = { name: string; surname: string } & Record<string, string>;
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