是的和用于“Yup.lazy()”的 Typescript 模式接口

如何解决是的和用于“Yup.lazy()”的 Typescript 模式接口

我正在努力为 yup 模式 (https://github.com/jquense/yup/blob/master/docs/typescript.md) 创建打字稿界面

给定打字稿接口：

interface TestInterface {
  name: string;
};

以下返回类型有效：

const testSchema: Yup.SchemaOf<TestInterface> = Yup.object().shape({
  name: Yup.string().required()
});

但在我的代码中我使用了 Yup.lazy()，但我无法找出我应该如何为以下内容创建接口：

const testSchema: Yup.SchemaOf<TestInterface> = Yup.object().shape({
  name: Yup.lazy(() => Yup.string().required())
});

我得到的错误是：

Type 'ObjectSchema<Assign<ObjectShape,{ name: Lazy<requiredStringSchema<string,Record<string,any>>,any>>; }>,Record<...>,TypeOfShape<...>,AssertsShape<...>>' is not assignable to type 'ObjectSchemaOf<TestInterface>'.
  Type 'Assign<ObjectShape,any>>; }>' is not assignable to type '{ name: BaseSchema<string,AnyObject,string>; }'.
    Types of property 'name' are incompatible.
      Type 'Lazy<requiredStringSchema<string,any>>' is missing the following properties from type 'BaseSchema<string,string>': deps,tests,transforms,conditions,and 35 more.ts(2322)