`Yup.lazy()的Yup和类型记录模式接口

Question

我很难为yup模式(https://github.com/jquense/yup/blob/master/docs/typescript.md)创建一个类型记录界面

给定类型记录界面：

interface TestInterface {
  name: string;
};

以下返回类型是有效的：

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

但是在我的代码中，我使用了Yup.lazy()，但我无法找到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>>, Record<string, any>>; }>, Record<...>, TypeOfShape<...>, AssertsShape<...>>' is not assignable to type 'ObjectSchemaOf<TestInterface>'.
  Type 'Assign<ObjectShape, { name: Lazy<RequiredStringSchema<string, Record<string, any>>, Record<string, any>>; }>' is not assignable to type '{ name: BaseSchema<string, AnyObject, string>; }'.
    Types of property 'name' are incompatible.
      Type 'Lazy<RequiredStringSchema<string, Record<string, any>>, Record<string, any>>' is missing the following properties from type 'BaseSchema<string, AnyObject, string>': deps, tests, transforms, conditions, and 35 more.ts(2322)

Answer 1

您的示例的工作原理是您希望使用Yup.ObjectSchema并添加要求您的对象。

一定要将.required()添加到.shape或.object链中，这样模式的字段就一定会从类型记录的角度存在。它实质上是充当NonNullable<TestInterface>的角色。

//    ✅ happy type
const testSchema1: Yup.ObjectSchema<TestInterface> = Yup.object().shape({
  name: Yup.string().required(),
}).required();

//    ✅ happy type
const testSchema2: Yup.ObjectSchema<TestInterface> = Yup.object().shape({
  name: Yup.lazy(() => Yup.string().required())
}).required();

注意事项：使用当前版本(截至本文) 0.32.11。在普通JS中有一个带有存根类型的runkit示例可以看到[这里]

资料来源：https://github.com/jquense/yup/#ensuring-a-schema-matches-an-existing-type