使用语法 Namespace['Mytype'] 获取类型

Question

我有一个生成的文件（我们称之为

generated.ts

），它给了我类似的东西：

...
export type GetCompanyForEmployeeQueryVariables = Exact<{
  companyId: Scalars['String'];
}>;
export type GetCompanyForEmployeeQuery = { __typename?: 'Query', getCompany?: { __typename?: 'QueryGetCompanyError', source: Source, reasons: Array<GetCompanyRefusalReasons> } | { __typename?: 'QueryGetCompanySuccess', data: { __typename?: 'Company', id: string, name: string, mdmPlatforms: Array<DevicePlatform> } } | null };
export function useGetCompanyForEmployeeLazyQuery(baseOptions?: Apollo.LazyQueryHookOptions<GetCompanyForEmployeeQuery, GetCompanyForEmployeeQueryVariables>) {
    const options = {...defaultOptions, ...baseOptions}
    return Apollo.useLazyQuery<GetCompanyForEmployeeQuery, GetCompanyForEmployeeQueryVariables>(GetCompanyForEmployeeDocument, options);
}
...

我想做的是“动态”访问内部的某种类型，但我有这个错误：

Cannot use namespace 'GeneratedTypes' as a type.ts(2709)

。

import type * as Generated from './generated.ts'

type WorkingVariables = GeneratedTypes.GetCompanyForEmployeeQueryVariables;

// error: Cannot use namespace 'GeneratedTypes' as a type.ts(2709)
type NotWorkingVariables = GeneratedTypes['GetCompanyForEmployeeQueryVariables'];


// My usecase looks like (it just for the example)
type ComputedNameFromOtherStrings = Compute<QueryName>
type Variables = GeneratedTypes[ComputedNameFromOtherStrings];

Answer 1

尝试使用

typeof

：

type NotWorkingVariables = typeof GeneratedTypes['GetCompanyForEmployeeQueryVariables'];

或者：

import { type GetCompanyForEmployeeQueryVariables } from "yourmodule";