如何開玩笑地模擬一個咖喱函式？-有解無憂

我一直遇到一個問題，即我的一個 curried 函式在根據 jest 模擬出來時不是一個函式。我在一個名為 httpRequest.js 的檔案中創建了一組 util httpRequest 函式，如下所示：

const httpRequest = (method) => {
  return (headers) => {
    return (data) => {
      return async (url) => {
        try {
          const result = await axios({ method, url, data, headers });
          const { data: axiosResult } = result;
          return axiosResult;
        } catch (err) {
          console.log(`${method}Data: `, err);
          throw err;
        }
      };
    };
  };
};

const getData = httpRequest('get')()();
const postData = httpRequest('post')();
const putData = httpRequest('put')();
const patchData = httpRequest('patch')();
const deleteData = httpRequest('delete')()();

const preBuiltGetRequest = httpRequest('get');
const preBuiltPostRequest = httpRequest('post');
const preBuiltPutRequest = httpRequest('put');
const preBuiltPatchRequest = httpRequest('patch');
const preBuiltDeleteRequest = httpRequest('delete');

module.exports = {
  httpRequest,
  getData,
  postData,
  putData,
  patchData,
  deleteData,
  preBuiltGetRequest,
  preBuiltPostRequest,
  preBuiltPutRequest,
  preBuiltPatchRequest,
  preBuiltDeleteRequest,
};

當我在測驗中模擬出這個檔案然后使用一個函式時，比如preBuiltGetRequest我在開玩笑時得到一個錯誤TypeError: preBuiltGetRequest(...) is not a function。這是一個實作的例子。

這是我正在測驗的代碼庫中的函式：

queryUser: async (accessToken, email) => {
      const query = `
        {
          getUsersByCriteria(criteria: Email, values: "${email}") {
            id
            groups {
              id
              name
              entitlements {
                id
                code
              }
              members {
                total
              }
            }
          }
        }
      `;
      const newUrl = new URL(`${BaseUrl}/v3/graphql`);
      newUrl.searchParams.append('query', papiQuery);

      console.log('From the Api ', preBuiltGetRequest);

      const getAuthenticatedData = preBuiltGetRequest({
        Authorization: `Bearer ${accessToken}`,
      })();

      const response = await getAuthenticatedData(newUrl.toString());

      const graphQlResult = response.data?.getUsersByCriteria;
      if (!graphQlResult || graphQlResult.length === 0) {
        throw new Error(`Could not find user with email=${email}`);
      }

      return graphQlResult[0];
    },

然后，當我使用以下代碼運行模擬 preBuiltGetRequest 的測驗代碼時：

jest.mock('/opt/httpRequest');

const { preBuiltGetRequest } = require('/opt/httpRequest');

我收到此錯誤：如何開玩笑地模擬一個咖喱函式？

uj5u.com熱心網友回復：

該preBuiltGetRequest函式有一個簽名，可以輸入為

declare const prebuiltGetRequest: (header: object) => (data: object) => (url: String) => Promise<never>;

你需要相應地模擬它，

jest.mock('/opt/httpRequest');

const { preBuiltGetRequest } = require('/opt/httpRequest');


const mockSig = jest.fn().mockReturnValue(
  jest.fn().mockResolvedValueOnce(error)
)
preBuiltGetRequest.mockReturnValue(mockSig)

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