玩笑單元測試在 soap.createClientAsync 返回函數中失敗

Question

我有一個肥皂 RestAPI 來調用從我使用帶有異步等待的肥皂庫的服務中執行一個函數。 代碼工作正常。 當進行單元測試時，測試用例在客戶端返回的回調方法處失敗。 代碼和 UT 錯誤如下。

調用 Soap 客戶端的函數 - 代碼 - Helper.ts

class soapHelper {
    public async registerInSoap(
            uploadRequest: ImportExperimentRequestDto,
        ): Promise<{ RegisterExperimentResult: IffManPublishResponseDto }> {
            const url = "http://test.com/Services/Req.svc?wsdl";
            const client = await soap.createClientAsync(flavourUrl);
            return client.RegisterExperimentAsync({ upload: uploadRequest});
        }
}

測試用例 - 代碼

describe("**** Register via soap service ****", () => {
        it("** should excecute register method **", async () => {
            const request = cloneDeep(MOCK.API_PAYLOAD);
            const clientResponse = {
                RegisterExperimentAsync: jest.fn(),
            };
            jest.spyOn<any, any>(soap, "createClientAsync").mockReturnValueOnce(() => Promise.resolve(clientResponse));
            const result= await soapHelper.registerInSoap(request);
            expect(result).toEqual({ Result: AFB_MOCK_RESPONSE.API_RESPONSE });
        });
    });

錯誤

TypeError: client.RegisterExperimentAsync is not a function

UT 錯誤

Answer 1

您告訴 Jest soapHelper.createClientAsync應該返回一個具有RegisterExperimentAsync方法的函數，而不是告訴它應該立即使用RegisterExperimentAsync方法返回對象的承諾。 這個片段應該修復它：

    describe("**** Register via soap service ****", () => {
        it("** should excecute register method **", async () => {
            const request = cloneDeep(MOCK.API_PAYLOAD);
            const clientResponse = {
                RegisterExperimentAsync: jest.fn(),
            };
            jest.spyOn<any, any>(soap, "createClientAsync").mockResolvedValue(clientResponse);
            const result= await soapHelper.registerInSoap(request);
            expect(result).toEqual({ Result: AFB_MOCK_RESPONSE.API_RESPONSE });
        });
    });

現在soapHelper.createClientAsync正確地返回了一個要await的承諾，並且承諾的解析值有一個RegisterExperimentAsync方法