如何在玩笑中强制函数的返回值？

Question

我正在尝试测试一些依赖于返回一些数据的 api 的代码。目前，我可以在这个函数中模拟 listNamespacedIngress 的调用：

async function getIngress(namespace) {
  try {
    const result = await k8sIngressApi.listNamespacedIngress(namespace, true);
    const resultSpec = result.body.items.filter(e => e.metadata.name === deploymentPrefix)[0];
    if (!resultSpec) {
      throw new TypeError('Ingress spec is undefined');
    }
    return resultSpec;
  } catch (e) {
    return Promise.reject(e);
  }
}

在这个库中使用 jest.mock 并模拟该函数的返回值，如下所示：

jest.mock('@kubernetes/client-node', () => ({
  KubeConfig: jest.fn().mockImplementation(() => ({
    loadFromCluster: jest.fn(),
    loadFromDefault: jest.fn(),
    makeApiClient: () => ({
      listNamespacedIngress: () =>
        Promise.resolve({
          body: {
            items: [
              {
                metadata: {
                  name: 'a',
                  namespace: 'b',
                },
                spec: {
                  rules: [
                    {
                      host: 'url.com',
                      http: {
                        paths: [
                          {
                            backend: {
                              serviceName: 'a',
                              servicePort: 80,
                            },
                          },
                        ],
                      },
                    },
                  ],
                },
              },

这样，如果 resultSpec 没有最终未定义，我可以测试初始返回值，如下所示（此测试通过）：

  it('Should return storybook-staging ingress details', async () => {
    // When
    const result = await getIngress();

    // Then
    expect(result.metadata.name).toEqual('a');
  });

但是，我不确定如何强制 listNamespacedIngress 返回未定义？

编辑：添加了对模块的充分利用

const kc = new k8s.KubeConfig();
kc.loadFromDefault();
const k8sDeploymentApi = kc.makeApiClient(k8s.AppsV1Api);
const k8sServiceApi = kc.makeApiClient(k8s.CoreV1Api);
const k8sIngressApi = kc.makeApiClient(k8s.NetworkingV1beta1Api);

const BRANCH_NAME = process.argv.slice(2)[0].toLowerCase();
const NAMESPACE = 'dev';
const deploymentPrefix = 'storybook-staging';

const DEPLOYMENT_CONFIG = getDeploymentConfig(deploymentPrefix, BRANCH_NAME);
const SERVICE_CONFIG = getServiceConfig(deploymentPrefix, BRANCH_NAME);
const INGRESS_CONFIG = getIngressConfig(deploymentPrefix);

const HTTP_CONFLICT = 409;

process.on('exit', code => {
  console.log(`About to exit with code: ${code}`);
});

async function getIngress(namespace) {
  try {
    const result = await k8sIngressApi.listNamespacedIngress(namespace, true);
    console.log(result);
    const resultSpec = result.body.items.filter(e => e.metadata.name === deploymentPrefix)[0];
    if (!resultSpec) {
      throw new TypeError('Ingress spec is undefined');
    }
    return resultSpec;
  } catch (e) {
    return Promise.reject(e);
  }
}

Answer 1

listNamespacedIngress 更改返回值的正确方法是使其成为 Jest 间谍。由于它是嵌套的并且在测试中不容易被引用，所以它应该在模块模拟之外定义：

// should be var because jest.mock is hoisted
var mockListNamespacedIngress;

jest.mock('@kubernetes/client-node', () => {
  mockListNamespacedIngress = jest.fn();

  return {
  KubeConfig: jest.fn().mockImplementation(() => ({
    loadFromCluster: jest.fn(),
    loadFromDefault: jest.fn(),
    makeApiClient: () => ({
      listNamespacedIngress: mockListNamespacedIngress
      ...

这也允许断言该函数被调用：

mockListNamespacedIngress.mockResolvedValue({ body: { items: [...] } });
await expect(getIngress()).resolves.toEqual('a');
expect(mockListNamespacedIngress).toBeCalledWith(...);

...

mockListNamespacedIngress.mockResolvedValue({ body: { items: [] } });
await expect(getIngress()).rejects.toThrow(TypeError);
expect(mockListNamespacedIngress).toBeCalledWith(...);