Reputation: 25172
Suppose I have a simple file exporting a default function:
// UniqueIdGenerator.js
const uniqueIdGenerator = () => Math.random().toString(36).substring(2, 8);
export default uniqueIdGenerator;
Which I would use like this:
import uniqueIdGenerator from './UniqueIdGenerator';
// ...
uniqueIdGenerator();
I want to assert in my test that this method was called while keeping the original functionality. I'd do that with jest.spyOn
however, it requires an object as well as a function name as parameters. How can you do this in a clean way? There's a similar GitHub issue for jasmine
for anyone interested.
Upvotes: 109
Views: 92373
Reputation: 429
one could also mock the import and pass the original implementation as mock implementation, like:
import uniqueIdGenerator from './UniqueIdGenerator'; // this import is a mock already
jest.mock('./UniqueIdGenerator.js', () => {
const original = jest.requireActual('./UniqueIdGenerator')
return {
__esModule: true,
default: jest.fn(original.default)
}
})
test(() => {
expect(uniqueIdGenerator).toHaveBeenCalled()
})
Upvotes: 32
Reputation: 1424
The solution by the user thisismydesign(the first answer) didn't work for me. I made a few modifications to get to work.
First, I exported a default object:
// UniqueIdGenerator.js
const uniqueIdGenerator = () => Math.random().toString(36).substring(2, 8);
export default {uniqueIdGenerator}
To use it, I did it as follows:
import generator from './UniqueIdGenerator';
// ...
const spy = jest.spyOn(generator, 'uniqueIdGenerator');
Upvotes: 0
Reputation: 369
I know I'm late to the party but I recently had this problem and wanted to share my solution as well ... though it seems a bit more unconventional but could be tweaked by someone with better knowledge.
I happen to have a file with the function that I would like to spy on.
// /foo/ModuleToBeMocked.ts
const fnToSpyOn = () => ...;
export default { fnToSpyOn }
This is then imported into a parent file that would bring, and export, alike functions. Sort of like a classification.
// /parent.ts
import fnToSpyOn from './foo/ModuleToBeMocked';
import someOtherFn from './foo/SomeOtherModule';
...
export { fnToSpyOn, someOtherFn, ... };
And this is how I test the fnToSpyOn
// /foo/ModuleToBeMocked.test.ts
import { ModuleToBeMocked } from '../parent';
const fnToSpyOnSpu = jest.spyOn(ModuleToBeMocked, 'fnToSpyOn');
Upvotes: 1
Reputation: 4098
Here it is even simpler.
Mock your exported module 'addDelay' (has the sleep
function in it) using jest
.
const { sleep } = require('../../src/utils/addDelay');
jest.mock('../../src/utils/addDelay', () => {
const delay = jest.requireActual('../../src/utils/addDelay');
return {
...delay,
sleep: jest.fn(),
};});
And the test is as follows and check if sleep
function was called with 1 sec as in arg.
test("Should delay 1 second if Okta user has no IDxM Roles", async () => {
// GIVEN
const MockSleep = sleep;
// WHEN
await getUser(req, res);
// THEN
expect(MockSleep).toHaveBeenCalledWith(1000);// sleep(1000): 1sec
});
Upvotes: 1
Reputation: 151
Use 'default' as the second argument in spyOn function.
import * as MyHelperMethod from '../myHelperMethod';
jest.spyOn(MyHelperMethod, 'default');
Upvotes: 15
Reputation: 2022
Here is a way of doing it for a default export without modifying the import (or even needing an import in the test at all):
const actual = jest.requireActual("./UniqueIdGenerator");
const spy = jest.spyOn(actual, "default");
Upvotes: 45
Reputation: 4893
Mock only the default export, or any other export, but keep remaining exports in module as original:
import myDefault, { myFunc, notMocked } from "./myModule";
jest.mock("./myModule", () => {
const original = jest.requireActual("./myModule");
return {
__esModule: true,
...original,
default: jest.fn(),
myFunc: jest.fn()
}
});
describe('my description', () => {
it('my test', () => {
myFunc();
myDefault();
expect(myFunct).toHaveBeenCalled();
expect(myDefault).toHaveBeenCalled();
myDefault.mockImplementation(() => 5);
expect(myDefault()).toBe(5);
expect(notMocked()).toBe("i'm not mocked!");
})
});
Upvotes: 11
Reputation: 452
What worked for me was a combination of the answer from Janne Annala and OP's own solution. All I wanted to test was that the helper method was called with the correct parameters as I had already written a test for the helper method and it didn't have any bearing on my subsequent test:
// myHelperMethod.js
export const myHelperMethod = (param1, param2) => { // do something with the params };
// someOtherFileUsingMyHelperMethod.js
import * as MyHelperMethod from '../myHelperMethod';
jest.mock('../myHelperMethod', () => ({
myHelperMethod: jest.fn(),
}));
let myHelperMethodSpy = jest.spyOn(MyHelperMethod, 'myHelperMethod');
// ...
// some setup
// ...
test(() => {
expect(myHelperMethodSpy).toHaveBeenCalledWith(param1, param2);
});
Upvotes: 5
Reputation: 28776
In some cases you have to mock the import to be able to spy the default export:
import * as fetch from 'node-fetch'
jest.mock('node-fetch', () => ({
default: jest.fn(),
}))
jest.spyOn(fetch, 'default')
Upvotes: 14
Reputation: 25172
I ended up ditching the default export:
// UniqueIdGenerator.js
export const uniqueIdGenerator = () => Math.random().toString(36).substring(2, 8);
And then I could use and spy it like this:
import * as UniqueIdGenerator from './UniqueIdGenerator';
// ...
const spy = jest.spyOn(UniqueIdGenerator, 'uniqueIdGenerator');
Some recommend wrapping them in a const object, and exporting that. I suppose you can also use a class for wrapping.
However, if you can't modify the class there's still a (not-so-nice) solution:
import * as UniqueIdGenerator from './UniqueIdGenerator';
// ...
const spy = jest.spyOn(UniqueIdGenerator, 'default');
Upvotes: 113