jest/no-untyped-mock-factory Style

🛠️ An auto-fix is available for this rule for some violations.

What it does

This rule triggers a warning if mock() or doMock() is used without a generic type parameter or return type.

Why is this bad?

By default, jest.mock and jest.doMock allow any type to be returned by a mock factory. A generic type parameter can be used to enforce that the factory returns an object with the same shape as the original module, or some other strict type. Requiring a type makes it easier to use TypeScript to catch changes needed in test mocks when the source module changes.

Examples

Examples of incorrect code for this rule:

jest.mock('../moduleName', () => {
    return jest.fn(() => 42);
});

jest.mock('./module', () => ({
    ...jest.requireActual('./module'),
    foo: jest.fn(),
}));

jest.mock('random-num', () => {
    return jest.fn(() => 42);
});

Examples of correct code for this rule:

// Uses typeof import()
jest.mock<typeof import('../moduleName')>('../moduleName', () => {
    return jest.fn(() => 42);
});

jest.mock<typeof import('./module')>('./module', () => ({
    ...jest.requireActual('./module'),
    foo: jest.fn(),
}));

// Uses custom type
jest.mock<() => number>('random-num', () => {
    return jest.fn(() => 42);
});

// No factory
jest.mock('random-num');

// Virtual mock
jest.mock(
    '../moduleName',
    () => {
        return jest.fn(() => 42);
    },
    { virtual: true },
);

How to use

To enable this rule using the config file or in the CLI, you can use:

Config (.oxlintrc.json)

{
    "plugins": ["jest"],
    "rules": {
        "jest/no-untyped-mock-factory": "error"
    }
}

CLI

oxlint --deny jest/no-untyped-mock-factory --jest-plugin

References

• Rule Source