Question

我用jest.fn嘲笑两个函数：

let first = jest.fn();
let second = jest.fn();

如何断言first之前调用的second？

我正在寻找的是sinon's .calledBefore断言。

更新我使用了这个简单的“临时”解决方法

it( 'should run all provided function in order', () => {

  // we are using this as simple solution
  // and asked this question here https://stackoverflow.com/q/46066250/2637185

  let excutionOrders = [];
  let processingFn1  = jest.fn( () => excutionOrders.push( 1 ) );
  let processingFn2  = jest.fn( () => excutionOrders.push( 2 ) );
  let processingFn3  = jest.fn( () => excutionOrders.push( 3 ) );
  let processingFn4  = jest.fn( () => excutionOrders.push( 4 ) );
  let data           = [ 1, 2, 3 ];
  processor( data, [ processingFn1, processingFn2, processingFn3, processingFn4 ] );

  expect( excutionOrders ).toEqual( [1, 2, 3, 4] );
} );

Answer 1

clemenspeters的solution（他想确保登录之前先退出登录）对我有用：

const logoutSpy = jest.spyOn(client, 'logout');
const loginSpy = jest.spyOn(client, 'login');
// Run actual function to test
await client.refreshToken();
const logoutOrder = logoutSpy.mock.invocationCallOrder[0];
const loginOrder = loginSpy.mock.invocationCallOrder[0];
expect(logoutOrder).toBeLessThan(loginOrder)

Answer 2

您可以安装jest-community的jest-extended软件包来代替您的解决方法，该软件包通过.toHaveBeenCalledBefore()为此提供支持，例如：

it('calls mock1 before mock2', () => {
  const mock1 = jest.fn();
  const mock2 = jest.fn();

  mock1();
  mock2();
  mock1();

  expect(mock1).toHaveBeenCalledBefore(mock2);
});

注意：根据他们的文档，您至少需要Jest v23才能使用此功能

https://github.com/jest-community/jest-extended#tohavebeencalledbefore

P.S。 -This feature was added a few months after you posted your question，所以希望这个答案仍然有帮助！

如何在jest中

2 个答案: