我用jest.fn嘲笑两个函数:
let first = jest.fn();
let second = jest.fn();
我如何断言first
在second
之前打电话?
我正在寻找的是像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] );
} );
您可以安装jest-community的jest-extended
软件包而不是您的解决方法,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doc,你需要至少v23的Jest来使用这个功能
https://github.com/jest-community/jest-extended#tohavebeencalledbefore
附: - 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!