Jest相当于RSpec惰性求值变量(let)？

Question

在rspec中你可以这样做:

let(:input) { 'foo' }
before_each do
   setup_some_thing(input)
end

context 'when input is bar do
  let(:input) { 'bar' }
  it 'does something different' do
  end
end

context 'when input is baz do
  let(:input) { 'baz' }
  it 'does something else different' do
  end
end

这允许您将大对象的方法调用或实例化定义为其较小部分的总和.然后,您可以在不同的上下文中覆盖那些单独的小部件.这个想法是在每次测试之前创建一个快乐路径,然后在上下文块中指定与快乐路径的偏差.

不幸的是,我似乎无法用Jest做到这一点.我尝试过以下方法:

beforeEach(() => {
  let input = 'foo';
  beforeEach(() => {
    setupSomeThing(input);
  });

  describe('when input is bar', () => {
    input = 'bar';
    it('does something different', () => {

    });
  });

  describe('when input is baz', () => {
    input = 'baz';
    it('does something different', () => {

    });
  });
});

因为jest在运行任何特定的描述块之前执行每个描述块,所以输入总是'baz'.有没有人知道一个工作,或一种方法来获得rspec行为？

提前致谢!

更新

您可以使用beforeAll获得类似的行为(尽管没有延迟评估).

beforeEach(() => {
  let input = 'foo';
  beforeEach(() => {
    setupSomeThing(input);
  });

  describe('when input is bar', () => {
    beforeAll(() => {
     input = 'bar';
    });

    it('does something different', () => {

    });
  });

  describe('when input is baz', () => {
    beforeAll(() => {
     input = 'baz';
    });        

    it('does something different', () => {

    });
  });
});

Answer 1

我发现的最好的解决方案是像这样的图书馆

https://github.com/stalniy/bdd-lazy-var

和

https://github.com/tatyshev/given2

如果您不想引入依赖项，您可以通过执行以下操作来获得类似的行为（尽管没有延迟评估）：

beforeEach(() => {
  let input = 'foo';
  beforeEach(() => {
    setupSomeThing(input);
  });

  describe('when input is bar', () => {
    beforeAll(() => {
     input = 'bar';
    });

    it('does something different', () => {

    });
  });

  describe('when input is baz', () => {
    beforeAll(() => {
     input = 'baz';
    });        

    it('does something different', () => {

    });
  });
});