javascript - 如何断言异步“抛出”？

Question

我想断言我的代码不会引发任何最终错误。

问题是：我的代码执行 DOM 操作，这会触发异步反应：

const slot = document.createElement('slot');
myElement.attachShadow({mode:'open'}).appendChild(slot);
// ...
  slot.addEventListener('slotchange', function badFunction(){
    throw "MyError";
    // const observer = new MutationObserver(()=>{});
    // observer.observe(null);
  });
// ...
it('when input child element is removed, should not throw an error', function() {
    function disconnect(){
        // this will throw once slotchange is emmited
        myElement.removeChild(myElement.firstElementChild);
    }
    expect(disconnect).not.to.throw();
});

• 工作片段：https ://glitch.com/edit/#!/how-to-assert-async-throw ？
• 现场样本：path=index.html:28:9

https://how-to-assert-async-throw.glitch.me/

上面的代码通过了测试，即使最终会抛出错误。

Answer 1

试试这个，让我知道这是否有效。

    const slot = document.createElement('slot');
    myElement.attachShadow({mode:'open'}).appendChild(slot);
    // ...
      slot.addEventListener('slotchange', function badFunction(){
        throw "MyError";
        // const observer = new MutationObserver(()=>{});
        // observer.observe(null);
      });
    // ...
   it('when input child element is removed, should not throw an error', async function() {
       async function disconnect(){
           // this will throw once slotchange is emmited
           await myElement.removeChild(myElement.firstElementChild);
       }
       expect(await disconnect()).not.to.throw();
   });