javascript - Try/Catch/Finally with ESLint 期望在异步箭头函数结束时返回一个值

Question

我的代码中有这个 ESLint 错误:

函数(productId:任何): promise 期望在异步箭头函数结束时返回一个值

export const getGooglePlayPayment = async (productId) => {
  await InAppBilling.close();
  try {
    await InAppBilling.open();

    if (!await InAppBilling.isSubscribed(productId)) {
      const details = await InAppBilling.subscribe(productId);
      console.log('You purchased: ', details);
      return details.purchaseState === PAYMENT_STATE.PurchasedSuccessfully;
    }
  } catch (err) {
    console.log(err);
    return false;
  } finally {
    await InAppBilling.consumePurchase(productId);
    await InAppBilling.close();
  }
};

有人可以帮我解决这个问题，而不必禁用 ESLing 规则:)

谢谢

Answer 1

这里的规则是consistent-return .

如果 try block 中的 if 语句未满足，则您不会返回任何内容。如果 isSubscribed 调用为真，您应该返回一些内容:

export const getGooglePlayPayment = async (productId) => {
  await InAppBilling.close();
  try {
    await InAppBilling.open();

    if (!await InAppBilling.isSubscribed(productId)) {
      const details = await InAppBilling.subscribe(productId);
      console.log('You purchased: ', details);
      return details.purchaseState === PAYMENT_STATE.PurchasedSuccessfully;
    }
    return 'Already subscribed';
  } catch (err) {
    console.log(err);
    return false;
  } finally {
    await InAppBilling.consumePurchase(productId);
    await InAppBilling.close();
  }
};

(当然，将 Already subscribed 替换为最有意义的内容。如果您只是想表明交易成功，也许 return true。重要的是是为了区别于catch中的return false。)