Javascript eslint警告意外的未命名异步函数，是否要清除它？

当我将异步函数定义为：

module.exports.sendSmtpMessage = async function(keys, mailOptions) {  // <== warning
    await sendSmtpMessage(keys, mailOptions);
}

var sendSmtpMessage = async function(keys, mailOptions) {
...
}

我收到一个eslint警告：

19:34  warning  Unexpected unnamed async function                          func-names

这是一篇糟糕的文章吗

谢谢你的反馈

****更新1**。根据安基特的评论…**

module.exports.sendSmtpMessage = async function sendSmtpMessage(keys, mailOptions) {  // <== warning
    await sendSmtpMessage(keys, mailOptions);
}

var sendSmtpMessage = async function(keys, mailOptions) { // <== warning and error !
...
}

23:7   error    'sendSmtpMessage' is assigned a value but never used  no-unused-vars
23:25  warning  Unexpected unnamed async function

---

要消除该错误，您可以为该

async

函数命名如下：

var sendSmtpMessage = async function sendSmtpMessage(keys, mailOptions)

---

命名导出使您可以轻松地组织代码

const doSomething = async function(args) {
  // ...
}

const doSomethingElse = async function() {
    await doSomething(args);
}

module.exports = {
  doSomething: doSomething,
  doSomethingElse: doSomethingElse
}

---

使用命名函数正确lint：定义然后导出

const sendMessage = async function sendMessage(keys, mailOptions) {
    ....
};

module.exports.sendSmtpMessage = async function sendSmtpMessage(keys, mailOptions) {
  await sendMessage(keys, mailOptions);
};

---

尝试

异步函数somename（键…

是否

返回等待发送消息（…）；

以允许使用代码等待结果？请修改您的问题并删除答案/解决方案（现在有重复的代码块用于更新）考虑在最后的解决方案中加入你自己的答案，或者用更详细的解释来改进现有答案。

const doSomething = async function(args) {
  // ...
}

const doSomethingElse = async function() {
    await doSomething(args);
}

module.exports = {
  doSomething: doSomething,
  doSomethingElse: doSomethingElse
}

const sendMessage = async function sendMessage(keys, mailOptions) {
    ....
};

module.exports.sendSmtpMessage = async function sendSmtpMessage(keys, mailOptions) {
  await sendMessage(keys, mailOptions);
};