javascript - 如何正确删除多个事件监听器？

Question

(显然这与 typescript 没有太大关系，只是示例代码是在 ts 中)。

import { createWriteStream, WriteStream } from "fs";

export class Util {
    public static openWrite(path: string): Promise<WriteStream> {
        return new Promise<WriteStream>((resolve, reject) => {
            const result = createWriteStream(path);

            const onError = (err: Error) => {
                // How to remove both listeners here?
                reject(err);
            }

            const onOpen = (fd: number) => {
                // How to remove both listeners here?
                resolve(result);
            };

            result.on("error", onError);
            result.on("open", onOpen);
        }); 
    }
}

代码应该说明一切。我很难理解应该如何编写该函数，以便正确处理成功和失败场景，同时确保在一切完成后删除所有添加的事件处理程序。

当然，总是有可能调用removeAllListeners，但这对我来说看起来像是一种黑客行为。

Answer 1

我认为这是 finally 的一个很好的用例。

let onError, onOpen;
const result = createWriteStream(path);
return new Promise<WriteStream>((resolve, reject) => {

    onError = (err: Error) => {
        // How to remove both listeners here?
        reject(err);
    }

    onOpen = (fd: number) => {
        // How to remove both listeners here?
        resolve(result);
    };

    result.on("error", onError);
    result.on("open", onOpen);
}).finally(() => {
    result.removeListener("error", onError);
    result.removeListener("open", onOpen);
});

或者，如果您知道没有其他监听器已附加到 EventEmitter，您可以按如下方式简化:

const result = createWriteStream(path);
return new Promise<WriteStream>((resolve, reject) => {
    result.on("error", reject);
    result.on("open", resolve);
}).then(() => {
    return result;
}).finally(() => {
    result.removeAllListeners("error");
    result.removeAllListeners("open");
});