Javascript 排除不适用于尝试从捆绑包中排除非节点_模块脚本

我试图在全局脚本中排除一些内部依赖项，全局脚本是共享的公共脚本，以后可以使用

文件夹结构

my-app
└────src
│    └───index.js
└────common
│    └───module.js
│
└────dist
│    └───bundle.js
│
└───gulpfile.js

index.js

import { someNodeFunction } from 'some-node-package';
import { someFunction } from '../common/module.js';

function main() {
...
...
someFunction();
...
someNodeFunction();
...
}

module.js

export function someFunction() {
...
}

gulpfile.js

var gulp = require('gulp');
const path = require('path');
var webpack = require('webpack');
var webpackGulp = require('webpack-stream');
var nodeExternals = require('webpack-node-externals');

gulp.task('bundle', function () {
    gulp.src('.')
        .pipe(webpackGulp({
            entry: './src/index.js',
            output: {
                filename: 'bundle.js'
            },
            mode: 'production',
            module: {
                rules: [
                    {
                        test: /\.js$/,
                        exclude: [/common/]
                    }
                ]
            },
            externals: [nodeExternals()]
        }, webpack))
        .pipe(gulp.dest('./dist'));
});

package.json

{
  "name": "scripts-utility",
  "version": "1.0.0",
  "description": "separate scripts utility",
  "dependencies": {
    "gulp": "^3.9.0",
    "webpack": "^4.29.6",
    "webpack-stream": "^5.2.1",
    "webpack-node-externals": "^1.7.2"
  },
  "main": "index.js",
  "author": "Jude Leander J",
  "license": "MIT"
}

在这里，我可以使用externals选项从bundle中排除node_modules函数

someNodeFunction

，这似乎很好，但是当我尝试从bundle中排除

someFunction

时，我无法实现同样的效果。希望，我错过了任何配置

我还尝试使用以下配置为

排除
选项提供
'absolute-path'

exclude: ['common/module.js', 'common', 'common/']

但似乎什么都不管用