从 gulp 编译中排除文件

Question

我有一个看起来像这样的 gulp 文件，

    const paths = {
  styles: {
    src: 'src/styles/**/*.scss',
    dest: 'dist/styles',
  },
  javascript: {
    src: 'src/scripts/**/*.js',
    dest: 'dist/scripts',
  },
  copy: {
    pluginsFile: 'src/scripts/plugins.js',
    pluginsFolder: 'src/scripts/plugins/**',
    mapsFolder: 'src/scripts/maps/**',
  },
  files: './**/*.php',
};

    function javascript() {
  gulp
    .src(paths.copy.pluginsFolder)
    .pipe(gulp.dest(`${paths.javascript.dest}/plugins`));

  gulp.src(paths.copy.pluginsFile).pipe(gulp.dest(paths.javascript.dest));

  gulp
    .src(paths.copy.mapsFolder)
    .pipe(gulp.dest(`${paths.javascript.dest}/maps`));

  const jsFiles = glob.sync(paths.javascript.src);

  return browserify({
    entries: jsFiles,
    transform: [babelify.configure({ presets: ['@babel/preset-env'] })],
  })
    .bundle()
    .pipe(source('bundle.js'))
    .pipe(buffer())
    .pipe(uglify())
    .pipe(gulp.dest(paths.javascript.dest))
    .pipe(browserSync.stream());
}

我在“src/scripts/visualiser.js”中有一个文件，我想将其排除在我的包之外，我该怎么做？

我已经将我的路径对象更改为以下内容，

const paths = {
  styles: {
    src: 'src/styles/**/*.scss',
    dest: 'dist/styles',
  },
  javascript: {
    src: ['src/scripts/**/*.js', '!src/scripts/three-d-scripts.js'],
    dest: 'dist/scripts',
  },
  copy: {
    pluginsFile: 'src/scripts/plugins.js',
    pluginsFolder: 'src/scripts/plugins/**',
    mapsFolder: 'src/scripts/maps/**',
  },
  files: './**/*.php',
};

我知道会出现以下吞咽错误，

类型错误：需要全局模式字符串

这是从这一行， const jsFiles = glob.sync(paths.javascript.src);

将此行更改为，

const jsFiles = paths.javascript.src;

返回以下错误，

无法行走依赖图：找不到模块“/Users/simon/Sites/website/website/wp-content/themes/storefront-child-website/!src/scripts/three-d-scripts.js”来自'/Users/simon/Sites/website/website/wp-content/themes/storefront-child-website/!src/scripts/_fake.js'