如何复制节点模块bootstrap.min.js->主文件夹

Question

我使用Gulp 4创建了一个项目。一切都进行得很详细。我想将我的js设置代码放置在Node Modules中，但不会发生

var gulp = require("gulp"),
sass = require("gulp-sass"),
browserSync = require("browser-sync").create();

var paths = {
styles: {
    src:"./*.html",
    src: "./scss/*.scss",
    dest: "./css",
}
};

function style() {
return gulp
    .src(paths.styles.src)
    .pipe(sass())
    .on("error", sass.logError)
    .pipe(gulp.dest(paths.styles.dest))
    .pipe(browserSync.stream(reload));
}
function reload() {
browserSync.reload();

}
function watch() {
browserSync.init({
    server: {
        baseDir: "./"
    }
});
gulp.watch(paths.styles.src, style);
gulp.watch("./*.html").on('change', browserSync.reload);
}
exports.watch = watch
exports.style = style;
var build = gulp.parallel(style, watch);
gulp.task('default', build);

我无法在上述代码中运行以下代码

function modules () {
  // Bootstrap
  var bootstrap = gulp.src ('./ node_modules / bootstrap / dist / ** / *')
    .pip A (gulp.dest ( './ Vendor / bootstrap'));
  // jQuery
  var jquery = gulp.src ([
      './Node_modules/jquery/dist/*'
      '! ./ node_modules / jQuery / dist / core.js'
    ])
    .pip A (gulp.dest ( './ Vendor / jquery'));
  return merge (bootstrap, jquery);
}