TypeError: expected a string — в чем проблема?

Question

[victoryankul]

Ошибка "TypeError: expected a string" появляется в функции 'watching()' строке

watch(['app/js/**/*.js',!'app/js/main.min.js'], scripts)

const {src, dest, watch, parallel} = require('gulp');

const scss_css = require('gulp-sass')(require('sass'));
const concat = require('gulp-concat');
const uglify = require('gulp-uglify');
const browserSync = require('browser-sync').create();

function browsersync() {
  browserSync.init({
    server: {
      baseDir: './app'
    }
  });
}

function styles() {
  return src('app/scss/style.scss')
    .pipe(scss_css({outputStyle: 'compressed'}))
    .pipe(concat('style.min.css'))
    .pipe(dest('app/css'))
    .pipe(browserSync.stream())
}

function scripts() {
  return src([
    'node_modules/jquery/dist/jquery.js',
    'app/js/main.js'
  ])
  .pipe(concat('main.min.js'))
  .pipe(uglify())
  .pipe(dest('app/js'))
  .pipe(browserSync.stream())
}

function watching() {
  watch(['app/scss/**/*.scss'], styles)
  watch(['app/js/**/*.js',!'app/js/main.min.js'], scripts)
  watch(['app/**/*.html']).on('change', browserSync.reload)
}

 exports.watching = watching;

Помогите побороть, пожалуйста.

Answer 1

[Сергей delphinpro]

Опечатка у вас

function watching() {
   watch(['app/scss/**/*.scss'], styles)
-  watch(['app/js/**/*.js',!'app/js/main.min.js'], scripts)
+  watch(['app/js/**/*.js','!app/js/main.min.js'], scripts)
   watch(['app/**/*.html']).on('change', browserSync.reload)
}

Comments

[victoryankul]

Сергей, большое спасибо Вам.