Webpack 我的吞咽文件赢了'；无法读取任何更改，但它运行时没有问题

问题是，当我开始编写新的SCS/css时，它不会像预期的那样关注更改。不 当然，我错在哪里了。我的文件是否被错误地监视，或者根本没有被监视。当我运行gulp时，一切运行正常，但当文件更改时，它不会自动更改

 [16:21:41] Using gulpfile ~\Documents\DeMarcusW\starterkit\gulpfile.js
 [16:21:41] Starting 'default'...
 [16:21:41] Starting 'sass'...   
 [16:21:45] Finished 'sass' after 3.47 s
 [16:21:45] Starting 'webpack'...       
 [16:22:18] Version: webpack 4.42.0
 Built at: 03/14/2020 4:22:18 PM
        Asset      Size     Chunks             Chunk Names
 firstComp.js  3.87 KiB  firstComp  [emitted]  firstComp
    vendor.js  80.5 KiB     vendor  [emitted]  vendor
 Entrypoint firstComp = firstComp.js
 Entrypoint vendor = vendor.js
 [16:22:18] Finished 'webpack' after 34 s
 [16:22:18] Starting 'browser-sync'...
 [Browsersync] Access URLs:
  -------------------------------------
        Local: http://localhost:3000
     External: http://192.168.1.11:3000
  -------------------------------------
           UI: http://localhost:3001
  UI External: http://localhost:3001
  -------------------------------------
 [Browsersync] Serving files from: ./public

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这两条语句在gulp v4中不起作用（它们是v3风格的语法）：

改为：

gulp.watch('./public/scss/**/*', 'sass')
gulp.watch('./public/js/**/*', 'webpack')

如果要运行多个任务

gulp.watch('./public/scss/**/*', gulp.series('sass', 'otherTask'));

---

此外，您也不能在以下内容的末尾执行

pipe

语句：

gulp.task( 'default', gulp.series('sass', 'webpack', 'browser-sync'), ()=>{
  gulp.watch('./public/scss/**/*', ['sass'])      // change as per above
  gulp.watch('./public/js/**/*', ['webpack'])     // change as per above
  .pipe(gulp.dest('./public/js'))                 // remove this
});

您还没有创建流，因此无法通过管道传输任何内容，只需删除该行即可。如果需要移动js文件，请在其他地方进行

gulp.watch('./public/scss/**/*', gulp.series('sass', 'otherTask'));

gulp.task( 'default', gulp.series('sass', 'webpack', 'browser-sync'), ()=>{
  gulp.watch('./public/scss/**/*', ['sass'])      // change as per above
  gulp.watch('./public/js/**/*', ['webpack'])     // change as per above
  .pipe(gulp.dest('./public/js'))                 // remove this
});