gulp 4不更新CSS和JS

我正在为WordPress开发自己的主题, 我了解Glup以及它如何简化我的工作流程, 下面的代码所面临的问题是我能够看到我对主页进行的即时更改(html或php), 但是我对css文件或java-script文件所做的任何更改都不会生效, 但仍然必须手动刷新页面：

var gulp = require('gulp'), settings = require('./settings'), webpack = require('webpack'), browserSync = require('browser-sync').create(), postcss = require('gulp-postcss'), rgba = require('postcss-hexrgba'), autoprefixer = require('autoprefixer'), cssvars = require('postcss-simple-vars'), nested = require('postcss-nested'), cssImport = require('postcss-import'), mixins = require('postcss-mixins'), colorFunctions = require('postcss-color-function');

gulp.task('styles', function() {
  return gulp.src(settings.themeLocation + 'css/style.css')
    .pipe(postcss([cssImport, mixins, cssvars, nested, rgba, colorFunctions, autoprefixer]))
    .on('error', (error) => console.log(error.toString()))
    .pipe(gulp.dest(settings.themeLocation));
});

gulp.task('scripts', function(callback) {
  webpack(require('./webpack.config.js'), function(err, stats) {
    if (err) {
      console.log(err.toString());
    }

    console.log(stats.toString());
    callback();
  });
});

gulp.task('watch', function() {
  browserSync.init({
    notify: false, proxy: settings.urlToPreview, ghostMode: false
  });

  gulp.watch('./**/*.php', function(done) {
    browserSync.reload();
    done();
  });

  gulp.watch(settings.themeLocation + 'css/**/*.css', gulp.parallel('waitForStyles'));
  gulp.watch([settings.themeLocation + 'js/modules/*.js', settings.themeLocation + 'js/scripts.js'], gulp.parallel('waitForScripts'));
});

gulp.task('waitForStyles', gulp.series('styles', function() {
  return gulp.src(settings.themeLocation + 'style.css')
    .pipe(browserSync.stream());
}))

gulp.task('waitForScripts', gulp.series('scripts', function(cb) {
  browserSync.reload();
  cb()
}))

#1

尝试这个：

gulp.task('styles', function() {
  return gulp.src(settings.themeLocation + 'css/style.css')
    .pipe(postcss([cssImport, mixins, cssvars, nested, rgba, colorFunctions, autoprefixer]))
    .on('error', (error) => console.log(error.toString()))
    .pipe(gulp.dest(settings.themeLocation))

     // added below
    .pipe(browserSync.stream());
});

// now this task is unnecessary:
// gulp.task('waitForStyles', gulp.series('styles', function() {
//  return gulp.src(settings.themeLocation + 'style.css')
//    .pipe(browserSync.stream());
// }))


 // cb added, called below       
gulp.task('watch', function(cb) {
  browserSync.init({
    notify: false, proxy: settings.urlToPreview, ghostMode: false
  });

  gulp.watch('./**/*.php', function(done) {
    browserSync.reload();
    done();
  });

      // change to gulp.series below 
  // gulp.watch(settings.themeLocation + 'css/**/*.css', gulp.series('waitForStyles'));

   // changed to 'styles' below        
 gulp.watch(settings.themeLocation + 'css/**/*.css', gulp.series('styles'));

  gulp.watch([settings.themeLocation + 'js/modules/*.js', settings.themeLocation + 'js/scripts.js'], gulp.series('waitForScripts'));

  cb();
});

我已经看到gulp4仅在单个任务中遇到了麻烦ala gulp.parallel(‘oneTaskHere’), 所以请尝试像上面的代码一样在watch语句中与series并行交换。

我进行了一些编辑以简化代码-试试看。无需’waitForStyles’, 只需将browserSync.stream()管道移至样式任务的末尾即可。

或者执行以下操作, 而不是移动browserSync.stream管道：

gulp.watch(settings.themeLocation + 'css/**/*.css', gulp.series('styles', browserSync.reload));

但是我本人似乎在”样式”任务版本末尾使用browserSync管道时运气更好。

因为你使用的是webpack插件, 所以我认为脚本任务必须与样式任务不同。你可以尝试：

gulp.watch([settings.themeLocation + 'js/modules/*.js', settings.themeLocation + 'js/scripts.js'], gulp.series('waitForScripts', browserSync.reload));

然后无需执行” waitForScripts”任务。