Gulp浏览器-同步只起一次作用

Question

我正在我的一个项目上试用Gulp，我想像以前使用Grunt watch那样运行它。这意味着，一旦所有这些都完成了，它就需要查看更少的文件和js文件，编译、合并、编译和刷新浏览器。

我设法让它与gulp-browser-sync一起工作，但由于某种原因，它只能工作一次。我对.less文件进行了更改，浏览器将重新加载。然后，第二次更改，它会编译，但不会发生重新加载。

日志如下：

[BS] Serving files from: ./
[09:47:26] Starting 'css-clean'...
[09:47:26] Finished 'css-clean' after 16 ms
[09:47:26] Starting 'styles'...
[BS] 1 file changed (styles.min.css)
[09:47:27] Finished 'styles' after 624 ms
[09:47:27] Starting 'styles-watch'...
[BS] Reloading Browsers...

当我第二次点击保存时：

[09:47:31] Starting 'css-clean'...
[09:47:31] Finished 'css-clean' after 3.39 ms
[09:47:31] Starting 'styles'...
[BS] 1 file changed (styles.min.css)
[09:47:32] Finished 'styles' after 362 ms

至于JS，它一直都在工作。没有任何问题，即使在样式任务完成之后，JS更改仍然会正确地触发重新加载。真正有问题的只是风格。

这是gulpfile.js

var gulp = require('gulp'),
    autoprefixer = require('gulp-autoprefixer'),
    less = require('gulp-less'),
    minifycss = require('gulp-minify-css'),
    concat = require('gulp-concat'),
    jshint = require('gulp-jshint'),
    uglify = require('gulp-uglify'),
    rename = require('gulp-rename'),
    notify = require('gulp-notify'),
    path = require('path'),
    streamqueue = require('streamqueue'),
    clean = require('gulp-clean'),
    browserSync = require('browser-sync').create(),
    reload = browserSync.reload;


// Clean the CSS folder
gulp.task('css-clean', function () {
    return gulp.src(['dist/css'], {read: false})
        .pipe(clean());
});

// Clean the CSS folder
gulp.task('js-clean', function () {
    return gulp.src(['dist/js'], {read: false})
        .pipe(clean());
});


// Generate the CSS styles, clean before hand
gulp.task('styles', ['css-clean'], function() {
  return streamqueue({ objectMode: true },
            gulp.src(['./bower_components/uikit/less/uikit.less']),
            gulp.src(['./src/less/main.less'])
        )
    .pipe(concat('styles.less'))
    .pipe(less({
      paths: [ path.join(__dirname, 'less', 'includes') ]
    }))
    .pipe(autoprefixer('last 2 version', 'safari 5', 'ie 8', 'ie 9', 'opera 12.1', 'ios 6', 'android 4'))
    .pipe(gulp.dest('dist/css'))
    .pipe(rename({suffix: '.min'}))
    .pipe(minifycss())
    .pipe(gulp.dest('dist/css'))
    .pipe(browserSync.reload({stream:true}));
});

// create a task that ensures the `styles` task is complete before
// reloading browsers
gulp.task('styles-watch', ['styles'], browserSync.reload);



// Lint Task
gulp.task('lint', function() {
    return gulp.src('src/js/*.js')
        .pipe(jshint())
        .pipe(jshint.reporter('default'));
});


// Generate the scripts, clean before hand
gulp.task('scripts', ['js-clean', 'lint'], function() {
  return streamqueue({ objectMode: true },
            gulp.src(['./bower_components/jquery/dist/jquery.js']),
            gulp.src(['./bower_components/modernizer/modernizr.js']),
            gulp.src(['./bower_components/uikit/js/uikit.js']),
            gulp.src(['./src/js/plugin.js']),
            gulp.src(['./src/js/main.js'])
        )
    .pipe(concat('scripts.js'))
    .pipe(gulp.dest('dist/js'))
    .pipe(rename({suffix: '.min'}))
    .pipe(uglify())
    .pipe(gulp.dest('dist/js'))
    .pipe(browserSync.reload({stream:true}));
});

// create a task that ensures the `scripts` task is complete before
// reloading browsers
gulp.task('scripts-watch', ['scripts'], browserSync.reload);


// Lint Task
gulp.task('alldone', ['scripts'], function() {
    return gulp.src('src/js/*.js')
               .pipe(notify({ message: 'Gulp tasks are completed!!' }));
});


// Static server
gulp.task('browsersync', function() {
    browserSync.init({
        server: {
            baseDir: "./"
        }
    });

    gulp.watch("src/less/*.less", ['styles-watch']);
    gulp.watch('src/js/*.js', ['lint', 'scripts-watch']);
    gulp.watch("*.html").on('change', reload);
});


// Default Task
gulp.task('default', ['css-clean', 'js-clean', 'styles', 'lint', 'scripts', 'alldone']);

// Build Task
gulp.task('build',   ['css-clean', 'js-clean', 'styles', 'lint', 'scripts', 'alldone']);

谢谢你的帮忙!

Answer 1

对于我在Gulp4中，匿名函数没有解决这个问题，我所做的是将browserSync.reload()包装在一个带有完成回调的函数中：

function reload(done) {
  browserSync.reload();
  done();
}

gulp.watch(['scripts/**/*.js','!scripts/main.min.js'], gulp.series(buildScripts, reload));

Answer 2

直接使用怎么样？

.pipe(browserSync.stream())

我曾经在更改'.jade‘文件时遇到过同样的问题，它只重新加载了一次。然后我将browserSync.reload包装在一个匿名函数中，比如

gulp.task('templates-watch', ['templates'], function() {
    browserSync.reload();
});

这对我很管用。我不知道browserSync.reload.有没有什么特别之处试试看怎么样？:)

Answer 3

我知道这不是正确的做法，但经过反复试验，我发现在browserSync.reload()之后使用未定义的函数

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

导致我的日志中出现引用错误，从而触发重新加载浏览器：

ReferenceError: done is not defined 
    at ...
    at ...
    at asyncRunner (...)
    at processTicksAndRejections (...)
[Browsersync] Reloading Browsers...