Gulp不会观察到任何变化 [英] Gulp wouldn't watch any changes

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问题描述



这简直就赢得了'我的头脑',但我不知道下面的gulpfile有什么不对, t观察少量变化,我尝试了所有 gulp gulp watch 。我必须在每次更改后手动运行gulp以编译它们。是否有任何错误导致手表无法正常工作?



Gulp版本
CLI版本1.4.0
本地版本3.9.1

节点v7.7.2

  var gulp = require('gulp'); 

var less = require('gulp-less');

// gulp编译路径

var paths = {
dist:'assets / dist'
};
$ b $ gulp.task('css',function(){
return gulp.src('assets / less / styles.less')
.pipe(less({
compress:true
)))
.pipe(gulp.dest(paths.dist))
});

gulp.task('watch',function(){
gulp.watch('assets / less / *。less',['css']); //观察所有.less文件,然后运行较少的任务
});

gulp.task('default',['watch']);


解决方案

确保您有.less文件的正确路径你也需要添加bowersync.reload才能工作。

  gulp.task('watch',function(){
gulp.watch('assets / less / * (''css']);
// init server
browserSync.init({
server:{
proxy:local.build,
baseDir:appPath.root
}
});

gulp.watch([appPath.root +'**'],browserSync.reload);
});

gulp.task('default',['watch']);


I am scratching my head around but can't seem to figure out whats wrong with the following gulpfile which simply watch and compile less file.

This simply won't watch less changes, I have tried all gulp, gulp watch. I have to manually run gulp after each change to compile them. Is there something wrong that causing watch to not work as expected?

Gulp Version CLI version 1.4.0 Local version 3.9.1

NPM 4.1.2 Node v7.7.2

var gulp = require('gulp');

var less = require('gulp-less');

// gulp compile paths

var paths = {
    dist: 'assets/dist'
};

gulp.task('css', function() {
    return gulp.src('assets/less/styles.less')
        .pipe(less({
            compress: true
        }))
        .pipe(gulp.dest(paths.dist))
});

gulp.task('watch', function() {
    gulp.watch('assets/less/*.less', ['css']);  // Watch all the .less files, then run the less task
});

gulp.task('default', ['watch']);

解决方案

Make sure you have correct path to the .less files also you need to add bowersync.reload for it to work.

gulp.task('watch', function () {
    gulp.watch('assets/less/*.less', ['css']);
    // init server
    browserSync.init({
        server: {
            proxy: "local.build",
            baseDir: appPath.root
        }
    });

    gulp.watch([appPath.root + '**'], browserSync.reload);
});

gulp.task('default', ['watch']);

这篇关于Gulp不会观察到任何变化的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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