Question

在按照记录的示例后，我无法点击gulp-watch或gulp-watch-less。我最初的问题是lazypipe（这里没有显示），但在我看来，我在使用插件的方式上做错了。这是我愚蠢的代码，但仍无效。

请注意，我尝试使用普通的gulp-watch，它表现出完全相同的问题：它不会在更改时触发后续管道。如果出现问题，我会在这里附上信息。

这是我的gulpfile。

var debug = require ( 'gulp-debug' );
var gulp = require ( 'gulp' );
var less = require ( 'gulp-less' );
var watchLess = require ( 'gulp-watch-less' ); 

gulp.task ( 'dev-watch', function () {
  // main.less just imports child less files
  gulp.src ( './app/styles/less/main.less' )
    .pipe ( watchLess ( './app/styles/less/main.less' ) )
    .pipe ( debug () );
    .pipe ( less () )
    .pipe ( gulp.dest ( './app/styles' ) )
  ;
});

当我开始任务时，它会执行并完美地生成预期的文件。我看到调试输出流信息也很好。

当我更改文件时，我发现watchLess正在接受更改：

 [10:49:54] LESS saw child.less was changed
 [10:49:54] LESS saw child.less was changed
 [10:49:54] LESS saw main.less was changed:by:import
 [10:49:54] LESS saw main.less was changed:by:import

但是，较少的任务不会执行。它似乎没有发出任何东西，因为调试不会触发。

这是相关的package.json信息：

"devDependencies": {
  "gulp": "^3.8.7",
  "gulp-less": "^1.3.6",
  "gulp-watch": "^1.2.0",
  "gulp-watch-less": "^0.2.1"
}

Answer 1

您的代码仅在管道中运行观察程序，但不会告诉您该怎么做。

工作示例应如下：

var
  gulp = require('gulp'),
  debug = require ('gulp-debug'),
  less = require ( 'gulp-less'),
  watchLess = require('gulp-watch-less');

gulp.task('dev-watch', function () {
  watchLess('./app/styles/less/main.less')
    .pipe (debug ())
    .pipe(less())
    .pipe(gulp.dest('./app/styles'))
});

但是，您也可以使用gulp-watch或gulp（gulp.watch）来做同样的事情。

Answer 2

这一定是最好的解决方案，我在gulp-less github中获得自述文件; https://github.com/plus3network/gulp-less https://github.com/gulpjs/gulp/blob/master/docs/recipes/combining-streams-to-handle-errors.md npm i stream-combiner2 --save-dev

var combiner = require('stream-combiner2');

var combined = combiner.obj([
    gulp.src(srcs),
    less(),
    autoprefixer({
        browsers: ['last 6 versions'],
        cascade: false
    }),
    isDev ? null : cleanCss(),
    gulp.dest(targetDir + 'css/multi/'),
].filter(v => v));

// any errors in the above streams will get caught
// by this listener, instead of being thrown:
combined.on('error', console.error.bind(console));
combined.on('end', () => {}); //done have been call when return combined;
return combined;

gulp-watch，gulp-watch-less don not f fire

2 个答案: