gulp-watch，gulp-watch-less不要开火

作者: 不想吃东西
发布时间: 2024-09-12 02:26:18 (9天前)
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那年 | 2019-08-31 10-32

<div class =“post-text”itemprop =“text”> <P> 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。 </p> <P> 工作示例应如下： </p> <pre> <code> 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')) }); </code> </pre> <P> 但是，您也可以使用gulp-watch或gulp（gulp.watch）来做同样的事情。 </p> </DIV>

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1#
回复此人
岁爵 | 2019-08-31 10-32

<div class =“post-text”itemprop =“text”> <P> 这一定是最好的解决方案，我在gulp-less github中获得自述文件; <a href="https://github.com/plus3network/gulp-less" rel="nofollow noreferrer"> https://github.com/plus3network/gulp-less </A> <a href="https://github.com/gulpjs/gulp/blob/master/docs/recipes/combining-streams-to-handle-errors.md" rel="nofollow noreferrer"> https://github.com/gulpjs/gulp/blob/master/docs/recipes/combining-streams-to-handle-errors.md </A> <code> npm i stream-combiner2 --save-dev </code> </p> <pre> <code> 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; </code> </pre> </DIV>

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