Gulp 监视所有文件，但只渲染一个(sass) [英] Gulp watch all files but render only one (sass)

问题描述

我想让 gulp 监视我的工作文件夹上的所有更改，但只生成一个文件.因为我使用 scss 导入所有需要的文件，所以不需要编译所有 .css 文件，只需要编译 main 一个.

I want to make gulp watch for all changes on my work folders but to generate only one file. Because I use scss which imports all required files, there is no need to compile all .css files, only main one.

现在，我的 gulpfile.js 包含:

var gulp    = require('gulp');
var util    = require('gulp-util');
var sass    = require('gulp-sass');

gulp.task('sass', function () {
  return gulp.src('./sass/style.scss')
  .pipe(sass().on('error', sass.logError))
  .pipe(gulp.dest('./dist/css'));
});

gulp.task('watch', function() {
  gulp.watch('./sass/**/*.scss', ['sass']);
});

我必须进入 ./sass/style.scss 并将其保存到触发 gulp watch.

And I have to go in ./sass/style.scss and save it to triger gulp watch.

我希望 gulp 观看所有文件(类似于 ./**/*.scss)但只渲染一个 - ./sass/style.scss.如何实现?

I want gulp to watch all files (something like ./**/*.scss) but to render only one - ./sass/style.scss. How to achieve that?

推荐答案

解决方案很简单，只需将 gulpfile.js 的 watch 部分编辑为:p>

Solution to this is simple, just edit watch part of the gulpfile.js to:

gulp.task('watch', function() {
  gulp.watch('./**/*.scss', ['sass']);
});

其中说:注意所有 .scss 并在更改时运行 'sass' taks.

Which says: watch for all .scss and on change run 'sass' taks.

'sass' taks 只编译 ./sass/style.scss'

这篇关于Gulp 监视所有文件，但只渲染一个(sass)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！