我做了什么?
gulp
(正在处理SCSS文件,我得到一个CSS文件)预期:
会发生什么?
命令行输出:
$ gulp
[09:24:28] Using gulpfile c:\Users\User\_dev\github\project\gulpfile.js
[09:24:28] Starting 'sass'...
[09:24:28] Finished 'sass' after 98 ms
[09:24:28] Starting 'default'...
[09:24:28] Finished 'default' after 7.31 μs
[09:24:35] sass-watch saw _base.scss was changed
[09:25:39] sass-watch saw _base.scss was changed
gulpfile.js:
gulp.task('sass', function() {
watch({ glob: 'css/**/*.{scss,sass}', name: 'sass-watch'})
.pipe(plumber())
.pipe(sass())
.pipe(gulp.dest('css'))
});
gulp.task('default', ['sass']);
注意:
源文件的输入方式并不重要。使用gulp.src()
答案 0 :(得分:0)
我不认为你的sass任务写得正确。
尝试这样的事情:
var gulp = require('gulp');
var sass = require('gulp-sass')
gulp.task('sass', function () {
gulp.src('PATH-TO-SASS-FILES/*.scss')
.pipe(sass())
.pipe(gulp.dest('./css'));
});