一边跑一边大口大口地说
var gulp = require('gulp');
var sass = require('gulp-sass');
var uglify = require('gulp-uglify');
var rename = require('gulp-rename');
var notify = require('gulp-notify');
var minifycss = require('gulp-minify-css');
var concat = require('gulp-concat');
var plumber = require('gulp-plumber');
var browserSync = require('browser-sync');
var reload = browserSync.reload;
/* Setup scss path */
var paths = {
scss: './assets/sass/*.scss'
};
/* Sass task */
gulp.task('sass', function () {
gulp.src('assets/scss/main.scss')
.pipe(plumber())
.pipe(sass({
includePaths: ['scss'].concat(neat)
}))
.pipe(gulp.dest('assets/css'))
.pipe(rename({suffix: '.min'}))
.pipe(minifycss())
.pipe(gulp.dest('assets/css'))
/* Reload the browser CSS after every change */
.pipe(reload({stream:true}));
});
我收到错误“ReferenceError:未定义整洁”。因为"neat“不是定义的变量。它应该被替换。
任何帮助都是非常感谢的。
发布于 2018-06-19 14:20:41
includePaths:‘scss’.concat(整形)
你在哪里定义了整形变量??
发布于 2018-06-19 15:06:33
var neat =请求(‘node-neat’).includePaths;
然后
includePaths:‘style’.concat(整形)
发布于 2018-06-19 15:11:55
也许吧
var neat =请求(‘node-neat’).includePaths;
然后
includePaths:‘style’.concat(整形)
??
var gulp = require('gulp');
var sass = require('gulp-sass');
gulp.task('sass', function () {
gulp.src('path/to/input.scss')
.pipe(sass({
// includePaths: require('node-neat').with('other/path', 'another/path')
// - or -
includePaths: require('node-neat').includePaths
}))
.pipe(gulp.dest('path/to/output.css'));
});
https://github.com/sass/node-sass上的文档
说
includePaths
类型:数组默认值:[]
一个路径数组,LibSass可以在其中查找以尝试解析@import声明。在使用数据时,建议您使用此选项。
所以这应该行得通吗??
var path = require('path');
gulp.task('sass', function () {
gulp.src('assets/scss/main.scss')
.pipe(plumber())
.pipe(sass({
includePaths: [path.resolve('./assets/sass')]
}))
.pipe(gulp.dest('assets/css'))
.pipe(rename({suffix: '.min'}))
.pipe(minifycss())
.pipe(gulp.dest('assets/css'))
/* Reload the browser CSS after every change */
.pipe(reload({stream:true}));
});
https://stackoverflow.com/questions/50930632
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