TypeError:需要glob模式字符串

Question

我正在尝试编译sass,gulp-ruby-sass但我得到了TypeError: glob pattern string required.

这就是我的gulpfile.js样子:

var gulp = require('gulp'),
    sass = require('gulp-ruby-sass');

var paths = {
    sassSrcPath: ['Content/css/**/*.scss'],
    sassDestPath: ['build/css/'],
    sassImportsPath: ['Content/styleguide/']
};

// Styles Task
gulp.task('styles', function () {
    gulp.src(paths.sassSrcPath)
        .pipe(sass({
            style: 'compressed',
            loadPath: [paths.sassImportsPath]
        }))
        .pipe(gulp.dest(paths.sassDestPath));
});

// Watch Task
gulp.task('watch', function () {
    gulp.watch(paths.sassSrcPath, ['styles']);
});

gulp.task('default', ['styles', 'watch']);

这是我的文件夹结构:

??? Content
?   ??? css
?   ?   ??? partials
?   ?   ?   ??_icons.scss
?   ?   ??? main.css.scss
?   ?   ??? styleguide.css.scss
?   ??? styleguide
?       ??? css
?       ?   ?? vendor
?       ?      ?? // Bunch of other folders that contain .sass files
?       ??? // Other .sass files
??? gulpfile.js

我是Gulp的新手,我也没有使用过Grunt,如果我犯了一个小错误,请原谅我.

Answer 1

看一下这些文档,看起来你似乎不想进入gulp-ruby-sass,而是直接调用它:

gulp.task('styles', function () {
    return sass(paths.sassSrcPath, {
            style: 'compressed',
            loadPath: [paths.sassImportsPath]
        })
        .pipe(gulp.dest(paths.sassDestPath));
});