gulp 使用单个命令从多个 sass 文件夹编译多个 css？

Question

我有以下文件结构：

themes
  folder1
    -scss
      -style.scss
  folder2
    -scss
      -style.scss
  foldern
    -scss
      -style.scss

  package.json
  gulpfile.js

需要这种格式的输出：

themes
  folder1
    -scss
      -style.scss
    css
      -style.css
  folder2
    -scss
      -style.scss
    css
      -style.css
  foldern
    -scss
      -style.scss
    css
      -style.css

  package.json
  gulpfile.js

gulpfile.js：

(function () {
  'use strict';

var gulp = require('gulp'),
    eslint = require('gulp-eslint'),
  sass = require('gulp-sass'),
  sourcemaps = require('gulp-sourcemaps')

  gulp.task('sass', function () {
    return gulp
      .src('./**/scss/**/*.scss')
      .pipe(sourcemaps.init())
      .pipe(sass({
        outputStyle: 'uncompressed'
      }).on('error', sass.logError))
      .pipe(sourcemaps.write('./'))
      .pipe(gulp.dest('./**/css'));
  });

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

})();

当我使用上面的 gulpfile 编译时，它会在主题目录中创建 **/css/folder1/stle.css **/css/folder2/style.css 文件夹。我怎样才能以这种格式编译？ gulp 中是否有任何特定的插件。 提前致谢。

Answer 1

试试这个：

var gulp = require('gulp'),
  sass = require('gulp-sass'),
  // etc.

  // two more plugins
  rename = require('gulp-rename')
  path = require('path');


  gulp.task('sass', function () {
    return gulp
      .src('./**/scss/**/*.scss')
      .pipe(sourcemaps.init())
      .pipe(sass({
        outputStyle: 'uncompressed'
      }).on('error', sass.logError))
      .pipe(sourcemaps.write('./'))

       // rename the current file's parent directory

      .pipe(rename(function (file) {

        // file.dirname = current folder, your "scss"
        // then get the parent of the current folder, e.g., "folder1", "folder2", etc.
        let parentFolder = path.dirname(file.dirname)

        // set each file's folder to "folder1/css", "folder2/css", etc.
        file.dirname = path.join(parentFolder, 'css');
      }))

      .pipe(gulp.dest('.'));
  });