gulp用单个命令从多个sass文件夹编译多个css？

作者: 早岁那知世事艰
发布时间: 2024-10-31 12:16:49 (13天前)
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polo | 2019-08-31 10-32

<div class =“post-text”itemprop =“text”> <P> 试试这个： </p> <pre> <code> 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('.')); }); </code> </pre> </DIV>

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