Switch output folder based on filename in gulp task
问题 I have different *.scss files in my src folder and I want one file to be compiled in its own separate folder. Lets assume I have the files normalFile_1.scss , specialFile.scss , normalFile_2.scss . I want the two normal files to be compiled to the folder Public/Css , the special file however should end up in the folder Public/Css/Special . I have tried to get the current filename in the task with gulp-tap , which works fine. .pipe($.tap(function (file, t) { filename = path.basename(file.path)