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问题描述
我有这样的:
gulp.task('default', ['css', 'browser-sync'] , function() { gulp.watch(['sass/**/*.scss', 'layouts/*.css'], function() { gulp.run('css'); }); });但它不工作,因为它监视两个目录,sass和layouts目录的更改。
but it does not work, because it watches two directories, the sass and the layouts directory for changes.
如何使其工作,以便gulp监视那些目录中发生的任何事情?
How do I make it work, so that gulp watches anything that happens inside those directories?
,
乔治
推荐答案gulp.task('default', ['css', 'browser-sync'] , function() { gulp.watch(['sass/**/*.scss', 'layouts/**/*.css'], ['css']); });
sass / ** / * scss 和 layouts / ** / *。css 将监视每个目录和子目录对 .scss 和 .css 更改的文件。如果要将其更改为任何文件,请将最后一位 *。*
sass/**/*.scss and layouts/**/*.css will watch every directory and subdirectory for any changes to .scssand .css files that change. If you want to change that to any file make the last bit *.*
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如何Gulp观看多个文件?
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