我有这样的事情:
gulp.task('default', ['css', 'browser-sync'] , function() {
gulp.watch(['sass/**/*.scss', 'layouts/*.css'], function() {
gulp.run('css');
});
});
但它不起作用,因为它会监视两个目录,即 sass 和 layouts 目录以进行更改.
but it does not work, because it watches two directories, the sass and the layouts directory for changes.
如何让它工作,以便 gulp 监视这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?
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<的任何更改/code> 和
.css
文件更改.如果您想将其更改为 any 文件,请在最后一位 *.*
sass/**/*.scss
and layouts/**/*.css
will watch every directory and subdirectory for any changes to .scss
and .css
files that change. If you want to change that to any file make the last bit *.*
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