I want to make gulp
watch for all changes on my work folders but to generate only one file. Because I use scss
which imports all required files, there is no need to compile all .css
files, only main one.
Now, my gulpfile.js
contains:
var gulp = require('gulp');
var util = require('gulp-util');
var sass = require('gulp-sass');
gulp.task('sass', function () {
return gulp.src('./sass/style.scss')
.pipe(sass().on('error', sass.logError))
.pipe(gulp.dest('./dist/css'));
});
gulp.task('watch', function() {
gulp.watch('./sass/**/*.scss', ['sass']);
});
And I have to go in ./sass/style.scss
and save it to triger gulp watch.
I want gulp to watch all files (something like ./**/*.scss
) but to render only one - ./sass/style.scss
. How to achieve that?
Solution to this is simple, just edit watch
part of the gulpfile.js
to:
gulp.task('watch', function() {
gulp.watch('./**/*.scss', ['sass']);
});
Which says: watch for all .scss
and on change run 'sass'
taks.
'sass'
taks compiles only ./sass/style.scss'
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