This answer has been appended to reflect recent changes to Gulp. I've retained the original response, for relevance to the OPs question. If you are using Gulp 2.x, skip to the second section
Original response, Gulp 1.x
You may change this default behavior by passing errLogToConsole: true
as an option to the sass()
method.
Your task might look something like this, right now:
gulp.task('sass', function () {
gulp.src('./*.scss')
.pipe(sass())
.pipe(gulp.dest('./'));
});
Change the .pipe(sass())
line to include the errLogToConsole: true
option:
.pipe(sass({errLogToConsole: true}))
This is what the task, with error logging, should look like:
gulp.task('sass', function () {
gulp.src('./*.scss')
.pipe(sass({errLogToConsole: true}))
.pipe(gulp.dest('./'));
});
Errors output will now be inline, like so:
[gulp] [gulp-sass] source string:1: error: invalid top-level expression
You can read more about gulp-sass
options and configuration, on nmpjs.org
Gulp 2.x
In Gulp 2.x errLogToConsole
may no longer be used. Fortunately, gulp-sass
has a method for handling errors. Use on('error', sass.logError)
:
gulp.task('sass', function () {
gulp.src('./sass/**/*.scss')
.pipe(sass().on('error', sass.logError))
.pipe(gulp.dest('./css'));
});
If you need more fine-grained control, feel free to provide a callback function:
gulp.task('sass', function () {
gulp.src('./sass/**/*.scss')
.pipe(sass()
.on('error', function (err) {
sass.logError(err);
this.emit('end');
})
)
.pipe(gulp.dest('./css'));
});
This is a good thread to read if you need more information on process-control: https://github.com/gulpjs/gulp/issues/259#issuecomment-55098512