I'm trying to pile a less file to css with Gulp. But it keeps giving me Process terminated with code 1.
errors... Does someone know what this means and why it occurs? And how I should fix it?
This is my gulpfile.js:
var gulp = require('gulp');
var less = require('gulp-less');
gulp.task('pile less simple', function () {
gulp.src('box-sizing.less')
.pipe(less())
.pipe(gulp.dest('output.min.css'));
});
I'm trying to pile a less file to css with Gulp. But it keeps giving me Process terminated with code 1.
errors... Does someone know what this means and why it occurs? And how I should fix it?
This is my gulpfile.js:
var gulp = require('gulp');
var less = require('gulp-less');
gulp.task('pile less simple', function () {
gulp.src('box-sizing.less')
.pipe(less())
.pipe(gulp.dest('output.min.css'));
});
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asked Aug 3, 2015 at 14:31
koelkastfilosoofkoelkastfilosoof
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1 Answer
Reset to default 6It looks right but you shouldn't use whitespaces in your task name declaration (https://github./gulpjs/gulp/blob/master/docs/API.md#gulptaskname-deps-fn).
So try:
var gulp = require('gulp');
var less = require('gulp-less');
gulp.task('pile:less', function () {
gulp.src('box-sizing.less')
.pipe(less())
.pipe(gulp.dest('output.min.css'));
});