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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'));
});

• javascript
• less
• gulp

Share Improve this question Follow asked Aug 3, 2015 at 14:31 koelkastfilosoofkoelkastfilosoof 2,24211 gold badge2121 silver badges2828 bronze badges

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1 Answer 1

Sorted by: Reset to default 6

It 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'));
});

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