gulp-uglify notify on failing build

Question

I am using gulp-uglify to compress & minify the javascript file. It runs perfectly. But the problem is, when there is javascript error in the file, it will fail to build and it doesn't show any sign of error informing the process is fail.

gulp.task('compressjs', function() {
    gulp.src(['public/**/*.js','!public/**/*.min.js'])
    .pipe(sourcemaps.init())
    .pipe(concat('all.js'))
    .pipe(wrap('(function(){"use strict"; <%= contents %>\n})();'))
    .pipe(uglify())
    .pipe(rename({
        extname: '.min.js'
    }))
    .pipe(sourcemaps.write('.'))
    .pipe(gulp.dest('public'));
});

How can I get the error message when it fail to minify javascript file?

@Ben Hi, I have updated with my gulp task. When there is error in JS, it fail to output the script without any sign of error. But if I disable `uglify`, it can output the file. So, how can I get informing of error when it fail? — user1995781, Feb 06 '15 at 01:17
@entre It would be great if you can show example how to use it. Thank you. — user1995781, Feb 06 '15 at 15:19

score 0 · Answer 1 · answered Feb 08 '15 at 17:32

below is example of how to use .on('error',...

var gulp, browserify, stringify, sass, concat, uglify, source, buffer;

gulp = require('gulp');
browserify = require('gulp-browserify');
stringify = require('stringify');
sass = require('gulp-sass');
concat = require('gulp-concat');
uglify = require('gulp-uglify');

function handleError(err) {
    console.log(err.toString());
}

gulp.task('browserify', function () {
    return gulp.src('./src/js/**/*.js', { read: false })
               .pipe(browserify({
                   transform: [stringify({
                       extensions: ['.html'],
                       minify: true
                   })]
               }))
               .on('error', handleError)
               .pipe(concat('main.js'))
               .pipe(gulp.dest('./bin/js'));
});

gulp-uglify notify on failing build

1 Answers1