I am trying to minify my script files for which i am using gulp task runner And I am trying gulp-uglify plugin
Code:
gulp.task(\'concat\', function() {
Try using this
var concat = require('gulp-concat');
var uglify = require('gulp-uglify');
var minifyJS = require('gulp-minify');
gulp.task('concat', function() {
return gulp.src('app/**/*.js')
.pipe(minifyJS())
.pipe(concat('bundle.min.js'))
.pipe(uglify({ mangle: false }))
.pipe(gulp.dest('./dist/'));
});
I think the top answers here are not explaining how to get the error. The docs have a section on error handling:
gulp-uglify emits an 'error' event if it is unable to minify a specific file
So, just capture the error and do whatever you want with it (such as logging to console) to see the filename, line number, and additional info:
uglify().on('error', console.error)
or in a larger context:
gulp.task('foo', () => {
return gulp.src([
'asset/src/js/foo/*.js',
'asset/src/js/bar/*.js',
])
.pipe(uglify().on('error', console.error))
.pipe(concat('bundle.js'))
.pipe(gulp.dest('./'));
});
This gives you a super helpful error!
{ GulpUglifyError: unable to minify JavaScript
at [stack trace])
cause:
{ SyntaxError: Continue not inside a loop or switch
[stack trace]
message: 'Continue not inside a loop or switch',
filename: 'ProductForm.js',
line: 301,
col: 37,
pos: 10331 },
plugin: 'gulp-uglify',
fileName:
'/asset/src/js/foo/ProductForm.js',
showStack: false }