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() {
return gulp.src('app/**/*.js')
// .pipe(concat('script.js'))
.pipe(uglify())
.pipe(gulp.dest('./dist/'))
});
but i am getting error as
when i try to run gulp task as gulp concat Any help would be appreciated
The main error is generated when you're using ES6 format. Use the gulp-uglify-es module instead of 'gulp-uglify' to overcome this error.
var uglify = require('gulp-uglify-es').default;
Note: gulp-uglify-es is no longer being maintained. You may want to use terser
/gulp-terser
instead:
To see the error in console:
var gutil = require('gulp-util');
gulp.task('concat', function() {
return gulp.src('app/**/*.js')
// .pipe(concat('script.js'))
.pipe(uglify())
.on('error', function (err) { gutil.log(gutil.colors.red('[Error]'), err.toString()); })
.pipe(gulp.dest('./dist/'))
});
To find the exact file, with line number of error register and run this task:
var pump = require('pump');
gulp.task('uglify-error-debugging', function (cb) {
pump([
gulp.src('app/**/*.js'),
uglify(),
gulp.dest('./dist/')
], cb);
});
Have you used ES6 format in your script file? If so try ES5 now because when you do gulp-uglify it doesnt understand ES6 format as of now and after that try your code
gulp.task('concat', function() {
return gulp.src('app/**/*.js')
.pipe(concat('script.js'))
.pipe(uglify())
.pipe(gulp.dest('./dist/'))
});
and run the task gulp concat
it will work
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 }
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/'));
});
For me, it was a deprecated option "preserveComments" that generated the error (and completely crashed the script).
Found the issue using:
gulp.task('concat', function() {
return gulp.src('app/**/*.js')
.pipe(uglify())
.on('error', function (err) { console.log( err ) })
.pipe(gulp.dest('./dist/'))
});
you may have syntax error or used ES6 syntax. you can try https://skalman.github.io/UglifyJS-online/ firstly.
The main error to Unable to minifies JavaScript is the path not found. You can use the task usemin For this you need:
$ sudo npm install gulp-usemin --save-dev
$ sudo npm install gulp-livereload --save-dev
$ sudo npm install gulp-util --save-dev
and requires :
var usemin = require('gulp-usemin');
var livereload = require('gulp-livereload');
var gutil = require('gulp-util'); //gutil for error display
gulp.task('usemin',['jshint'], function(){
return gulp.src('./app/index.html')
.pipe(usemin({
css: [minifycss(),rev()],
scripts: [uglify().on('error', function(err) {gutil.log(gutil.colors.red('[Error]'), err.toString());this.emit('end');}),rev()]
}))
.pipe(gulp.dest('dist'))
.pipe(livereload());
});
Change the js: [uglify(), rev()] to scripts: [uglify(), rev()]
来源:https://stackoverflow.com/questions/40521506/gulpuglifyerrorunable-to-minify-javascript