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JavaScript Gulp Basics Gulp your JavaScript Workflow Minifying JavaScript Files

Daniel Fang
Daniel Fang
3,489 Points
Posted by Daniel Fang
Daniel Fang
3,489 Points

Is necessary the last line of code '.pipe(gulp.dest('js'))' in the minifyScripts task?

// Minify Js
gulp.task('minifyScripts', () => {
    gulp.src('js/app.js')
        .pipe(uglify())
        .pipe(rename('app.min.js'))
        .pipe(gulp.dest('js'))
});

When I run the code above with and without the line .pipe(gulp.dest('js')) return the expect result (app.min.js) so I think this line is not necessary because the app.js file was already generated in the concat task, am I right?

2 Answers

Tom Geraghty
Tom Geraghty
24,174 Points
Tom Geraghty
Tom Geraghty
24,174 Points

It's not required to have it save the newly created file in the same directory. But this will rarely be useful in production where you want to separate your source/development code from your build/production code. Typically you will provide a destination folder in your build/ or public/ or whatever you call it for production folder.

Sebastian H
Sebastian H
19,905 Points
Sebastian H
Sebastian H
19,905 Points

From my understanding it isn't necessary in this situation because your gulp.src path and gulp.dest path are the same. Outputting the result to the same path is most probably the default behaviour, though I haven't checked the docs to be sure.

Gulp.dest is a method used to output the result of your above pipes in a different folder. For example, I used gulp.dest to output minified scripts from a public folder into a new 'dist' (distribution) folder. 

.pipe(gulp.dest('src/dist'));

The documentation on the gulp site also have great examples. Give 'em a read for more info.

https://github.com/gulpjs/gulp/blob/master/docs/API.md#gulpdestpath-options