Change root within zip?

[bartlewis]

gulp.task('default', function () {
    gulp.src('src/*')
        .pipe(zip('archive.zip'))
        .pipe(gulp.dest('dist'));
});

The above results in the "src" folder being the top level within the zip file. How do I make the contents of "src" the top level of the zip instead?

[bartlewis]

I tried the following, but this resulted in an empty zip, like #1.

gulp.task('default', function () {
    gulp.src('*', {cwd:'build'})
        .pipe(zip('archive.zip'))
        .pipe(gulp.dest('dist'));
});

[bartlewis]

Answer:

var path = require('path');

gulp.task('default', function () {
  gulp.src('*', {cwd: process.cwd() + path.sep + 'build'})
    .pipe(zip('archive.zip'))
    .pipe(gulp.dest('dist'));
});

[sindresorhus]

Ya, cwd probably expects an absolute path. I would tweak it slightly though:

var path = require('path');

gulp.task('default', function () {
  gulp.src('*', {cwd: path.join(process.cwd(), 'build')})
    .pipe(zip('archive.zip'))
    .pipe(gulp.dest('dist'));
});

@contra @phated is this the correct way to handle this?

[yocontra]

@bartlewis In 3.4 I'll make sure cwd accepts relative paths. Should be out soon

[bartlewis]

Thanks guys.

[kaelteschutz]

For me on OSX it is impossible to create an archive without the parent folder in it:

var scormArchiveTask = function () {
  gulp.src('**', {cwd: 'public'})
  .pipe(zip('archive.zip'))
  .pipe(gulp.dest('./public'))
}

[warengonzaga]

Anything doesnt work! I've tried the code but unlucky to achieve.