gulp js src - copy and usage of base

Question

I have gulp task to copy js files

This doesn't work

gulp.src('./**/*.js', {base: '../src/main/'})
    .pipe(gulp.dest('../target/dist'));

This works:

gulp.src('../src/main/**/*.js', {base: '../src/main/'})
        .pipe(gulp.dest('../target/dist'));

So whats the use of base here ? if i have to put whole path in first param, why should i use base ?

is there any official documentation about gulp src ? is it worth using gulp over grunt with limited documentation ?

[UPDATE BASED ON COMMENT]
Why am i using base ?

and moreoever gulp.src can take array of paths so i would need base.

Why are you using base in the first place? It's not in the docs because it's not used except in special circumstances. gulp.src is in the API docs, base is documented via glob-stream. — OverZealous
– OverZealous, Commented Mar 27, 2014 at 2:25

tim-montague · Accepted Answer · 2017-04-21 02:43:14Z

21

The use of .src() is documented on the vinyl-fs github repo: https://github.com/wearefractal/vinyl-fs

The base property is used to determine the file names when saving in .dest().

I think you need to set the current working directory:

gulp
  .src('./**/*.js', {cwd: '../src/main/'})
  .pipe(gulp.dest('../target/dist'))
;

answered Mar 13, 2015 at 1:37

tim-montague

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Comments

Nico Toub · Accepted Answer · 2014-04-17 14:38:31Z

-3

You should try to use 'root' param instead:

 gulp.src('./**/*.js', {root: '../src/main/'})
     .pipe(gulp.dest('../target/dist'));

answered Apr 17, 2014 at 14:38

Nico Toub

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