user3147268 user3147268 - 4 months ago 33
Javascript Question

List all files in array with gulp.src()

I'm trying to create an index of all file(path)s within a given folder. So far I worked with

gulp.src(filePath)
to achieve that. According to this blog post, it should work:


gulp.src(files) is a string or array containing the file(s) / file paths.


My current code:

gulp.task("createFileIndex", function(){
var index = gulp.src(['./content/**/*.*']);
console.log("INDEX:", index[0]);
});


By outputing the returned values of
gulp.src()
with
index[0]
I get
undefined
and the whole
index
outputs only a large dictionary without any filepaths.

Answer

According to the gulp documentation on gulp.src (https://github.com/gulpjs/gulp/blob/master/docs/API.md#gulpsrcglobs-options)

gulp.src(globs[, options])

Emits files matching provided glob or an array of globs. Returns a stream of Vinyl files that can be piped to plugins.

gulp.src('client/templates/*.jade')
  .pipe(jade())
  .pipe(minify())
  .pipe(gulp.dest('build/minified_templates'));

glob refers to node-glob syntax or it can be a direct file path.

globs

Type: String or Array

Glob or array of globs to read.

options

Type: Object

Options to pass to node-glob through glob-stream.

gulp adds some additional options in addition to the options supported by node-glob and glob-stream

So it seems you need to look in further on this. Otherwise this maybe helpful Get the current file name in gulp.src()

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