Question

141

List all files in array with gulp.src()

rated 0 times [ 143] [ 2] / answers: 1 / hits: 33597 / 9 Years ago, sun, august 2, 2015, 12:00:00

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.

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bretd

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answered 9 Years ago keyanah · Accepted 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()