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Rename file based on regex in Gulp

Say I have a LESS CSS directory structure like this:

less/
    core/
        _main.less
        header.less
        body.less
        footer.less
    contact/
        _main.less
        form.less
        details.less

I want to write a Gulp task that will look for the _main.less files in each subdirectory of the less/ dir, pass this to gulp-less and write the output to the css/ dir like this:

css/
    core.css
    contact.css

Because _main.less includes the other files in it's directory, only the _main.less files will have to be parsed, but I want the output file to have the name of the directory it's in.

So far I have this:

gulp.src('less/*/*/_main.less')
    .pipe(less())
    .pipe(gulp.dest('css'));

But this will create a directory structure like this:

css/
    core/
        _main.css
    contact/
        _main.css

But that's not what I want. I was thinking to use a regex to match the directory name, get this back in a matches var, and rename the file accordingly. Something like this:

gulp.src(/less\/[^\/]+\/([^\/]+)\/_main.less/)
    .pipe(less())
    .pipe(rename(function(context) {
        context.src.matches[1] + '.css'
     }))
    .pipe(gulp.dest('css'));

This code is just an example. I can't figure out how to do something like this, or if it's even possible.

Is this possible? If so, how?

like image 862
gitaarik Avatar asked Aug 28 '14 15:08

gitaarik


1 Answers

You look like you already have this, but maybe didn't see the gulp-rename plugin?

I don't even think you'll need to use a RegExp, something like this should work:

var rename = require('gulp-rename'),
    path = require('path'); // node Path

//...

gulp.src('less/**/_main.less')
    .pipe(less())
    .pipe(rename(function(filepath) {
        // replace file name to that of the parent directory
        filepath.basename = path.basename(filepath.dirname);
        // remove parent directory from relative path
        filepath.dirname = path.dirname(filepath.dirname);
        // leave extension as-is
     }))
    .pipe(gulp.dest('css'));
like image 195
OverZealous Avatar answered Nov 05 '22 18:11

OverZealous