I'm trying to write a gulp plugin for the first time. The plugin is later supposed to validate some styleguide requirements for all css and js files in a project.
In order to properly debug the plugin while testing, I want to use gutil.log() for printout. To get started, I created a gulp plugin which simply outputs the path of the tested file.
module.exports = function(){...};
function validate(file) {
gutil.log(gutil.colors.yellow(file.path));
};
However, only one filename is printed in the gulp output. Debug shows, that all the correct files are listed with gulp.src(), but there is always only log output for a single file.
I then found the following here: Handle multiple files in a Gulp plugin
I tried using gulp-foreach with the following code:
gulp.task('head', function () {
return gulp.src('./src/app/**/*.{css,js}')
.pipe(foreach(function (stream, file) {
return stream
.pipe(validate());
}));
});
But the result remains the same,there is only one log entry. I feel like I'm missing something very obvious here...