Gulp - pass a parameter to an internal function

Question

I have a watch task that goes through an array. The lines inside the array are used to get different file paths for viewing. This is a line with a watch function. The on-function function looks for changes inside these files. If there is a change, a new function is called that starts compiling the sass files.

function setWatchPipe(groupName) {
  gulp
    .watch(watchStyles[groupName])
    .on('change', function(e, groupName) {
        console.log(groupName);
        return gulp.src(appFiles.styles.src + groupName)
            .pipe($.plumber({ errorHandler: function (error) {}}))
            .pipe($.rubySass({
                style: sassStyle,
                compass: true,
                noCache: true
            }))
            .pipe(isProduction ? $.combineMediaQueries({ log: true }) : _.noop())
            .pipe(isProduction ? $.minifyCss({ keepSpecialComments: 1 }) : _.noop())
            .pipe($.plumber.stop())
            .pipe(gulp.dest(paths.styles.dest))
            .pipe($.size({
                showFiles: true
            }));

    });
}

var watchArray = ['base', 'front'];

gulp.task('watch', ['build-styles'], function() {
  for(var i = 0; i < watchArray.length; i++)
    setWatchPipe(watchArray[i]);
});

My problem is the "groupName" parameter inside. The watch task is responsive, but console.log is undefined. I need to get the output of strings from an array, but I don’t know how to pass the variable to this position (I needed to get the correct source file that needs to be written).

Hello lars

Answer 1

function(e) function(e, groupName). groupName undefined . change , .