CODEWITHSUNDEEP
javascriptgulp
Batman
Batman

Reputation: 6353

Why am I getting "Task is not in your gulpfile" when creating a task with arguments?

I have created a simple task that takes an argument:

gulp.task("js", () => {
    let stream = gulp
        .src("./src/**/*.js", { base: "./src/" })
        .pipe($.plumber())
        .pipe($.changed("temp"))
        .pipe(babel());

    // minify is min argument provided
    if (args.min == true) stream = stream.pipe($.uglify());

    // Update paths
    stream
        .pipe(
            $.preprocess({
                context: {
                    PATH: save.dest,
                    COMMIT: pkg.version
                }
            })
        )
        .pipe(gulp.dest("temp"));

    return stream;
});

I want to minify based on whether I'm pushing to QA or PROD:

gulp.task("push-dev", ["js"], function() {})

gulp.task("push-prod", ["js --min"], function() {})

But I get the error:

[11:51:32] Task 'js --min' is not in your gulpfile

Is what I'm trying possible? I'm trying to avoid having to create a whole new task just to handle minification in different environments.

Upvotes: 1

Views: 63

Answers (1)

crashmstr
crashmstr

Reputation: 28563

Define your function separately from your task instead of directly in your gulp.task.

let jsFunc = (doMininfy) => {
  let stream = gulp
    .src("./src/**/*.js", { base: "./src/" })
    .pipe($.plumber())
    .pipe($.changed("temp"))
    .pipe(babel());

  // assuming you meant the parameter to go here?
  if (doMininfy) stream = stream.pipe($.uglify());

  // Update paths
  stream
    .pipe(
        $.preprocess({
            context: {
                PATH: save.dest,
                COMMIT: pkg.version
            }
        })
    )
    .pipe(gulp.dest("temp"));

  return stream;
});

gulp.task("js", () => jsFunc());
gulp.task("js-min", () => jsFunc(true));
gulp.task("push-dev", ["js"], function() {});
gulp.task("push-prod", ["js-min"], function() {});

Upvotes: 1

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