Gulp-concat根本不起作用-写入所有基本文件而不是一个

Question

I'm pulling out my hairs on this one, everything seems to be OK - as docs stated

gulp.task('compress-js-inline', function () {
    let rtask = gulp.src([
        'templates/js/jquery-1.7.min.js',
        'templates/js/bootstrap.js',
        'templates/js/js.inline.js'
    ]);
    rtask.pipe(gulpuglify());
    rtask.pipe(gulpconcat('js.inline.min.js'));
    rtask.pipe(gulp.dest('templates/out/'));
    return rtask;
});

In my templates/out as a result I get jquery-1.7.min.js , bootstrap.js , js.inline.js and no js.inline.min.js
Tried switching pipes places but not working.

Everything is newest version from npm.

Answer 1

Try to chain .pipe calls as follows:

gulp.task('compress-js-inline', function () {
  return gulp.src([
    'templates/js/jquery-1.7.min.js',
    'templates/js/bootstrap.js',
    'templates/js/js.inline.js'
  ])
    .pipe(gulpconcat('js.inline.min.js'))
    .pipe(gulp.dest('templates/out/'));
});