javascript - Gulp.js: task based on forEach loop

I have an array of objects that looks like the following.

var bundles = [
  {
    src: 'js/my-ponent/*.js',
    bundleName: 'my-ponent.js'
  },
  {
    src: 'js/my-other-ponent/*.js',
    bundleName: 'my-other-ponent.js'
  }
]

I want the gulp task to process/concat every entry in the array, but it doesn't seem to work.

gulp.task('bundlejs', function(){
    return bundles.forEach(function(obj){
      return gulp.src(obj.src)
      .pipe(concat(obj.bundleName))
      .pipe(gulp.dest('js/_bundles'))
    });
});

I have an array of objects that looks like the following.

var bundles = [
  {
    src: 'js/my-ponent/*.js',
    bundleName: 'my-ponent.js'
  },
  {
    src: 'js/my-other-ponent/*.js',
    bundleName: 'my-other-ponent.js'
  }
]

I want the gulp task to process/concat every entry in the array, but it doesn't seem to work.

gulp.task('bundlejs', function(){
    return bundles.forEach(function(obj){
      return gulp.src(obj.src)
      .pipe(concat(obj.bundleName))
      .pipe(gulp.dest('js/_bundles'))
    });
});

Share Improve this question edited Aug 17, 2015 at 9:03 asked Aug 17, 2015 at 8:58 gkiely 3,0071 gold badge25 silver badges37 bronze badges

Add a ment |

2 Answers 2

Sorted by: Reset to default 13

You should probably be merging the streams and returning the result, so that the task will plete at the appropriate time:

var es = require('event-stream');

gulp.task('bundlejs', function () {
  return es.merge(bundles.map(function (obj) {
    return gulp.src(obj.src)
      .pipe(concat(obj.bundleName))
      .pipe(gulp.dest('js/_bundles'));
  }));
});

This solution does work, I had the wrong directory name. Doh.

https://github./adamayres/gulp-filelog helped me find the problem.

科技改变生活-雨落星辰 - 所有的伟大,都源于一个勇敢的开始

javascript - Gulp.js: task based on forEach loop - Stack Overflow

2 Answers 2

与本文相关的文章

评论列表(0)