Gulp Concat + Uglify命令JS错误

时间:2018-04-24 12:54:27

标签: javascript gulp gulp-uglify gulp-concat

我试图按特定顺序连接和uglify javascript文件。当我禁用uglify时,它按照我的订单运行良好。但是当我打开uglify时,JS命令被忽略了。我做错了什么?

Script_1.js

let script_one = () => {
   console.log(1);
};

Script_2.js

let script_two = () => {
   console.log(2);
};

Gulp任务

gulp.task('js', function() {
    return gulp.src(['js/Script1.js', 'js/Script2.js'])
      .pipe(rigger())
      .pipe(concat('main.js'))
      .pipe(babel({presets: ['env']}))
      .pipe(uglify())
      .pipe(gulp.dest('dist/js'));
});

输出

  

var script_two = function(){console.log(2)}; var script_one = function(){console.log(1)};

预期输出

  

var script_one = function(){console.log(1)}; var script_two = function(){console.log(2)};

没有Uglify的输出

  var script_one = function(){
     console.log(1)
  };
  var script_two = function(){
     console.log(2)
  };

谢谢!

0 个答案:

没有答案