吞咽手表不看 [英] gulp watch doesn't watch

问题描述

以下是我的gulpfile.js。其中有几个任务，并且所有工作都正常 - 但最后一项任务 watch 不会。

我已经尝试过所有可能的路径和文件组合，但仍然没有运气。我在这里阅读了很多答案，但无法解决我的问题。我试图运行gulp.watch，不需要大量的观看，尝试了几种不同的方法来设置任务，等等......

  var gulp = require（'gulp'）; 
 var browserify = require（'browserify'）; 
 var babelify = require（'babelify'）; 
 var source = require（'vinyl-source-stream'）; 
 var watch = require（'gulp-watch'）; 
 
 gulp.task（'application'，function（）{
 return browserify（'./ public / resources / jsx / application.js'）
 .transform（babelify， {stage：0}）
 .bundle（）
 .on（'error'，function（e）{
 console.log（e.message）; 
 
 this.emit（'end'）; 
}）
 .pipe（source（'appBundle.js'））
 .pipe（gulp.dest（'./ public / resources / jsx'））; 
}）; 
 
 gulp.task（'watch'，function（）{
 gulp.watch（'./ public / resources / jsx / project / *。js'，['application']） 
}）; 
  

有人可以提出一个解决方案吗？

编辑：

以下是控制台输出：

  michael @ michael-桌面：/ opt / PhpstormProjects / app_april_2015 $ gulp watch 
 [23:05:03]使用gulpfile /opt/PhpstormProjects/app_april_2015/gulpfile.js 
 [23:05:03]开始观看。 .. 
 [23:05:03] 13 ms后完成'watch'
  

解决方案

您应该返回手表：

  gulp.task（'watch'，function（）{
 return gulp.watch（'./ public / resources / jsx / project / *。js'，['application']）
}） ; 
  

watch 是 async 方法，所以Gulp可以知道发生的事情的唯一方法就是如果你返回一个 promise ，该手表就是这样。

编辑

正如@JMM所述， watch 不会返回Promise。它返回一个 EventEmitter 。

Following is my gulpfile.js. There are a couple of more tasks in it and all are working fine - but the last task, watch doesn't.

I've tried every possible combination of paths and files and what so ever, but still I don't have luck. I've read many answers on this here, but couldn't solve my problem. I tried to run gulp.watch with and without requiring gulp-watch, tried several different approaches on how to set up the task and so on and so on...

var gulp = require('gulp');
var browserify = require('browserify');
var babelify = require('babelify');
var source = require('vinyl-source-stream');
var watch = require('gulp-watch');

gulp.task('application', function() {
    return browserify('./public/resources/jsx/application.js')
        .transform(babelify, { stage: 0 })
        .bundle()
        .on('error', function(e){
            console.log(e.message);

            this.emit('end');
        })
        .pipe(source('appBundle.js'))
        .pipe(gulp.dest('./public/resources/jsx'));
});

gulp.task('watch', function() {
    gulp.watch('./public/resources/jsx/project/*.js',['application'])
});

Can someone suggest a solution?

EDIT:

Here's the console output:

michael@michael-desktop:/opt/PhpstormProjects/app_april_2015$ gulp watch
[23:05:03] Using gulpfile /opt/PhpstormProjects/app_april_2015/gulpfile.js
[23:05:03] Starting 'watch'...
[23:05:03] Finished 'watch' after 13 ms

解决方案

You should return watch:

gulp.task('watch', function() {
    return gulp.watch('./public/resources/jsx/project/*.js',['application'])
});

watch is an async method, so the only way Gulp can know that something is happening is if you return a promise, which watch does.

Edit

As @JMM stated, watch doesn't return a Promise. It returns an EventEmitter.

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