node.js - gulp监听目录内无监听文件时，监听将失效的问题

Question

问题 示例代码 {代码...} 描述 如上方代码中，watch任务，监听了./Style/JS/目录下的scss文件，但当JS目录中无任何scss文件时开启任务，则之后的监听文件add、change事件都不会触发。 该如何解决这个问题？ 使用...

巴扎黑 · Answer

The watch event of Gulp4.0 is based on the chokidar module. This problem will be fixed after upgrading Gulp to 4.0.

PHPz · Answer

The problem is that the wildcard character you use is ./. If you use gulp.watch('Style/JS'), then there will be no problem.

大家讲道理 · Answer

<p>Remove<code>./</code></p>