gulp.run()已被弃用。改为使用任务依赖关系或gulp.watch任务触发

Question

我的gulpfile有一个错误。有人能帮我吗？

它以前是有效的。但是在我添加了gulp watch插件之后，它就崩溃了。

那么我需要使用什么呢？

这是我的Gulpfile.js。

这是一个日志：

azat@pc:~/git/lasttest$ gulp
[18:01:25] Using gulpfile ~/git/lasttest/gulpfile.js
[18:01:25] Starting 'default'...
gulp.run() has been deprecated. Use task dependencies or gulp.watch task triggering instead.
[18:01:25] Starting 'server'...
[18:01:25] Finished 'server' after 18 ms
[18:01:25] Finished 'default' after 37 ms
events.js:85
  throw er; // Unhandled 'error' event
        ^
Error: listen EADDRINUSE
at exports._errnoException (util.js:746:11)
at Server._listen2 (net.js:1156:14)
at listen (net.js:1182:10)
at Server.listen (net.js:1267:5)
at ConnectApp.server (/home/azat/git/lasttest/node_modules/gulp-connect/index.js:57:19)
at new ConnectApp (/home/azat/git/lasttest/node_modules/gulp-connect/index.js:37:10)
at Object.module.exports.server (/home/azat/git/lasttest/node_modules/gulp-connect/index.js:170:12)
at Gulp.<anonymous> (/home/azat/git/lasttest/gulpfile.js:112:11)
at module.exports (/home/azat/git/lasttest/node_modules/gulp/node_modules/orchestrator/lib/runTask.js:34:7)
at Gulp.Orchestrator._runTask (/home/azat/git/lasttest/node_modules/gulp/node_modules/orchestrator/index.js:273:3)

Answer 1

只需在watch方法中使用gulp的依赖系统即可。

// default depends on 'server' so that the server is started before
// the watches are started
gulp.task('default', ['server'], function() {
  // run jade whenever files in src/jade change
  gulp.watch('src/jade/**', ['jade']);

  // run images whenever files in src/images change
  gulp.watch('src/images/**/*', ['images']);
});

我不确定您是否需要在这里使用gulp batch，除非您正在运行一次更改多个图像的操作。