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我的 gulpfile 有错误。有谁能够帮我？

它以前有效。但是在我添加 gulp-watch 插件后它崩溃了。

那我需要用什么？

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)
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1 回答 1

2

只需使用 gulp 内置于 watch 方法的依赖系统即可。

// 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 gulp-batch ，除非您正在运行一次更改许多图像的东西。

于 2015-09-08T19:45:20.907 回答