Jupyter Lab Cron作业失败KeyError：用户

Question

我正在使用Ancaconda的Jupyter Lab安排一些Jupyter笔记本。

我安装了jupyterlab_scheduler，当我尝试安排一个cronjob时，我会遇到“有错误提交作业”的错误。

我的cron语法是1 14 * * * *，该作业的命令是jupyter nbconvert - to Notebook -ececute c：/users/users/charl/conversion率sdr.ipynb

我的日志看起来像这样：

[E 2022-06-06 14:21:15.410 ServerApp] Uncaught exception POST /jupyterlab_scheduler/add?1654539675407 (::1)
HTTPServerRequest(protocol='http', host='localhost:8888', method='POST', uri='/jupyterlab_scheduler/add?1654539675407', version='HTTP/1.1', remote_ip='::1')
Traceback (most recent call last):
  File "c:\users\charl\anaconda3\lib\site-packages\tornado\web.py", line 1702, in _execute
    result = method(*self.path_args, **self.path_kwargs)
  File "c:\users\charl\anaconda3\lib\site-packages\tornado\web.py", line 3173, in wrapper
    return method(self, *args, **kwargs)
  File "c:\users\charl\anaconda3\lib\site-packages\jupyterlab_scheduler\handlers.py", line 114, in post
    with CronTab(user=os.environ["USER"]) as cron:
  File "c:\users\charl\anaconda3\lib\os.py", line 675, in __getitem__
    raise KeyError(key) from None
KeyError: 'USER'
[W 2022-06-06 14:21:15.412 ServerApp] Unhandled error
[E 2022-06-06 14:21:15.414 ServerApp] {
  "Host": "localhost:8888",
  "Accept": "*/*",
  "Referer": "http://localhost:8888/lab",
  "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/102.0.5005.63 Safari/537.36"
}
[E 2022-06-06 14:21:15.415 ServerApp] 500 POST /jupyterlab_scheduler/add?1654539675407 (::1) 
3.99ms referer=http://localhost:8888/lab

Answer 1

堆栈跟踪说，环境变量用户找不到。 用户如果使用Linux，则默认情况下定义，但如果您使用的是Windows。

看来该软件包是为Linux用户设计的。您可以通过定义用户环境变量手动。