Python使用命令promp传递参数

Question

我对Python还不熟悉。但是，我尝试使用命令提示符运行代码（Found here），出现以下错误：

usage: vogon.py [-h] [--auth_host_name AUTH_HOST_NAME]
                [--noauth_local_webserver]
                [--auth_host_port [AUTH_HOST_PORT [AUTH_HOST_PORT ...]]]
                [--logging_level {DEBUG,INFO,WARNING,ERROR,CRITICAL}]
                [--youtube_upload] [--preview_line PREVIEW_LINE]
                config_file
vogon.py: error: argument --youtube_upload: ignored explicit argument 'Yes'

我用来执行它的命令是：

>python vogon.py --config_file=sample_win.json --auth_host_name=http://localhost/ --auth_host_port=8080 --logging_level=DEBUG --youtube_upload=Yes --preview_line=2

我怎样才能解决这个问题？你知道吗

更新

> File "vogon.py", line 351, in <module>
>     main()   File "vogon.py", line 348, in main
>     generate_videos(args.config_file, args.youtube_upload, args.preview_line, args)   File "vogon.py", line 68, in
> generate_videos
>     video = generate_video(config, row, (i + 1))   File "vogon.py", line 100, in generate_video
>     filters = filter_strings(image_overlays, text_overlays)   File "vogon.py", line 127, in filter_strings
>     ovr['y'], ovr['h_align'], ovr['start_time'], KeyError: 'h_align

'

Answer 1

选项定义为

parser.add_argument(" youtube_upload",
        help="Upload generated videos to YouTube",
        action="store_true")

youtube_upload不接受任何参数；删除=YES。你知道吗

完整的命令（使用shell行继续分割多行）看起来是

python vogon.py  auth_host_name=http://localhost/ \
                 auth_host_port=8080 \
                 logging_level=DEBUG \
                 youtube_upload \ 
                 preview_line=2 \
                sample_win.json