Question

如何让用户命名＆＃34; copied_holter.ecg＆＃34;他们自己？我提供了这个名字，但我希望用户选择他们想要的任何东西。

class MyParser(argparse.ArgumentParser):
def error(self, message):
    sys.stderr.write('error: %s\n' % message)
    self.print_help()
    sys.exit(2)



parser = MyParser(util_help)
parser.add_argument('filename', help='The full path/to/holter_file that you would like to parse.', action="store")
parser.add_argument('packet_start', help='The offset location of the start packet ID in base 10 decimal', action="store", type = int)
parser.add_argument('packet_end', help='The offset location of the end packet ID in base 10 decimal', action = "store", type = int)
parser.add_argument('new_filename', help='The name of the new file with the copied holter data chosen by user.', action = "store")
args = parser.parse_args()

start = args.packet_start
end = args.packet_end

if start % 5 != 0:
  start = int(5*round(float(start)/5))
if end % 5 != 0:
  end = int(5*round(float(end)/5))

try:
  print("Beginning copying of holter data...")

# Output the specific holter data
output_file = open(new_filename+".ecg", 'w')

Answer 1

从arg解析file_name，然后使用类似：

的内容

output_file = open(file_name+".ecg", "w")

Answer 2

你需要做的就是上线

output_file = open(file_name+".ecg", "w")

替换为

output_file = open(args.new_filename, 'w')

允许用户命名新文件

2 个答案: