使用'argparse'而不是sys.argv [英] Using 'argparse' as opposed to sys.argv

问题描述

我的脚本当前使用sys.argv来检查提供给程序的输入文件.

My script currently uses sys.argv to check for an input file provided to the program.

我正尝试使用argparse，但是我似乎无法使其正常工作.我能够设置它并添加一个参数，但是当我解析一个参数并打印该解析的参数时，我得到了一个命名空间.我怎样才能得到一个字符串?基本上，我想将参数作为字符串，然后打开具有该名称的文件.

I am trying to utilise argparse instead but I cant seem to get it to work. I was able to set it up and add an argument, but when I parse an argument, and print that parsed argument, I get a namespace. How can I get a string? Basically, I want to take the argument as a string, and open a file with that name.

当前，我的sys.argv是:

filename = sys.argv[1]
f = open(filename, 'r')

我的argparse如下打印Namespace:

arg = parser.parse_args()
print arg

如何使用它打开文件?我想使用argparse，因为参数的错误处理容易得多.

How can I use that to open a file? I want to use argparse since the error handlign for arguments there is a lot easier.

推荐答案

考虑使用with语句打开文件的首选方法(或类似方法):

think its preferable (or something!) to use the with statement to open the file like this:

# printfile.py
import argparse

parser = argparse.ArgumentParser(description="Opens a file and does cool stuff ^^")
parser.add_argument('filename', type=str, help="Path to file to open")
args = parser.parse_args()

with open(args.filename) as f:
    print '   my uber cool file:'
    print f.readlines()

指定这些关键字args还有助于使-h帮助文本选项更加简洁(整洁)

specifying those keyword args also helps make a pretty -h help text option (which is neat neat)

[dlam@dlam-63221:~] $ python printfile.py -h
usage: printfile.py [-h] filename

Opens a file and does cool stuff ^^

positional arguments:
    filename    Path to file to open

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