从 Python 命令行获取字符串参数

Take string argument from Python command line

提问by Kreuzade

I need to take an optional argument when running my Python script:

运行 Python 脚本时，我需要采用可选参数：

python3 myprogram.py afile.json

or

或者

python3 myprogram.py

This is what I've been trying:

这是我一直在尝试的：

filename = 0
parser = argparse.ArgumentParser(description='Create Configuration')
parser.add_argument('filename', type=str,
                   help='optional filename')

if filename is not 0:
    json_data = open(filename).read()
else:
    json_data = open('defaultFile.json').read()

I was hoping to have the filename stored in my variable called "filename" if it was provided. Obviously this isn't working. Any advice?

如果提供了文件名，我希望将文件名存储在名为“filename”的变量中。显然这是行不通的。有什么建议吗？

采纳答案by dm03514

Please read the tutorial carefully. http://docs.python.org/howto/argparse.html

请仔细阅读教程。http://docs.python.org/howto/argparse.html

i believe you need to actually parse the arguments:

我相信你需要真正解析参数：

parser = argparse.ArgumentParser()
args = parser.parse_args()

then filename will be come available args.filename

然后文件名将可用 args.filename

回答by Junuxx

Check sys.argv. It gives a list with the name of the script and any command line arguments.

检查sys.argv。它给出了一个包含脚本名称和任何命令行参数的列表。

Example script:

示例脚本：

import sys
print sys.argv

Calling it:

调用它：

> python script.py foobar baz
['script.py', 'foobar', 'baz']

回答by Damian Schenkelman

If you are looking for the first parameter sys.argv[1]does the trick. More info here.

如果您正在寻找第一个参数sys.argv[1]就可以了。更多信息在这里。