如何打开命令行中传递的文件 arguments

Question

I have my config encoded here:

@staticmethod
def getConfig(env):
    pwd=os.getcwd()
    if "win" in (platform.system().lower()):
        f = open(pwd+"\config_"+env.lower()+"_data2.json")
    else:
        f = open(pwd+"/config_"+env.lower()+"_data2.json")
    config = json.load(f)
    f.close()
    return config

@staticmethod
def isWin():
    if "win" in (platform.system().lower()):
        return True
    else:
        return False

I have 2 JSON files I want my script to read, but the way it's written above it only reads 1 of them. I want to know how to change it to something like:

f = open(pwd+"\config_"+env.lower()+"_data_f'{}'.json")

so it can read either dataset1.config or dataset2.config. I'm not sure if this is possible, but I want to do that so I can specify which file to run in the command line: python datascript.py -f dataset1.config or python datascript.py -f dataset2.config . Do I assign that entire open() call to a variable?

Answer 1

All you need to do is parse sys.argv to get the argument of the -f flag, then concatenate the strings and pass the result to open() . Try this:

import sys

### ... more code ...

    @staticmethod
    def getConfig(env):
        pwd = os.getcwd()
        file = None
        try:
            file = sys.argv[sys.argv.index('-f')+1]
        except ValueError:
            file = "data2.json"

        if "win" in (platform.system().lower()):
            f = open(pwd+"\config_"+env.lower()+"_" + file)
        else:
            f = open(pwd+"/config_"+env.lower()+"_" + file)
        config = json.load(f)
        f.close()
        return config

sys.argv.index('-f') gives the index of -f in the command line arguments, so the argument must be filename. The try-except statement will provide a default value if no -f argument is given.