Python3。x调用函数与sys.argv[]-Python3.xcallfunctionwithsys.argv[]

7  

something like this:

是这样的:

#!/usr/bin/python3

import sys


def myFirstFunction():
    return open(sys.argv[1], 'r')

openFile = myFirstFunction()

for line in openFile:
    print (line.strip()) #remove '\n'? if not remove .strip()
    #do other stuff

openFile.close() #don't forget to close open file

then I would call it like the following:

然后我将其命名为:

./readFile.py temp.txt

/ readFile。py temp.txt

which would output the contents of temp.txt

它将输出temp.txt的内容?

sys.argv[0] outputs the name of script. In this case ./readFile.py

sys。argv[0]输出脚本的名称。在这种情况下。/ readFile.py

Updating My Answer
because it seems others want a try approach

更新我的答案，因为其他人似乎想尝试一下。

How do I check whether a file exists using Python? is a good question on this subject of how to check if a file exists. There appears to be a disagreement on which method to use, but using the accepted version it would be as followed:

如何检查是否存在使用Python的文件?这是一个关于如何检查文件是否存在的好问题。在使用哪种方法上似乎存在分歧，但使用公认的版本将会是:

 #!/usr/bin/python3

import sys


def myFirstFunction():
    try:
        inputFile = open(sys.argv[1], 'r')
        return inputFile
    except Exception as e:
        print('Oh No! => %s' %e)
        sys.exit(2) #Unix programs generally use 2 for 
                    #command line syntax errors
                    # and 1 for all other kind of errors.


openFile = myFirstFunction()

for line in openFile:
    print (line.strip())
    #do other stuff
openFile.close()

which would output the following:

它将输出以下内容:

$ ./readFile.py badFile
Oh No! => [Errno 2] No such file or directory: 'badFile'

you could probably do this with an if statement, but I like this comment on EAFP VS LBYL

你可以用if语句来做这个，但是我喜欢这个关于EAFP和LBYL的评论。

2  

For Python 3 you can use the context manager.

对于Python 3，您可以使用上下文管理器。

# argv[0] is always the name of the program itself.
try:
    filename = sys.argv[1]
except IndexError:
    print "You must supply a file name."
    sys.exit(2)

def do_something_with_file(filename):    
    with open(filename, "r") as fileobject:
        for line in fileobject:
            do_something_with(line)

do_something_with_file(filename)

2  

This is more than you asked for, but here's a common idiom I use for using command line arguments:

这比您所要求的要多，但这里有一个常用的习惯用法，用于使用命令行参数:

def do_something_with_file(filename):    
    with open(filename, "r") as fileobject:
        for line in fileobject:
            pass    # Replace with something useful with line.

def main(args):
    'Execute command line options.'
    try:
        src_name = args[0]
    except IndexError:
        raise SystemExit('A filename is required.')

    do_something_with_file(src_name)


# The following three lines of boilerplate are identical in all my command-line scripts.
if __name__ == '__main__':
    import sys
    main(sys.argv[1:])  # Execute 'main' with all the command line arguments (excluding sys.argv[0], the program name).