main中的python命令行参数，跳过脚本名称
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python command line arguments in main, skip script name
提问by CQM
This is my script
这是我的脚本
def main(argv):
if len(sys.argv)>1:
for x in sys.argv:
build(x)
if __name__ == "__main__":
main(sys.argv)
so from the command line I write python myscript.py commandlineargument
所以从命令行我写 python myscript.py commandlineargument
I want it to skip myscript.pyand simply run commandlineargumentthrough commandlineargument(n)
我希望它跳过myscript.py和简单的运行commandlineargument通过commandlineargument(n)
so I understand that my for loop doesn't account for this, but how do I make it do that?
所以我知道我的 for 循环没有考虑到这一点，但我如何让它做到这一点？
采纳答案by alecxe
Since sys.argvis a list, you can use slicing sys.argv[1:]:
由于sys.argv是一个列表，您可以使用切片sys.argv[1:]：
def main(argv):
for x in argv[1:]:
build(x)
if __name__ == "__main__":
main(sys.argv)
But, if you can only have one script parameter, just get it by index: sys.argv[1]. But, you should check if the length of sys.argvis more than 1 and throw an error if it doesn't, for example:
但是，如果您只能有一个脚本参数，则只需通过 index: 获取它sys.argv[1]。但是，您应该检查的长度sys.argv是否大于 1，如果不是则抛出错误，例如：
def main(argv):
if len(argv) == 1:
print "Not enough arguments"
return
else:
build(argv[1])
if __name__ == "__main__":
main(sys.argv)
回答by jhermann
The real answer is to learn about and use argparse, though.
不过，真正的答案是了解和使用argparse。
