main中的python命令行参数,跳过脚本名称 [英] python command line arguments in main, skip script name
问题描述
这是我的剧本
def main(argv):
if len(sys.argv)>1:
for x in sys.argv:
build(x)
if __name__ == "__main__":
main(sys.argv)
所以从命令行我写了python myscript.py commandlineargument
我希望它跳过myscript.py
并仅通过commandlineargument(n)
I want it to skip myscript.py
and simply run commandlineargument
through commandlineargument(n)
所以我知道我的for循环并不能解决这个问题,但是我如何使它做到这一点呢?
so I understand that my for loop doesn't account for this, but how do I make it do that?
推荐答案
由于 sys.argv 是一个列表,您可以使用切片sys.argv[1:]
:
Since sys.argv is a list, you can use slicing sys.argv[1:]
:
def main(argv):
for x in argv[1:]:
build(x)
if __name__ == "__main__":
main(sys.argv)
但是,如果您只能有一个脚本参数,只需按索引:sys.argv[1]
即可获取.但是,您应该检查sys.argv
的长度是否大于1,如果不是,则抛出错误,例如:
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.argv
is more than 1 and throw an error if it doesn't, for example:
def main(argv):
if len(argv) == 1:
print "Not enough arguments"
return
else:
build(argv[1])
if __name__ == "__main__":
main(sys.argv)
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