正则表达式在方括号中保留顺序

Question

The context of the question is as follows. I wish to grab the file that wget is attempting to download, but need to ignore flags which may or may not appear. eg. wget -qO http://google.com/myfile.sh . The expected output would be: http://google.com/myfile.sh . For this example the regex:

r'wget\s-\w+\s([^\s]*)'

seems to do the trick. However, will not work when there is no flag.

In order for the (possibly absent) flag to work I attempted: r'wget\\s[-\\w+\\s]?([^\\s]*)' which I was hoping would say that "you can expect 0 or 1 instance of a dash followed by some characters", however it seems to think that the order of the -\\w+\\s is optional, atleast that is my explanation of the following results:

import re
re.search(r'wget\s-\w+\s([^\s]*)','wget -qO http://google.com/myfile.sh').group(1)
>>> 'http://google.com/myfile.sh'
re.search(r'wget\s[-\w+\s]?([^\s]*)','wget -qO http://google.com/myfile.sh').group(1)
>>> 'q0'
re.search(r'wget\s[-\w+\s]*([^\s]*)','wget -qO http://google.com/myfile.sh').group(1)
>>> '://google.com/myfile.sh'

Can someone explain the last two results, and show how to make sure that it matches 0 or more flags?

Answer 1

Try the following:

 wget\s*(?:-\w+)?\s*(.*)

https://regex101.com/r/aDWM3X/1 for reference

The reason why your example was not working is because you are using brackets which means "any of the following characters or range" while also using + after \\w (which does not mean 1 or more of \\w, it means look for any \\w, any + ....if you use a group then you can make the group optional with ? (zero or 1) or * if it can be zero or unlimited