Python regex not matching all patterns in string

Question

I want to match a curl command's header pattern like:

-H 'key: value'

or

-H "key: value"

This switch can appear in the middle somewhere or at the end of string.

My pattern:

>>> header_pattern = re.compile(' \-H (?:\'|\").+?:.+?(?:\'|\")(?:\s+|$)')

My string:

>>> a = " -H 'Authorization: Bearer xxx' -H 'Content-Type: text/plain' "

Now I attempted to find all instances of this pattern but its matching just the first pattern.

>>> headers = header_pattern.findall(a)
>>> headers
[" -H 'Authorization: Bearer xxx' "]

Yes because your pattern matches a space at the end and at the beginning. Change (?:\s+|$) to (?!\S) (not followed by a character that is not a whitespace) — Casimir et Hippolyte
– Casimir et Hippolyte, Commented Jun 28, 2016 at 17:53

alecxe · Accepted Answer · 2016-06-28 17:57:35Z

4

Why don't use argparse module instead of regular expressions:

import argparse
import shlex


parser = argparse.ArgumentParser()
parser.add_argument('command')
parser.add_argument('url')
parser.add_argument('-d', '--data')
parser.add_argument('-b', '--data-binary', default=None)
parser.add_argument('-H', '--header', action='append', default=[])
parser.add_argument('--compressed', action='store_true')


curl_command = "curl https://google.com  -H 'Authorization: Bearer xxx' -H 'Content-Type: text/plain'"

tokens = shlex.split(curl_command)
parsed_args = parser.parse_args(tokens)
print(parsed_args.header)

Prints ['Authorization: Bearer xxx', 'Content-Type: text/plain'].

(Inspired by uncurl package).

answered Jun 28, 2016 at 17:57

alecxe

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1 Comment

MohitC Over a year ago

Thanks a lot. Though this is not answer to question's concern but really gives me a lot better approach to achieve what I am trying to do.!

anubhava · Accepted Answer · 2016-06-28 17:54:28Z

2

You can use this regex to match all header options:

header_pattern = re.compile(r'-H\s*([\'"])(.+?)\1')

RegEx Demo

There is no real need to assert for whitespace before -H but you need you can use:

header_pattern = re.compile(r'(?<=\s)-H\s*([\'"])(.+?)\1')

answered Jun 28, 2016 at 17:54

anubhava

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