URL Query String in PY
Extract the query string portion of a URL (everything between `?` and `#` or end-of-string).
Try it in the PY tester →Pattern
regexPY
\?([^#\s]+) (flags: g)Python (re) code
pyPython
import re
pattern = re.compile(r"\?([^#\s]+)")
input_text = "https://example.com/page?utm_source=google&id=42"
for m in pattern.finditer(input_text):
print(m.group(0))Stdlib `re` module — no third-party dependency. Works on Python 3.6+.
How the pattern works
\? matches the literal question mark. ([^#\s]+) captures everything up to (but not including) a `#` (fragment) or whitespace (URL boundary). Use URL parsing libraries for production, but this regex is handy for quick log scraping.
Examples
Input
https://example.com/page?utm_source=google&id=42Matches
?utm_source=google&id=42
Input
/api?token=abc#sectionMatches
?token=abc
Input
no query string hereNo match
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