Webflags: g

URL Query String

Extract the query string portion of a URL (everything between `?` and `#` or end-of-string).

Available in

JavaScript / ECMAScript code

jsJavaScript

const re = new RegExp("\\?([^#\\s]+)", "g");
const input = "https://example.com/page?utm_source=google&id=42";
const matches = [...input.matchAll(re)];
console.log(matches.map(m => m[0]));

Uses `String.prototype.matchAll` for global iteration (Node 12+ / all modern browsers).

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+.

Go (RE2) code

goGo

package main

import (
	"fmt"
	"regexp"
)

func main() {
	re := regexp.MustCompile(`\?([^#\s]+)`)
	input := `https://example.com/page?utm_source=google&id=42`
	for _, match := range re.FindAllString(input, -1) {
		fmt.Println(match)
	}
}

Uses `regexp.MustCompile` (panics on bad patterns at startup) and `FindAllString` for all matches.

Pattern

regexengine-agnostic

\?([^#\s]+)   (flags: g)

Raw source: \?([^#\s]+)

How it 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.