CVE-2007-6738
pyftpdlib before 0.1.1 does not choose a random value for the port associated with the PASV command, which makes it easier for remote attackers to obtain…
published October 19, 2010
What it is
pyftpdlib before 0.1.1 does not choose a random value for the port associated with the PASV command, which makes it easier for remote attackers to obtain potentially sensitive information about the number of in-progress data connections by reading the response to this command.
The record
Technical detail
- CVSS
- Not scored
- CVSS v4.0
- Not supplied
- EPSS
- 0.01127 · 65.9th percentile
- Published
- 2010-10-19T19:00Z
EPSS history
Timeline