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 potentially sensitive i…
Published: 2010-10-19 · CVSS: Unscored
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.
Source: CVE Program · Revision 8b16575c39f1e1810c5e863010da073d2d3285e7