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Ramesh_M
New Contributor

Created on ‎07-28-2012 12:44 PM

Create Ip range if 3rd octect varies

Hi all, Could you please help me to create a address IP range 192.168.x.25 x may be 10 to 252. Kindly help us to do the same same. Regards Ramesh M

Ramesh M Technical Specialist - CCNA(Security), FCNSP, ACE, ASE, ITIL blogs.itzecuriry.in

Ramesh M Technical Specialist - CCNA(Security), FCNSP, ACE, ASE, ITIL blogs.itzecuriry.in
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ede_pfau
SuperUser
SuperUser

Created on ‎07-30-2012 03:49 AM

hi, I think you mean to create an address object covering that range? Then I' d specify an IP range as start IP: 192.168.10.1 end IP: 192.168.252.254 You cannot use the shorthand ' 192.168.[10-252].0/24' for this.

Ede

"Kernel panic: Aiee, killing interrupt handler!"
Ede"Kernel panic: Aiee, killing interrupt handler!"
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Ramesh_M
New Contributor
In response to ede_pfau

Created on ‎08-01-2012 10:00 PM

Hi, thanks for your post. I want to to achieve like ' 192.168.[10-252].0/24' . please let me know f any other options are there. Regards Ramesh M

Ramesh M Technical Specialist - CCNA(Security), FCNSP, ACE, ASE, ITIL blogs.itzecuriry.in

Ramesh M Technical Specialist - CCNA(Security), FCNSP, ACE, ASE, ITIL blogs.itzecuriry.in
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ede_pfau
SuperUser
SuperUser

Created on ‎08-02-2012 12:44 AM

You can use ' 192.168.0.0/16' which covers 192.168.0.0-192.168.254.254.

Ede

"Kernel panic: Aiee, killing interrupt handler!"
Ede"Kernel panic: Aiee, killing interrupt handler!"
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