Get information only by part of it's name?

RouterOS Scripting
1

Hello ,
I want to be able to print only the name of the vendor
to do something like this :

/system resource usb print where vendor="*Linux"

and then it will print me just :

DEVICE VENDOR NAME SPEED

0 2:1 Linux 3.3.5 ohci_hcd RB400 OHCI 12 Mbps
1 1:1 Linux 3.3.5 ehci_hcd RB400 EHCI 480 Mbps

and not the all list that include the usb modems

DEVICE VENDOR NAME SPEED

0 2:1 Linux 3.3.5 ohci_hcd RB400 OHCI 12 Mbps
1 1:1 Linux 3.3.5 ehci_hcd RB400 EHCI 480 Mbps
2 1:2 Sierra Wireless, Inco… MC7304 480 Mbps

how can I do this ?

Thanks ,

2

Use ~ instead of = to do pattern matching!
And of course the pattern should in your case be Linux* not *Linux

3

this is working

/system resource usb print where vendor~"Linux*" 
 # DEVICE VENDOR                   NAME                   SPEED                  
 0 2:1    Linux 3.3.5 ohci_hcd     RB400 OHCI             12 Mbps                
 1 1:1    Linux 3.3.5 ehci_hcd     RB400 EHCI             480 Mbps

now I want it to print me not the Linux and when I do this

/system resource usb print where vendor~!"Linux*"
 # DEVICE VENDOR                   NAME                   SPEED   
 
/system resource usb print where vendor!~"Linux*"
# DEVICE VENDOR                   NAME                   SPEED

and I don’t see the modem
why?

Thanks ,

4

! operator does not work with regexp.
What you can do is

print where !(vendor~"Linux")
5

thanks !

6

Yes of course I forgot to point out that “Linux*” does not mean what it would mean in filename usage, i.e. Linux with any characters after it,
but instead it means Linu plus 0 or more x characters, so Linu Linux Linuxx Linuxxx etc.
As there are no “anchoring” characters ^ or $ in the regexp this has no meaning, and the Linux* regexp effectively just searches for Linu
appearing anwhere in the string.