extracting parameter from file

Hi
I have file

128 BES-Exchange [perf08:storage1] BES-Exchange/BES-Exchange.vmx winNetEnterpriseGuest vmx-04
144 BES-SVR [perf08:storage1] BES-SVR/BES-SVR.vmx winNetEnterpriseGuest vmx-04
176 BES AD [perf08:storage1] BES-AD/BES AD.vmx winNetEnterpriseGuest vmx-04

so in above file i want to extract only BES AD
the BES AD is seperated by tab i tried with gawk but not getting exact what i want
I want following output

BES-Exchange
BES-SVR
BES AD

Thanks
Brahma

awk -F"[" '{print $1}' infile|cut -d' ' -f1 something like this
or instead of cut you can use sed also...
awk -F"[" '{print $1}'|sed -e 's/^ [^ ] //'

cat File.txt
128 BES-Exchange [perf08:storage1] BES-Exchange/BES-Exchange.vmx winNetEnterpriseGuest vmx-04
144 BES-SVR [perf08:storage1] BES-SVR/BES-SVR.vmx winNetEnterpriseGuest vmx-04
176 BES AD [perf08:storage1] BES-AD/BES AD.vmx winNetEnterpriseGuest vmx-04

awk -F"/|[.]" '{print $2}' File.txt

when i do this it gives me output as

128 BES-Exchange

144 BES-SVR
176 BES AD

i want only BES and things

the pipe it to cut or sed to remove first word as i suggested..

bp_vardhaman,

du tried awk -F"/|[.]" '{print $2}' File.txt this ?

But i even want tab also but here its coming as without tab :slight_smile:

BES AD

vmware-vim-cmd vmsvc/getallvms| grep -v Vmid | awk -F"[" '{print $1}' | sed -e 's/[ !0-9]*//'

Got the right one after doing this

Thanks everybody who helped

#!/bin/bash
while read -r line
do
    line=${line%%[*}
    set -- ${line}
    echo $2
done < file
cat file | sed 's/\(.*\/\)\(.*\)\(\.vmx.*\)/\2/'