print the last field of a record using awk

[Lesleyb]

Hi,

My file looks something like the following

/data/udb/sync/exp/hrzsync.exp.1
/data/udb/sync/exp/hrzsync.exp.2
/data/udb/sync/exp/hrzsync.exp.3
etc............

I need to print the last field (FS="/&quot ie just the file names, removing the preceeding full path so that my output is

hrzsync.exp.1
hrzsync.exp.2
hrzsync.exp.3

Running this within a script the exact number of entries in the directory string will vary so I can't simply use $5 with FS="/"

Can anyone help

Thanks
Lesley

 

[Ygor]

You could use

awk -F/ '{print $NF}' file1

Maybe 'man basename' will help too.

 

[PHV]

Another way is to use sed:

sed -e 's!.*/!!' files.lst

Hope This Help
PH.

 

[dickiebird]

Not awk, but ksh solution :-

while read filens
do
print `basename $filens`
done < yourfile

Dickie Bird (-)))

 

[PHV]

Yet another way, with ksh:

while read path
do echo ${path##*/}
done<files.lst

Hope This Help
PH.