hi all,

not sure if this has been posted b4 but i try to search but not valid.

this is my question:

when i do a ls -ltr there will be a list generated as follows:

-rw-r--r-- 1 root sys 923260 Jan 10 04:38 FilePolling.41025.083TL021.xml
-rw-r--r-- 1 root sys 1761337 Jan 10 04:40 FilePolling.41073.114TL020.xml
-rw-r--r-- 1 root sys 76 Jan 11 04:32 FilePolling.41423.024TL022.xml
-rw-r--r-- 1 root sys 82 Jan 11 04:40 FilePolling.41681.110TL022.xml
-rw-r--r-- 1 root sys 1136429 Jan 12 04:36 FilePolling.42235.068TL020.xml
-rw-r--r-- 1 root sys 82 Jan 12 04:40 FilePolling.42301.110TL022.xml
-rw-r--r-- 1 root sys 791807 Jan 13 04:31 FilePolling.42629.018TL024.xml

my question here is how do i grep the latest file (in this case the file will be the 13 Jan)?

can it be done?

thanks.

Simplest could be

ls -latr

If you looking for an xml file then, ls -latr | grep '.xml$'

grep "something" `ls -tr | tail -1`

Cheers
ZB

hi there,

thanks for the solution but dont think it's working.

what i meant is i want the latest file to be shown not the whole list of them ie:

-rw-r--r-- 1 root sys 791807 Jan 13 04:31 FilePolling.42629.018TL024.xml

Now I see what you want.

ls -latr | tail -1
-or-
ls -latr | sed -n '$p'

Cheers
ZB

I think the option a can be done away with. It would be ls -ltr | tail -1

hey guys,

its working...many thanks.

wee