I want to print only the name of one file which is last modified.
I use ::

$ls -ltr S07* |cut -f 8-9

but it is not working.

it gives the outut

-rw-r--r-- 1 aaaaa aaaaa 0 May 29 00:32 S070528D
-rw-r--r-- 1 aaaaa aaaaa 0 May 29 00:41 S070528S

I want the output

S070528S

please help me.

ls-command-here | awk '{print $NF}'

I could not understand.
Please explain what is "ls-command-here" and what is "$NF"
please tell.

ls-command-here is your ls command. Replace "ls-command-here" and put your ls command there

"print $NF" tells awk to print the field that is referenced by the total number of fields. i.e. if your ls output has 9 fields, NF will be 9, and hence $NF will print field 9 - the last field.

Cheers
ZB

thanx.
it works but it print All file name which are similar to that pattren.
that means it print
S070528D
S070528S

both file name. but I want to print only last modified file name.
If I have to change that ls command then plz tell me what is it.

ls-command-here | sed -n '$p' | awk '{print $NF}'

sed -n '$p' tells sed not to print anything (-n) except what is explicitly requested, and that is to print (p) the last line ($).

Cheers
ZB

Try this

ls -1 |tail -1

NOTE: ls option is not l (el), but 1 (one)