UNIX for Dummies Questions & Answers

Hi All,

I am using the below command to find the latest file in a dir:

Code:

ls -tr $v_sftphomedir/$v_sourcefile |tail -1

or

Code:

ls -t1 $v_sftphomedir/$v_sourcefile |head -1

and the outpur returned is below:

Code:

/home/cobr_sftp/var/controllingload/Backup/Dbrwds_Div_1796050246.txt

I need only the filename(Dbrwds_Div_1796050246.txt) and not path. Who do i fetch only the filename from it?

Please help

Moderator's Comments:

Video tutorial on how to use code tags in The UNIX and Linux Forums.

Hi abhi_123,

Use sed. Add it at the end of the pipe chain:

Code:

... | sed -e 's|^.*/||'

What OS are you using since your output is not what I would expect?
IMHO using ls without -l should not give the absolute path...

A simpler ways include:

Code:

basename "$v_sourcefile"

and when using POSIX compliant shells

Code:

${v_sourcefile##*/}

---------- Post updated at 08:08 ---------- Previous update was at 07:58 ----------

If $v_sftphomedir/$v_sourcefile names a regular file (not a directory), then ls (no matter what options are given) will definitely print the full pathname of the file. If $v_sourcefile is named appropriately, it should name a single file; not a directory. Note that in this case the |tail -1 and |head -1, respectively, in the code examples given in the 1st message in this thread were no-ops.

Exact but the title was: and if it were a file then the -t is useless...

thanks all it worked.