[Solved] Finding the latest file in a directory


 
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# 1  
Old 11-15-2012
[Solved] Finding the latest file in a directory

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

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Last edited by abhi_123; 11-15-2012 at 07:50 AM..
# 2  
Old 11-15-2012
Hi abhi_123,

Use sed. Add it at the end of the pipe chain:
Code:
... | sed -e 's|^.*/||'

# 3  
Old 11-15-2012
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...
# 4  
Old 11-15-2012
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 ----------

Quote:
Originally Posted by vbe
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...
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.
This User Gave Thanks to Don Cragun For This Post:
# 5  
Old 11-15-2012
Quote:
If $v_sftphomedir/$v_sourcefile names a regular file (not a directory)
Exact but the title was:
Quote:
Finding the latest file in a directory
and if it were a file then the -t is useless...Smilie
# 6  
Old 11-16-2012
thanks it worked

thanks all it worked. Smilie
 
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