UNIX Desktop Questions & Answers

Hi,

i have some files in one directory(say some sample dir) whose names will be like the following.
some_file1.txt
some_file2.txt.

i need to get the last modified file size based on file name pattern like some_

here i am able to get the value of the last modified file size using the following command:
echo `ls -lt /sample/some_* |awk '{print $5}'` | `awk '{ F = " " ; print $1 }'


but the problem is, i need to store this value in one variable, and needs to print that variable value.

i tried several ways like the following.

SIZE=`echo `ls -lt /sample/some_* |awk '{print $5}'` | `awk '{ F = " " ; print $1 }'`
echo $SIZE

but it is not successful.


Your immediate help will be greatly appreciated

Reagrds,
Eswar

Looking for this

Code:

filename=`ls -lt /sample/some_* |awk '{print $5}' | awk '{ F = " " ; print $1 }'`
echo $filename

You may try Perl, to avoid the pipeline:

Code:

size=$(perl -e'
  print -s (sort {-M $a <=> -M $b } glob "some_*")[0]
  ')

Thanks matrixmadhan for your immediate respone.

but i am getting the file size as

2580 111

i.e, both of the file sizes seperated by space.
Note: I have 2 files with name some_ under sample dir, whose file sizes are 2580 and 111.


i need only the last updated file size only.
i.e like 2580 (or) 111


Thanks,
eswar

Thanks matrixmadhan for your immediate respone.

but i am getting the file size as

2580 111

i.e, both of the file sizes seperated by space.
Note: I have 2 files with name some_ under sample dir, whose file sizes are 2580 and 111.


i need only the last updated file size only.
i.e like 2580 (or) 111


Thanks,
eswar

Code:

SIZE=`ls -lrt | awk 'END{ print $5 }'`
echo $SIZE

ls -lrt => display files with file that is last modified at the end
awk 'END{ print $5 }' => list only the file size of the last displayed file

SIZE => will contain the size of the file that is modified latest