I am trying to get the filenames which is $9 when i use ls -lrt
but i am getting only some of the name using below command.
ls -lrt *.edf
is it bcs the filename has spaces in between them?
How can I get the complete file name in cases like this?
y dont u try ..
ls -a *.edf > output
I also need file size along with that. So i can't use ls
ls -a *.edf | cut -d" " -f5,9-
This is not working. I am not getting the filename and size correctly. All my files have spaces in them and i have to get their full name along with size
instead of
ls -a *.edf | cut -d" " -f5,9-
try
ls -a *.edf | tr -s " " | cut -d" " -f5,9-
This will produce wrong results if the file name contains multiple sequential spaces.
Two stage process. Generate the filename(s) with "ls" (or "find") then find out the size.
ls -1 *\.edf | while read filename
do
filesize=`ls -lad "${filename}"|awk '{print $5}'`
echo "filename : ${filename}"
echo "filesize : ${filesize}"
done
or:
#!/bin/ksh
ls -l *\.edf| while read -r a a a a s a a a n
do
echo "[$s] [$n]"
done
ls -lrt *.edf | cut -d" " -f7- | sed 's/[A-Z][a-z][a-z] [0-9][0-9 ] [0-9][0-9]:[0-9][0-9] //'
Does not work with a normal "ls" listing (though the one posted is strangely spaced).
methyl,
did you run my command and it did not work for you? There was a small error, because the initial regex was for systems which have 00 Mon format in the ls -l output (where 00 represents the day and Mon month). I've adjusted the regex for Mon 00 format. Maybe you want to run it again?
However "ls -lrt *.edf | cut -d" " -f7-" always worked pretty nice for me 
How about this?:
ls -lrt -D %\ *.edf | awk '{print substr($0, index($0,$5))}'
Why not just use sed?
ls -lrt *.edf | sed 's/.*:.. //g'
Because he needs the size field as well 
Else he could simply run ls *.edf
... that and:
-rw-rw-rw- 1 user group 407 Oct 14 2009 foo bar
-rw-rw-rw- 1 user group 407 Oct 14 2009 foo:ba .txt
1 Like
Thanks Vgersh. Your solution worked fine. Many thanks to all who gave wonderful responses. But I needed file size and name and hence vgresh solution worked for me.
---------- Post updated at 02:32 PM ---------- Previous update was at 02:28 PM ----------
vgersh,
Can you please explain the script?
On behalf of vgersh99. The "while read -r" statement reads each matching line from the "ls -l *\.edf" statement in turn. It places the results into environment variables called $a, $s, $n . $a is a variable to use unwanted fields, $s is the file size, $n is the file name. The clever bit of the script is that $n contains the whole filename whether or not it contains spaces or special characters. This is because the last field in a "read" statement is filled with any remaining characters from the input line. See "man read".
The square brackets in the output echo are not special syntax, they just help to show where each field starts and stops when testing the script.
Note the backslash in "ls -l *\.edf". This is good practice because unix is not MSDOS and a fullstop in a filename is just another character but it can have special meaning in a regular expression (which this isn't).
if you have du
du -s *.edf
@trey85stang
The file sizes output from "du -s" are quite different from "ls -la", but they are the same as those from "ls -las".
The file sizes from "du -s" and "ls -las" are in units of disc blocks and reflect the actual space taken by the files. A disc block is usually 512 bytes ... but it depends on the system.
The output from "ls -l" is how much data there is in each file up to the EOF marker. The EOF marker may be part-way through a disc block.
On one of my systems a single byte text file takes 8,192 bytes of disc space.