I have a requirement to sort all directories of current directory in descended order by size.
I tried following
du -sh * | sort -rg
It is listing all the folders by size but it's just listing by size of folder by values. However it's not sorting correcting. 100 MB Dir should be listed before 200KB.
Any help will be appreciable.
-g is for floats. For human-readable output use human-readable sort:
du -sh * | sort -rh
If you have numfmt utility from coreutils, you can use numeric sort with numfmt formatting:
du -s * | sort -rn | numfmt --to=iec -d$'\t' --field=1
I prefer to just go straight to comparing bytes.
du -b * | sort -nr
du -b reports bytes.
sort -n sorts numerically. Obviously, -r reverses.
My /tmp before I clean it -
104857600 wbxtra_RESIDENT_07202018_075931.wbt
815372 wbxtra_RESIDENT_07192018_075744.wbt
215310 Slack Crashes
148028 wbxtra_RESIDENT_07182018_162525.wbt
144496 wbxtra_RESIDENT_07182018_163507.wbt
141688 wbxtra_RESIDENT_07182018_161957.wbt
56617 Notification Cache
20480 ~DFFA6E4895E749B423.TMP
16384 ~DF543949D7B4DF074A.TMP
13254 AdobeARM.log
3614 PhishMeOutlookReporterLoader.log
3448 msohtmlclip1/01
3448 msohtmlclip1
512 ~DF92FFF2C02995D884.TMP
28 ExchangePerflog_8484fa311d504d0fdcd6c672.dat
0 WPDNSE
0 VPMECTMP
0 VBE
Don't ask the machine to process human data:
du -s * | sort -rg
I want to get size of image. The image is in folder by name encodedImage.jpc
a="$(ls -s encodedImage.jpc | cut -d " " -f 1)"
temp="$(( $a*1024 * 8))"
echo "$a"
The output is not correct. How to get size? Thank You
Better than parsing ls output, the proper way is to use the command stat like this :
stat -c '%s' file
Check
man stat | less +/'total size, in bytes'
If by size you mean bytes or pretty bytes can you just use
ls -lh
-h When used with the -l option, use unit suffixes: Byte, Kilobyte, Megabyte, Gigabyte, Terabyte and Petabyte in order to reduce the number of digits to three or less using base 2 for sizes.
I guess the more complete answer if you're just trying to tear off the file size alone (I added the file name as well you can remove ,$9 to drop it)
ls -lh | awk '{print $5,$9}'
U can use this command
du -sh your_file
I have a file containing some file paths like:
./file1
./dir/file2
./dir3/dir4/fil3
etc
How can i find the total filesize of all of them? I know about "du" for getting the filesize of single file but no idea how to use a file.
Thank you
you can use du to give total size of multiple files
cat file | tr "\n" "\0" | du -ch --files0-from=- | tail -n1
Use awk for getting file size
cat file | awk '{system("ls -l " $0)}' | awk '{ TOTAL += $5} END { print TOTAL}'
GNU coreutils du only suggestion
EDIT: the named option --files0-from is a GNU extension, so this suggested solution won't work with any non GNU coreutils du version. As you don't semm to have it the awk version posted by Vivek Goel is the one you should try instead.
You already answered your own question. Using du is the key. The "missing" option you're looking for might be this one found in the manual pages. (man du)
--files0-from=F
summarize disk usage of the NUL-terminated file names specified in file F; If F is - then read names from standard input
Usage would be like this:
tr "\n" "\0" <file-list | du --files0-from=-
I am doing a homework which ask me to find the smallest file and biggest file under the directory, I have done that. But my output is something like
"the smallest file is xxx (xxxx -'filename' bytes)
I wish I could print something without the filename part.
I am using du -b $filename to get the size.
du -b | sort -rh | head -n 1 | awk '{print "The smallest file is " $1 " bytes"}'
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When I use ls or du, I get the amount of disk space each file is occupying.
I need the sum total of all the data in files and subdirectories I would get if I opened each file and counted the bytes. Bonus points if I can get this without opening each file and counting.
If you want the 'apparent size' (that is the number of bytes in each file), not size taken up by files on the disk, use the -b or --bytes option (if you got a Linux system with GNU coreutils):
% du -sbh <directory>
Use du -sb:
du -sb DIR
Optionally, add the h option for more user-friendly output:
du -sbh DIR
cd to directory, then:
du -sh
ftw!
Originally wrote about it here:
https://ao.ms/get-the-total-size-of-all-the-files-in-a-directory/
Just an alternative:
ls -lAR | grep -v '^d' | awk '{total += $5} END {print "Total:", total}'
grep -v '^d' will exclude the directories.
stat's "%s" format gives you the actual number of bytes in a file.
find . -type f |
xargs stat --format=%s |
awk '{s+=$1} END {print s}'
Feel free to substitute your favourite method for summing numbers.
If you use busybox's "du" in emebedded system, you can not get a exact bytes with du, only Kbytes you can get.
BusyBox v1.4.1 (2007-11-30 20:37:49 EST) multi-call binary
Usage: du [-aHLdclsxhmk] [FILE]...
Summarize disk space used for each FILE and/or directory.
Disk space is printed in units of 1024 bytes.
Options:
-a Show sizes of files in addition to directories
-H Follow symbolic links that are FILE command line args
-L Follow all symbolic links encountered
-d N Limit output to directories (and files with -a) of depth < N
-c Output a grand total
-l Count sizes many times if hard linked
-s Display only a total for each argument
-x Skip directories on different filesystems
-h Print sizes in human readable format (e.g., 1K 243M 2G )
-m Print sizes in megabytes
-k Print sizes in kilobytes(default)
For Win32 DOS, you can:
c:> dir /s c:\directory\you\want
and the penultimate line will tell you how many bytes the files take up.
I know this reads all files and directories, but works faster in some situations.
When a folder is created, many Linux filesystems allocate 4096 bytes to store some metadata about the directory itself.
This space is increased by a multiple of 4096 bytes as the directory grows.
du command (with or without -b option) take in count this space, as you can see typing:
mkdir test && du -b test
you will have a result of 4096 bytes for an empty dir.
So, if you put 2 files of 10000 bytes inside the dir, the total amount given by du -sb would be 24096 bytes.
If you read carefully the question, this is not what asked. The questioner asked:
the sum total of all the data in files and subdirectories I would get if I opened each file and counted the bytes
that in the example above should be 20000 bytes, not 24096.
So, the correct answer IMHO could be a blend of Nelson answer and hlovdal suggestion to handle filenames containing spaces:
find . -type f -print0 | xargs -0 stat --format=%s | awk '{s+=$1} END {print s}'
There are at least three ways to get the "sum total of all the data in files and subdirectories" in bytes that work in both Linux/Unix and Git Bash for Windows, listed below in order from fastest to slowest on average. For your reference, they were executed at the root of a fairly deep file system (docroot in a Magento 2 Enterprise installation comprising 71,158 files in 30,027 directories).
1.
$ time find -type f -printf '%s\n' | awk '{ total += $1 }; END { print total" bytes" }'
748660546 bytes
real 0m0.221s
user 0m0.068s
sys 0m0.160s
2.
$ time echo `find -type f -print0 | xargs -0 stat --format=%s | awk '{total+=$1} END {print total}'` bytes
748660546 bytes
real 0m0.256s
user 0m0.164s
sys 0m0.196s
3.
$ time echo `find -type f -exec du -bc {} + | grep -P "\ttotal$" | cut -f1 | awk '{ total += $1 }; END { print total }'` bytes
748660546 bytes
real 0m0.553s
user 0m0.308s
sys 0m0.416s
These two also work, but they rely on commands that don't exist on Git Bash for Windows:
1.
$ time echo `find -type f -printf "%s + " | dc -e0 -f- -ep` bytes
748660546 bytes
real 0m0.233s
user 0m0.116s
sys 0m0.176s
2.
$ time echo `find -type f -printf '%s\n' | paste -sd+ | bc` bytes
748660546 bytes
real 0m0.242s
user 0m0.104s
sys 0m0.152s
If you only want the total for the current directory, then add -maxdepth 1 to find.
Note that some of the suggested solutions don't return accurate results, so I would stick with the solutions above instead.
$ du -sbh
832M .
$ ls -lR | grep -v '^d' | awk '{total += $5} END {print "Total:", total}'
Total: 583772525
$ find . -type f | xargs stat --format=%s | awk '{s+=$1} END {print s}'
xargs: unmatched single quote; by default quotes are special to xargs unless you use the -0 option
4390471
$ ls -l| grep -v '^d'| awk '{total = total + $5} END {print "Total" , total}'
Total 968133
du is handy, but find is useful in case if you want to calculate the size of some files only (for example, using filter by extension). Also note that find themselves can print the size of each file in bytes. To calculate a total size we can connect dc command in the following manner:
find . -type f -printf "%s + " | dc -e0 -f- -ep
Here find generates sequence of commands for dc like 123 + 456 + 11 +.
Although, the completed program should be like 0 123 + 456 + 11 + p (remember postfix notation).
So, to get the completed program we need to put 0 on the stack before executing the sequence from stdin, and print the top number after executing (the p command at the end).
We achieve it via dc options:
-e0 is just shortcut for -e '0' that puts 0 on the stack,
-f- is for read and execute commands from stdin (that generated by find here),
-ep is for print the result (-e 'p').
To print the size in MiB like 284.06 MiB we can use -e '2 k 1024 / 1024 / n [ MiB] p' in point 3 instead (most spaces are optional).
Use:
$ du -ckx <DIR> | grep total | awk '{print $1}'
Where <DIR> is the directory you want to inspect.
The '-c' gives you grand total data which is extracted using the 'grep total' portion of the command, and the count in Kbytes is extracted with the awk command.
The only caveat here is if you have a subdirectory containing the text "total" it will get spit out as well.
This may help:
ls -l| grep -v '^d'| awk '{total = total + $5} END {print "Total" , total}'
The above command will sum total all the files leaving the directories size.