Check folder size in Bash
I’m trying to write a script that will calculate a directory size and if the size is less than 10GB, and greater then 2GB do some action. Where do I need to mention my folder name?
# 10GB SIZE="1074747474" # check the current size CHECK="`du /data/sflow_log/`" if [ "$CHECK" -gt "$SIZE" ]; then echo "DONE" fi Since this is a popular question — If any beginner is encountering the answers on this question and wants to learn more about what the heck du is and how everyone knows all these commands: You can type man du in your terminal to lookup the du command in the manual. This will display an output which you can view, and will summarize all the flags like -h, -c, -s, -b, -B, —apparent-size, etc. that answers are you suggesting you use. Then, you can decide for yourself how you best want to use du for your specific use case.
8 Answers 8
which will give you a brief output of the size of your target directory. Using a wildcard like * can select multiple directories.
If you want a full listing of sizes for all files and sub-directories inside your target, you can do:
- Add the argument -c to see a Total line at the end. Example: du -hcs or du -hc .
- Remove the argument -h to see the sizes in exact KiB instead of human-readable MiB or GiB formats. Example: du -s or du -cs .
With no directory path specified it will default to the current working directory. so du -hs == du -hs . .
if you just want to see the folder size and not the sub-folders, you can use:
You should know that du shows the used disk space; and not the file size.
You can use —apparent-size if u want to see sum of actual file sizes.
--apparent-size print apparent sizes, rather than disk usage; although the apparent size is usually smaller, it may be larger due to holes in ('sparse') files, internal fragmentation, indirect blocks, and the like And of course theres no need for -h (Human readable) option inside a script.
Instead You can use -b for easier comparison inside script.
But You should Note that -b applies —apparent-size by itself. And it might not be what you need.
-b, --bytes equivalent to '--apparent-size --block-size=1' so I think, you should use —block-size or -B
#!/bin/bash SIZE=$(du -B 1 /path/to/directory | cut -f 1 -d " ") # 2GB = 2147483648 bytes # 10GB = 10737418240 bytes if [[ $SIZE -gt 2147483648 && $SIZE -lt 10737418240 ]]; then echo 'Condition returned True' fi How to get the summarized sizes of directories and their subdirectories?
Let’s say I want to get the size of each directory of a Linux file system. When I use ls -la I don’t really get the summarized size of the folders. If I use df I get the size of each mounted file system but that also doesn’t help me. And with du I get the size of each subdirectory and the summary of the whole file system. But I want to have only the summarized size of each directory within the ROOT folder of the file system. Is there any command to achieve that?
The —total flag was helpful for me. E.g. du -sh —total applications/* . askubuntu.com/a/465436/48214
9 Answers 9
This does what you’re looking for:
- -s to give only the total for each command line argument.
- -h for human-readable suffixes like M for megabytes and G for gigabytes (optional).
- /* simply expands to all directories (and files) in / . Note: dotfiles are not included; run shopt -s dotglob to include those too.
Also useful is sorting by size:
If you have dot-directories in the root directory, you can use shopt -s dotglob to include them in the count.
It’s very usefull, because it’s simple and you can place what path you want instead of /* , e.g. ./ for current directory or ./* for each item in current directory.
@c1phr If your sort doesn’t have -h , you need to leave it off from du as well, otherwise the sorting will mix up kilo/mega/gigabytes. du -s /* | sort -nr .
I often need to find the biggest directories, so to get a sorted list containing the 20 biggest dirs I do this:
du -m /some/path | sort -nr | head -n 20 In this case the sizes will be reported in megabytes.
@Xedecima the problem with using h is the sort doesn’t know how to handle different sizes. For example 268K is sorted higher than 255M, and both are sorted higher than 2.7G
The -h (human readable) argument on the ‘sort’ command should properly read these values. Just like du’s -h flag exports them. Depending on what you’re running I’m guessing.
I like to use Ncdu for that, you can use the cursor to navigate and drill down through the directory structure it works really well.
The existing answers are very helpful, maybe some beginner (like me) will find this helpful as well.
- Very basic loop, but for me this was a good start for some other size related operations:
for each in $(ls) ; do du -hs "$each" ; done The following du invocation should work on BSD systems:
Right portable option combination on BSD/*NIX is du -sk /* . I hate the -k stuff soooo much. Linux’ -h totally rocks.
This isn’t easy. The du command either shows files and folders (default) or just the sizes of all items which you specify on the command line (option -s ).
To get the largest items (files and folders), sorted, with human readable sizes on Linux:
This will bury you in a ton of small files. You can get rid of them with —threshold (1 MB in my example):
The advantage of this command is that it includes hidden dot folders (folders which start with . ).
If you really just want the folders, you need to use find but this can be very, very slow since du will have to scan many folders several times:
find . -type d -print0 | sort -z | xargs --null -I '<>' du -sh '<>' | sort -h @podarok It’s available on OpenSUSE 13.2 Linux. Try to find a more recent version of your distribution or compile a more recent version of the package yourself.
Caching might have been a bad term. I was thinking of something like done in this port superuser.com/a/597173/121352 where we scan the disks contents once into a mapping and then continue using data from that mapping rather than hitting the disk again.
You might also want to check out xdiskusage. Will give you the same information, but shown graphically, plus allows to drill down (very useful). There are other similar utilities for KDE and even Windows.
Be aware, that you can’t compare directories with du on different systems/machines without getting sure, both share the same blocksize of the filesystem. This might count if you rsync some files from a linux machine to a nas and you want to compare the synced directory on your own. You might get different results with du because of different blocksizes.
You could use ls in conjunction with awk :
The output of ls is piped to awk . awk starts processing the data. Standard delimiter is space. The sum variable tot is initialised to zero; the following statement is executed for each row/line outputted by ls . It merely increments tot with the size. $5 stands for fifth column (outputted by ls ). At the end we divide by (1024*1024) to sum in megabytes.
If you would convert this into a script or function (.bashrc) you can also use it to get the size of certain subsets of directories, according to filetypes.
If you want system wide information, kdirstat may came in handy!
How do I determine the total size of a directory (folder) from the command line?
The -h flag on sort will consider «Human Readable» size values.
If want to avoid recursively listing all files and directories, you can supply the —max-depth parameter to limit how many items are displayed. Most commonly, —max-depth=1
du -h --max-depth=1 /path/to/directory I use du -sh or DOOSH as a way to remember it (NOTE: the command is the same, just the organization of commandline flags for memory purposes)
There is a useful option to du called the —apparent-size. It can be used to find the actual size of a file or directory (as opposed to its footprint on the disk) eg, a text file with just 4 characters will occupy about 6 bytes, but will still show up as taking up ~4K in a regular du -sh output. However, if you pass the —apparent-size option, the output will be 6. man du says: —apparent-size print apparent sizes, rather than disk usage; although the apparent size is usually smaller, it may be larger due to holes in (‘sparse’) files, internal fragmentation, indirect blocks
This works for OS X too! Thanks, I was really looking for a way to clear up files, both on my local machine, and my server, but automated methods seemed not to work. So, I ran du -hs * and went into the largest directory and found out which files were so large. This is such a good method, and the best part is you don’t have to install anything! Definitely deserved my upvote
@BandaMuhammadAlHelal I think there are two reasons: rounding ( du has somewhat peculiar rounding, showing no decimals if the value has more than one digit in the chosen unit), and the classical 1024 vs. 1000 prefix issue. du has an option -B (or —block-size ) to change the units in which it displays values, or you could use -b instead of -h to get the «raw» value in bytes.
How can I calculate the size of a directory?
The -s option means that it won’t list the size for each subdirectory, only the total size.
Actually du ‘s default unit is 512-byte blocks according to POSIX, and kilobytes on Linux (unless the environment variable POSIXLY_CORRECT is set) or with du -k .
if the directory is very big and have lots of subdirectories, it takes lots of time. almost 1 min.. is that normal? is there a way to get the size more rapidly?
While using a separate package such as ncdu may work well, the same comparison of many folders can be done, to some degree, by just giving du a list of folders to size up. For example to compare top-level directories on your system.
will list in human-readable format the sizes of all the directories, e.g.
656K ./rubberband 2.2M ./lame 652K ./pkg-config See the man page and the info page for more help:
-b , —bytes is equivalent to —apparent-size —block-size=1
The -c doesn’t make sense to use together with -s , right? -s only displays the size of the specified directory, that is the total size of the directory.
du -ahd 1 | sort -h will have a better visualization that sorted the items.
$ du -ahd 1 | sort -h 2.1M ./jinxing.oxps 2.1M ./jx.xps 3.5M ./instances_train2014_num10.json 5.9M ./realsense 7.8M ./html_ppt 8.5M ./pytorch-segmentation-toolbox 24M ./bpycv 24M ./inventory-v4 26M ./gittry 65M ./inventory 291M ./librealsense 466M . ls -ldh /etc drwxr-xr-x 145 root root 12K 2012-06-02 11:44 /etc
-l is for long listing ; -d is for displaying dir info, not the content of the dir, -h is for displaying size in huma readable format.
This isn’t correct, the person asking is clearly looking for footprint of a directory and it’s contents on disk. @sepp2k’s answer is correct.
The ls -ldh command only shows the size of inode structure of a directory. The metric is a reflection of size of the index table of file names, but not the actual size of the file content within the directory.
du -hax --max-depth=1 / | grep '9G' | sort -nr This helps find large directories to then sift through using du -sh ./*
You can use «file-size.sh» from the awk Velour library:
This gives a more accurate count than du. Unpack a tarball on two servers and use «du -s» (with or without —bytes) and you will likely see different totals, but using this technique the totals will match.
The original question asked the size, but did not specify if it was the size on disk or the actual size of data.
I have found that the calculation of ‘du’ can vary between servers with the same size partition using the same file system. If file system characteristics differ this makes sense, but otherwise I can’t figure why. The ‘ls|awk» answer that Steven Penny gave yields a more consistent answer, but still gave me inconsistent results with very large file lists.
Using ‘find’ gave consistent results for 300,000+ files, even when comparing one server using XFS and another using EXT4. So if you want to know the total bytes of data in all files then I suggest this is a good way to get it:
find /whatever/path -type f -printf "%s\n"|awk ' END '