I tried using tree command but I didn't know how .(I wanted to use tree because I don't want the files to show up , just the number)
Let's say c is the code for permission
For example I want to know how many files are there with the permission 751
Use find with the -perm flag, which only matches files with the specified permission bits.
For example, if you have the octal in $c, then run
find . -perm $c
The usual find options apply—if you only want to find files at the current level without recursing into directories, run
find . -maxdepth 1 -perm $c
To find the number of matching files, make find print a dot for every file and use wc to count the number of dots. (wc -l will not work with more exotic filenames with newlines as #BenjaminW. has pointed out in the comments. Source of idea of using wc -c is this answer.)
find . -maxdepth 1 -perm $c -printf '.' | wc -c
This will show the number of files without showing the files themselves.
If you're using zsh as your shell, you can do it natively without any external programs:
setopt EXTENDED_GLOB # Just in case it's not already set
c=0751
files=( **/*(#qf$c) )
echo "${#files[#]} files found"
will count all files in the current working directory and subdirectories with those permissions (And gives you all the names in an array in case you want to do something with them later). Read more about zsh glob qualifiers in the documentation.
Related
I have a directory which contains data for several years and several months.
Filenames have the format yy/mm/dd, f.e.
20150415,
20170831,
20121205
How can I find all data with month = 3?
F.e.
20150302,
20160331,
20190315
Thanks for your help!
ls -ltra ????03??
A question mark is a wildcard which stands for one character, so as your format seems to be YYYYmmDD, the regular expression ????03?? should stand for all files having 03 as mm.
Edit
Apparently the files have format YYYYmmDDxxx, where xxx is the rest of the filename, having an unknown length. This would correspond with regular expression *, so instead of ????03?? you might use ????03??*.
As far as the find is concerned: the same regular expression holds here, but as you seem to be working inside a directory (no subdirectories, at first sight), you might consider the -maxdepth switch):
find . -name "????03??*" | wc -l // including subdirectories
find . -maxdepth 1 -name "????03??*" | wc -l // only current directory
I would highly advise you to check without wc -l first for checking the results. (Oh, I just see the switch -type f, that one might still be useful too :-) )
I have a folder where different files can be located. I would like to check if it contains other files than .gitkeep and delete them, keeping .gitkeep at once. How can I do this ? (I'm a newbie when it comes to bash)
As always, there are multiple ways to do this, I am just sharing what little I know of linux :
1)find <path-to-the-folder> -maxdepth 1 -type f ! -iname '\.gitkeep' -delete
maxdepth of 1 specifies to search only the current directory. If you remove maxdepth, it will recursively find all files other than '.gitkeep' in all directories under your path. You can increase maxdepth to however deep you want find to go into directories from your path.
'-type f' specifies that we are just looking for files . If you want to find directories as well (or links, other types ) then you can omit this option.
-iname '.gitkeep' specifies a case insensitive math for '.gitkeep', the '\' is used for escaping the '.', since in bash, '.' is a regular expression.
You can leave it to be -name instead of -iname for case sensitive match.
The '!' before -iname, is to do an inverse match, i.e to find all files that don't have the name '.gitkeep', if you remove the '!', then you will get all files that match '.gitkeep'.
finally, '-delete' will delete the files that match this specification.
If you want to see what all files will be deleted before executing -delete, you can remove that flag and it will show you all the files :
find <path-to-the-folder> -maxdepth 1 -type f ! -iname '\.gitkeep'
(you can also use -print at the end, which is just redundant)
2) for i in `ls -a | grep -v '\.gitkeep'` ; do rm -rf $i ; done
Not really recommended to do it this way, since rm -rf is always a bad idea (IMO). You can change that to rm -f (to ensure it just works on file and not directories).
To be on the safe side, it is recommended to do an echo of the file list first to see if you are ready to delete all the files shown :
for i in `ls -a | grep -v '\.gitkeep'` ; do echo $i ; done
This will iterate thru all the files that don't match '.gitkeep' and delete them one by one ... not the best way I suppose to delete files
3)rm -rf $(ls -a | grep -v '\.gitkeep')
Again, careful with rm -rf, instead of rm -rf above, you can again do an echo to find out the files that will get deleted
I am sure there are more ways, but just a glimpse of the array of possibilities :)
Good Luck,
Ash
================================================================
EDIT :
=> manpages are your friend when you are trying to learn something new, if you don't understand how a command works or what options it can take and do, always lookup man for details.
ex : man find
=> I understand that you are trying to learn something out of your comfort zone, which is always commendable, but stack overflow doesn't like people asking questions without researching.
If you did research, you are expected to mention it in your question, letting people know what you have done to find answers on your own.
A simple google search or a deep dive into stack overflow questions would have provided you with a similar or even a better answer to your question. So be careful :)
Forewarned is forearmed :)
You can use find:
find /path/to/folder -maxdepth 1 ! -name .gitkeep -delete
I understand that using something like [^a]* will output all the files that do not start with "a".
If I want to echo files that contain at least 5 characters that do not start with "abc" (but can contain "abc" in the middle of the filename), how should I go about doing so?
I have
echo [^abc]?????*
but the output also removes files like "123abc", which I don't quite understand.
You don't indicate which OS your question applies to, but one way to determine the set of matching files on Mac OS X or Linux would be:
find . -maxdepth 1 -type f -name "?????*" | egrep -v "./abc"
Note that this will list only files in the current directory. If you want to include files in subdirectories, you'll need to remove the maxdepth argument.
Also note that these commands are case-sensitive. You'll need to use -iname and -i to make them case-insensitive.
EDIT:
If you really need to use the echo command, the following will work:
echo `find . -maxdepth 1 -type f -name "?????*" | egrep -v "./abc"`
Given a directory i'm looking for a bash one-liner to get a recursive list of all files with their size and modified time tab separated for easy parsing. Something like:
cows/betsy 145700 2011-03-02 08:27
horses/silver 109895 2011-06-04 17:43
You can use stat(1) to get the information you want, if you don't want the full ls -l output, and you can use find(1) to get a recursive directory listing. Combining them into one line, you could do this:
# Find all regular files under the current directory and print out their
# filenames, sizes, and last modified times
find . -type f -exec stat -f '%N %z %Sm' '{}' +
If you want to make the output more parseable, you can use %m instead of %Sm to get the last modified time as a time_t instead of as a human-readable date.
find is perfect for recursively searching through directories. The -ls action tells it to output its results in ls -l format:
find /dir/ -ls
On Linux machines you can print customized output using the -printf action:
find /dir/ -printf '%p\t%s\t%t\n'
See man find for full details on the format specifiers available with -printf. (This is not POSIX-compatible and may not be available on other UNIX flavors.)
find * -type f -printf '%p\t%s\t%TY-%Tm-%Td %Tk:%TM\n'
If you prefer fixed-width fields rather than tabs, you can do things like changing %s to %10s.
I used find * ... to avoid the leading "./" on each file name. If you don't mind that, use . rather than * (which also shows files whose names start with .). You can also pipe the output through sed 's/^\.\///'.
Note that the output order will be arbitrary. Pipe through sort if you want an ordered listing.
You could try this for recursive listing from current folder called "/from_dir"
find /from_dir/* -print0 | xargs -0 stat -c “%n|%A|%a|%U|%G” > permissions_list.txt
Lists files and directories passes through to stat command and puts all the info into a file called permissions_list.txt
“%n|%A|%a|%U|%G” will give you the following result in the file:
from_
dir|drwxr-sr-x|2755|root|root
from_dir/filename|-rw-r–r–|644|root|root
Cheers!
Is there any method in Linux to calculate the number of files in a directory (that is, immediate children) in O(1) (independently of the number of files) without having to list the directory first? If not O(1), is there a reasonably efficient way?
I'm searching for an alternative to ls | wc -l.
readdir is not as expensive as you may think. The knack is avoid stat'ing each file, and (optionally) sorting the output of ls.
/bin/ls -1U | wc -l
avoids aliases in your shell, doesn't sort the output, and lists 1 file-per-line (not strictly necessary when piping the output into wc).
The original question can be rephrased as "does the data structure of a directory store a count of the number of entries?", to which the answer is no. There isn't a more efficient way of counting files than readdir(2)/getdents(2).
One can get the number of subdirectories of a given directory without traversing the whole list by stat'ing (stat(1) or stat(2)) the given directory and observing the number of links to that directory. A given directory with N child directories will have a link count of N+2, one link for the ".." entry of each subdirectory, plus two for the "." and ".." entries of the given directory.
However one cannot get the number of all files (whether regular files or subdirectories) without traversing the whole list -- that is correct.
The "/bin/ls -1U" command will not get all entries however. It will get only those directory entries that do not start with the dot (.) character. For example, it would not count the ".profile" file found in many login $HOME directories.
One can use either the "/bin/ls -f" command or the "/bin/ls -Ua" command to avoid the sort and get all entries.
Perhaps unfortunately for your purposes, either the "/bin/ls -f" command or the "/bin/ls -Ua" command will also count the "." and ".." entries that are in each directory. You will have to subtract 2 from the count to avoid counting these two entries, such as in the following:
expr `/bin/ls -f | wc -l` - 2 # Those are back ticks, not single quotes.
The --format=single-column (-1) option is not necessary on the "/bin/ls -Ua" command when piping the "ls" output, as in to "wc" in this case. The "ls" command will automatically write its output in a single column if the output is not a terminal.
The -U option for ls is not in POSIX, and in OS X's ls it has a different meaning from GNU ls, which is that it makes -t and -l use creation times instead of modification times. -f is in POSIX as an XSI extension. The manual of GNU ls describes -f as do not sort, enable -aU, disable -ls --color and -U as do not sort; list entries in directory order.
POSIX describes -f like this:
Force each argument to be interpreted as a directory and list the name found in each slot. This option shall turn off -l, -t, -s, and -r, and shall turn on -a; the order is the order in which entries appear in the directory.
Commands like ls|wc -l give the wrong result when filenames contain newlines.
In zsh you can do something like this:
a=(*(DN));echo ${#a}
D (glob_dots) includes files whose name starts with a period and N (null_glob) causes the command to not result in an error in an empty directory.
Or the same in bash:
shopt -s dotglob nullglob;a=(*);echo ${#a[#]}
If IFS contains ASCII digits, add double quotes around ${#a[#]}. Add shopt -u failglob to ensure that failglob is unset.
A portable option is to use find:
find . ! -name . -prune|grep -c /
grep -c / can be replaced with wc -l if filenames do not contain newlines. ! -name . -prune is a portable alternative to -mindepth 1 -maxdepth 1.
Or here's another alternative that does not usually include files whose name starts with a period:
set -- *;[ -e "$1" ]&&echo "$#"
The command above does however include files whose name starts with a period when an option like dotglob in bash or glob_dots in zsh is set. When * matches no file, the command results in an error in zsh with the default settings.
I used this command..works like a charm..only to change the maxdepth..that is sub directories
find * -maxdepth 0 -type d -exec sh -c "echo -n {} ' ' ; ls -lR {} | wc -l" \;
I think you can have more control on this using find:
find <path> -maxdepth 1 -type f -printf "." | wc -c
find -maxdepth 1 will not go deeper into the hierarchy of files.
-type f allows filtering to just files. Similarly, you can use -type d for directories.
-printf "." prints a dot for every match.
wc -c counts the characters, so it counts the dots created by the print... which means counting how many files exist in the given path.
For the number of all file in a current directory try this:
ls -lR * | wc -l
As far as I know, there is no better alternative. This information might be off-topic to this question and you may already know this that under Linux (in general under Unix) directories are just special file which contains the list of other files (I understand that the exact details will be dependent on specific file system but this is the general idea). And there is no call to find the total number of entries without traversing the whole list. Please make me correct if I'm wrong.
use ls -1 | wc -l