I'm struggling to make a script to find every file in your home directory that is less than 3 days old and then get a count of the total number of characters in all of these files.
Any suggestions?
Thanks.
Below command should work from the current directory.
find ./ -ctime -3 | xargs wc -c
Below should work for home directory.
find ~ -ctime -3 | xargs wc -c
Related
I want recursively count files in a Linux directory using this
find DIR_NAME -type f | wc -l
my question is , how to stop the find execution above if more than 1000 files are found in that folder ?
Is it possible ? Or do I need to wait find execution ?
You can use head to limit the number of lines returned by find:
find DIR_NAME -type f | head -n 1000 | wc -l
The head program will exit after the first thousand lines, find will receive SIGPIPE and exit as well.
so I just added the extension .txt to all files in a directory, I want to go beyond that and now count the number of files whose extension I just changed. Any help is appreciated!
To know the number of .txt files, you can simply do ls | grep '.txt$' | wc -l
To know the number of file you change, you need to either count them while you change the extension, or count the number before, the number after, and substract them.
This last method can be done like this:
oldnum="$(ls | grep '.txt$' | wc -l)"
# Do the rename here
newnum="$(ls | grep '.txt$' | wc -l)"
result=$((newnum - oldnum)) # $result now hold the number of renamed files
I hope you didn't forget an hour when you had modified files.
For example, if you have modified files 1 hour ago, just run in working directory:
find . -maxdepth 1 -type f -name '*\.txt' -cmin -65
This code will print all the files with *.txt name who were modified less than 65 minutes ago.
I'm looking for a specific directory file count that returns a number. I would type it into the terminal and it can give me the specified directory's file count.
I've already tried echo find "'directory' | wc -l" but that didn't work, any ideas?
You seem to have the right idea. I'd use -type f to find only files:
$ find some_directory -type f | wc -l
If you only want files directly under this directory and not to search recursively through subdirectories, you could add the -maxdepth flag:
$ find some_directory -maxdepth 1 -type f | wc -l
Open the terminal and switch to the location of the directory.
Type in:
find . -type f | wc -l
This searches inside the current directory (that's what the . stands for) for all files, and counts them.
The fastest way to obtain the number of files within a directory is by obtaining the value of that directory's kMDItemFSNodeCount metadata attribute.
mdls -name kMDItemFSNodeCount directory_name -raw|xargs
The above command has a major advantage over find . -type f | wc -l in that it returns the count almost instantly, even for directories which contain millions of files.
Please note that the command obtains the number of files, not just regular files.
I don't understand why folks are using 'find' because for me it's a lot easier to just pipe in 'ls' like so:
ls *.png | wc -l
to find the number of png images in the current directory.
I'm using tree, this is the way :
tree ph
while I was trying to practice my linux skills, but I could not solve this question.
So its basically saying "Write a bash script that takes a name of
directory as a command argument and printf the name of subdirectories
that has more than 5 files in it."
I thought we will use the find command but ı still could not figure it out. My code is:
find directory -type d -mindepth5
but it's not working.
You can use find twice:
First you can use find and wc to count the number of files in a given directory:
nb=$(find directory -maxdepth 1 -type f -printf "x\n" | wc -l)
This just asks find to output an x on a line for each file in the directory directory, proceeding non-recursively, then wc -l counts the number of lines, so, really, nb is the number of files in directory.
If you want to know whether a directory contains more than 5 files, it's a good idea to stop find as soon as 6 files are found:
nb=$(find directory -maxdepth 1 -type f -printf "x\n" | head -6 | wc -l)
Here nb has an upper threshold of 6.
Now if for each subdirectory of a directory directory you want to output the number of files (threshold at 6), you can do this:
find directory -type d -exec bash -c 'nb=$(find "$0" -maxdepth 1 -type f -printf "x\n" | head -6 | wc -l); echo "$nb"' {} \;
where the $0 that appears is the 0-th argument, namely {} that find will replaced by the subdirectory of directory.
Finally, you only want to display the subdirectory name if the number of files is more than 5:
find . -type d -exec bash -c 'nb=$(find "$0" -maxdepth 1 -type f -printf "x\n" | head -6 | wc -l); ((nb>5))' {} \; -print
The final test ((nb>5)) returns success or failure whether nb is greater than 5 or not, and in case of success, find will -print the subdirectory name.
This should do the trick:
find directory/ -type f | sed 's/\(.*\)\/.*/\1/g' | sort | uniq -c | sort -n | awk '{if($1>5) print($2)}'
Using mindpeth is useless here since it only lists directories at at least depth 5. You say you need subdirectories with more then 5 files in it.
find directory -type f prints all files in subdirectories
sed 's/\(.*\)\/.*/\1/g' removes names of files leaving only list of subdirecotries without filenames
sort sorts that list so we can use uniq
uniq -c merges duplicate lines and writes how many times it occured
sort -n sorts it by number of occurences (so you end up with a list:(how many times, subdirectory))
awk '{if($1>5) print($2)}' prints only those with first comlun 1 > 5 (and it only prints the second column)
So you end up with a list of subdirectories with at least 5 files inside.
EDIT:
A fix for paths with spaces was proposed:
Instead of awk '{if($1>5) print($2)}' there should be awk '{if($1>5){ $1=""; print(substr($0,2)) }}' which sets first part of line to "" and then prints whole line without a leading space (which was delimiter). So put together we get this:
find directory/ -type f | sed 's/\(.*\)\/.*/\1/g' | sort | uniq -c | sort -n | awk '{if($1>5){ $1=""; print(substr($0,2)) }}'
I'd like to construct a Linux command to list all files (with their full paths) within a specific directory (and subdirectories) ordered by access time.
ls can order by access time, but doesn't give the full path. find gives the full path, but the only control you have over the access time is to specify a range with -atime N (accessed at least 24*N hours ago), which isn't what I want.
Is there a way to order by access time and get the full path at once? I could just write a script, but it seems there should be a way to do this with the standard Linux programs.
find . -type f -exec ls -l {} \; 2> /dev/null | sort -t' ' -k +6,6 -k +7,7
This will find all files, and sort them by date and then time. You can then use awk or cut to extract the dates and files name from the ls -l output
you could try:
ls -l $(find /foo/bar -type f )
you can add other options (e.g. -t for sorting) to ls command to achieve your goal.
also you could add your searching criteria to find cmd
find . -type f | xargs ls -ldt should do the trick as long as there's not so many files that you hit the command like argument limit and spawn 2 instances of ls.
pwd | xargs -I % find % -type f
find . -type f -exec ls -l --full-time {} \; 2> /dev/null | sort -t' ' -k +6,6 -k +7,7
Alex's answer did not work for me since I had files older than one year and the sorting got messed up. The above adds the --full-time parameter which nuetralizes the date/time values and makes them sortable regardless of how old they are.