I have been trying to get to count all the empty folders in a certain directory. sub-directories excluded. i used the code below but i don't know how to define empty folders or folders that contain files.
echo "$(ls -l | egrep -l $1/* | wc -l)"
the $1 will be the user argument in the command line. example: ./script.sh ~/Desktop/backups/March2021.
Edit - im not allowed to use find command
Edit 2 - ls -l * | awk '/total 0/{print last}{last=$0}' | wc -l this script works but lists all folders even if the directory contains files and data or if the directory is empty.
What about this:
grep -v "." *
I mean the following: "." means any character (I'm not sure the syntax is correct), so basically you look for every file which not even contain any character.
You should not parse ls (directories or file names with newlines), so this solution is only for the assignment:
ls -d */ */* | cut -d/ -f1 | sort | uniq -u | wc -l
Explanation:
ls -d */ shows all directories. This is combined with ls -d */* which will also show contents in the directories.
The resulting output will show all directories.
Empty directories will be shown only once, so you want to look for unique lines.
With the cut you only see the name of the directory, not the files in the directory.
The sort could be skipped here, the ls will give sorted output. When you change the solution to find (next assignment?) the sort might be needed.
uniq can look for lines that occur once. The flag -u removes all lines that have duplicates, so it will show the unique lines in the output.
Related
I'm trying to list some files, but I only want the file names, in order of file date. I've tried a few commands but they don't see to work.
I know that using this code I can list only the file names:
ls -f *
And I know that using this command I can list the files sorted by date:
ls -ltr *
So I have tried using this command to list the file names only, sorted by file date, but it doesn't sort by date:
ls -ltr -f *
That last command simply lists the file names, but sorted by file name, not date.
Any ideas how I can do this with a simple ls command?
FYI, once I get this working my ultimate goal is to only list the most recently created 10 file names, using something like this:
ls -ltr -f * | tail -10
You could try the following command:
ls -ltr | awk '{ print $9 }' | tail -n +2
It extracts the file names from the ls -ltr command.
According to the manual for ls, the -f flag is used to,
-f do not sort, enable -aU, disable -ls --color
One way of extracting only files would be,
ls -p | grep -v /
The option -p is used to append a '/' to a directory name, we can grep for lines not containing a '/'.
To extract 10 most recently used files you could do the following
ls -ptr * | grep -v / | tail -10
My UNIX is quite rusty but what I want is to search a location in UNIX for files containing the two separate words in their text of "generate" and "process", but both words on the SAME LINE?
I know there are script files that contain details of the script author and its function noted at the top of the script. For example, the start of one such script contains the following;
function: generate sales overtime process
I have tried things like the following (again my UNIX is rusty)
grep -rwl . -e "generate" | "process"
But this gives errors such unrecognised commands
What I want is a list of Progress files like;
salesovertime1.p
salestravel1.p
salesexpenses1.p
salesexpenses2.p
If you search for file then find is appropriate, and then you may filter with grep:
find . -exec grep -H generate {} \; 2> /dev/null | grep process
will find recursively every file form the current directory, then filter ones that contains word "generate" and the filter again with ones that contains word "process". Filenames will be produced on output with option -H (GNU grep) and error messages begin redirected to /dev/null.
Now if you want filenames only, you can use :
find . -exec grep -H generate {} \; 2> /dev/null | grep process | cut -f1 -d\:
If you want "generate" and "process" in the same file but on different lines, the following will do it:
grep process `find . -exec grep -H generate {} \; 2> /dev/null | cut -f1 -d\:` 2> /dev/null | cut -f1 -d\:
The find generates a file list that is used to grep against and extract filenames with redirecting errors again.
I have thousands of files named "DOCUMENT.PDF" and I want to rename them based on a numeric identifier in the path. Unfortunately, I don't seem to have access to the rename command.
Three examples:
/000/000/002/605/950/ÐÐ-02605950-00001/DOCUMENT.PDF
/000/000/002/591/945/ÐÐ-02591945-00002/DOCUMENT.PDF
/000/000/002/573/780/ÐÐ-02573780-00002/DOCUMENT.PDF
To be renamed as, without changing their parent directory:
2605950.pdf
2591945.pdf
2573780.pdf
Use a for loop, and then use the mv command
for file in *
do
num=$(awk -F "/" '{print $(NF-1)}' file.txt | cut -d "-" -f2);
mv "$file" "$num.pdf"
done
You could do this with globstar in Bash 4.0+:
cd _your_base_dir_
shopt -s globstar
for file in **/DOCUMENT.PDF; do # loop picks only DOCUMENT.PDF files
# here, we assume that the serial number is extracted from the 7th component in the directory path - change it according to your need
# and we don't strip out the leading zero in the serial number
new_name=$(dirname "$file")/$(cut -f7 -d/ <<< "$file" | cut -f2 -d-).pdf
echo "Renaming $file to $new_name"
# mv "$file" "$new_name" # uncomment after verifying
done
See this related post that talks about a similar problem: How to recursively traverse a directory tree and find only files?
I have a directory full of files with names such as:
file_name_is_001
file_name_001
file_name_is_002
file_name_002
file_name_is_003
file_name_003
I want to copy only the files that don't contain 'is'. I'm not sure how to do this. I have tried to search for it, but can't seem to google the right phrase to find the results.
Details depend on operating system, shell, etc.
For a unix system a quite verbose but easy to understand approach could look like this (please mind that I didn't test it):
mkdir some_temporary_directory
mv *_is_* some_temporary_directory
cp * where_ever_you_want_to_copy_it
mv some_temporary_directory/* .
rmdir some_temporary_directory
You can do this using bash. First, here's a command to get you a list of files that don't contain the text _is_:
ls | grep -v "_is_"
This takes the output of ls and matches all values with DO NOT contain _is_ using grep -v.
In order to then copy these files, we need to turn the lines output by grep into arguments of cp. We can do this using xargs:
ls | grep -v "_is_" | xargs -J % cp % new_folder
From the xargs man page, it is a tool to "build and execute command lines from standard input".
I'm very new to Unix, and currently taking a class learning the basics of the system and its commands.
I'm looking for a single command line to list off all of the user home directories in alphabetical order from the /etc/passwd directory. This applies only to the home directories, and not the contents within them. There should be no duplicate entries. I've tried many permutations of commands such as the following:
sort -d | find /etc/passwd /home/* -type -d | uniq | less
I've tried using -path, -name, removing -type, using -prune, and changing the search pattern to things like /home/*/$, but haven't gotten good results once. At best I can get a list of my own directory (complete with every directory inside it, which is bad), and the directories of the other students on the server (without the contained directories, which is good). I just can't get it to display the /home/user directories and nothing else for my own account.
Many thanks in advance.
/etc/passwd is a file. the home directory is usually at field/column 6, where ":" is the delimiter. When you are dealing with file structure that has distinct characters as delimiters, you should use a tool that can break your data down into smaller chunks for easier manipulation using fields and field delimiters. awk/cut etc, even using the shell with IFS variable set can do the job. eg
awk -F":" '{print $6}' /etc/passwd | sort
cut -d":" -f6 /etc/passwd |sort
using the shell to read the file
while IFS=":" read -r a b c d e home_dir g
do
echo $home_dir
done < /etc/passwd | sort
I think the tools you want are grep, tr and awk. Grep will give you lines from the file that actually contain home directories. tr will let you break up the delimiter into spaces, which makes each line easier to parse.
Awk is just one program that would help you display the results that you want.
Good luck :)
Another hint, try ls --color=auto /etc, passwd isn't the kind of file that you think it is. Directories show up in blue.
In Unix, find is a command for finding files under one or more directories. I think you are looking for a command for finding lines within a file that match a pattern? Look into the command grep.
sed 's|\(.[^:]*\):\(.[^:]*\):\(.*\):\(.[^:]*\):\(.[^:]*\)|\4|' /etc/passwd|sort
I think all this processing could be avoided. There is a utility to list directory contents.
ls -1 /home
If you'd like the order of the sorting reversed
ls -1r /home
Granted, this list out the name of just that directory name and doesn't include the '/home/', but that can be added back easily enough if desired with something like this
ls -1 /home | (while read line; do echo "/home/"$line; done)
I used something like :
ls -l -d $(cut -d':' -f6 /etc/passwd) 2>/dev/null | sort -u
The only thing I didn't do is to sort alphabetically, didn't figured that yet