So I have a huge folder full subfolders with tons of files, and I add files to it all the time.
I need a subfolder in the root of that folder with a symlink of the last 10-20 files added so that I can quickly find the things I recently added. This is located on a NAS, but I have a linux box running Arch connected through NFS, so I assume the best way is to run a bash script with a find command followed by a loop of ln -sf, but I can't do it safely without help.
Something like this is required:
mkdir -p subfolder
find /dir/ -type f -printf '%T# %p\n' | sort -n | tail -n 10 | cut -d' ' -f2- | while IFS= read -r file ; do ln -s "$file" subfolder ; done
Which will create symlinks in subfolder pointing to the 10 most recently modified files in the directory tree rooted at /dir/
You could just create a shell function like:
recent() { ls -lt ${1+"$#"} | head -n 20; }
which will give you a listing of the 20 most recent items in the specified directories, or the current directory if no arguments are given.
Related
I have 4 folders (named W1, W3, W5, W7) and each one of those folders has approximately 30 subfolders (named M1 - M30). Each subfolder contains 24 .tif files (named Image_XX.tif).
I need to randomly "sample" each subfolder, more specifically, I need to get rid of 14 .tif files while keeping 10 .tif files in each subfolder.
I figure that deleting 14 files at random is easier than choosing 10 files at random and copying them to new subfolders within folders.
I thought that writing a bash script to do so would be the way, but I'm fairly new to programming and I'm stuck.
Below is one of the several scripts I've tried:
#!/bin/bash
for dir in /Users/Fer/Subsets/W1/; do
if [ -d "$dir" ]; then
cd "$dir"
gshuf -zn14 -e *.tif | xargs -0 rm
cd ..
fi
done
It runs for a second, but nothing seems to happen. Any help is appreciated.
For every subdirectory.
Find all files.
Choose a random number of files from the list.
Delete.
I think something along:
for dir in /Users/Fer/Subsets/W*/M*/; do
printf "%s\n" "$dir"/*.tif |
shuf -z -n 14 |
xargs -0 -t echo rm -v
done
Used some of the suggestions above and the code below worked:
for dir in /Users/Fer/Subsets/W*/M*; do
gshuf -zn14 -e "$dir"/*.tif | xargs -0 rm
done
This is the script that I'm using to move files with the string "john" in them (124334_john_rtx.mp4 , 3464r64_john_gty.mp4 etc) to a certain folder
find /home/peter/Videos -maxdepth 1 -type f -iname '*john' -print0 | \
xargs -0 --no-run-if-empty echo mv --target-directory=/home/peter/Videos/john/
Since I have a large amount of videos with various names written in the files, I want to make a bash script which moves videos with a string between the underscores to a folder named based on the string between the underscores. So for example if a file is named 4345655_ben_rts.mp4 the script would identify the string "ben" between the underscores, create a folder named as the string between the underscores which in this case is "ben" and move the file to that folder. Any advice is greatly appreciated !
My way to do it :
cd /home/peter/Videos # Change directory to your start directory
for name in $(ls *.mp4 | cut -d'_' -f2 | sort -u) # loops on a list of names after the first underscore
do
mkdir -p /home/peter/Videos/${name} # create the target directory if it doesn't exist
mv *_${name}_*.mp4 /home/peter/Videos/${name} # Moving the files
done
This bash loop should do what you need:
find dir -maxdepth 1 -type f -iname '*mp4' -print0 | while IFS= read -r -d '' file
do
if [[ $file =~ _([^_]+)_ ]]; then
TARGET_DIR="/PARENTPATH/${BASH_REMATCH[1]}"
mkdir -p "$TARGET_DIR"
mv "$file" "$TARGET_DIR"
fi
done
It'll only move the files if it finds a directory token.
I used _([^_]+)_ to make sure there is no _ in the dir name, but you didn't specify what you want if there are more than two _ in the file name. _(.+)_ will work if foo_bar_baz_buz.mp4 is meant to go into directory bar_baz.
And this answer to a different question explains the find | while logic: https://stackoverflow.com/a/64826172/3216427 .
EDIT: As per a question in the comments, I added mkdir -p to create the target directory. The -p means recursively create any part of the path that doesn't already exist, and will not error out if the full directory already exists.
I have a directory and there a multiple sub directories under that, i want to display the last 10 files recursively from all the subdirectories or if i can mention some date parameters to list also will be helpful
Save the name of all directories.
ls -R $PWD/* | grep ./ > allDirectories
The next line show 10 files of each directory (only if the directory don't have spaces in their name). You can add more options to the ls command (i.e sort by time using -c)
for directory in $(cat allDirectories); do echo '\n\n\n'$directory; ls $directory[1,-2] | head -n 10; done 2>> /dev/null
I am writing a shell script where it is checking if the bin directory is present under all the users directory under /home directory. The bin directory can be present directly under user directory or under the child directory of the user directory.
I mean let say I have a user as amit under /home. So the bin directory can be present directly as /amit/bin or can be present as /amit/jash/bin
Now my requirement is that I should have a list of users directories where the bin directory is not present either directly under user directory or under the child directory of the user directory. I tried the command as :
find /home -type d ! -exec test -e '{}/bin' \; -print
but it is not working. However when I am replacing the bin directory with some file, the command is working fine. Looks like this command is particularly for files. Is there any similar command for directories?? Any help on this will be greatly appreciated.
You're on the right track. The catch is that your test of "does the following directory NOT exist in this target" can't be expressed within find's conditions in such a way as to return only the top-level directory. So you need to nest, one way or another.
One strategy would be to use a for loop in bash:
$ mkdir foo bar baz one two
$ mkdir bar/bin baz/bin
$ for d in /home/*/; do find "$d" -type d -name bin | grep -q . || echo "$d"; done
foo/
one/
two/
This uses pathname expansion (globbing) to generate the list of directories to test, and then checks for the existence of "bin". If that check fails (i.e. find outputs nothing), the directory is printed. Note the trailing slash on /home/*/, which ensures that you will only be searching within directories, rather than files that might accidentally exist in /home/.
Another possibility might be to use nested finds, if you don't want to depend on bash:
$ find /home/ -type d -depth 1 -not -exec sh -c "find {}/ -type d -name bin -print | grep -q . " \; -print
/home/foo
/home/one
/home/two
This roughly duplicates the effect of the bash for loop above, but by nesting find within find -exec. It uses grep -q . to convert the output of find into an exit status that can be used as a condition for the outer find.
Note that since you're looking for a bin directory, we want to use test -d rather than test -e (which would also check for a bin file, which probably does not matter to you.)
Another option is to use bash process redirection. On multiple lines for easier reading:
cd /home/
comm -3 \
<(printf '%s\n' */ | sed 's|/.*||' | sort) \
<(find */ -type d -name bin | cut -d/ -f1 | uniq)
This unfortunately requires you to change to the /home directory before running, because of the way it strips off subdirectories. You can of course collapse this into a big long one-liner if you feel so inclined.
This comm solution also has the risk of failing on directories with special characters in their names, like newlines.
One last option is bash-only but more than a one-liner. It involves subtracting the directories containing "bin" from the full list. It uses an associative array and globstar, so it depends on bash version 4.
#!/usr/bin/env bash
shopt -s globstar
# Go to our root
cd /home
# Declare an associative array
declare -A dirs=()
# Populate the array with our "full" list of home directories
for d in */; do dirs[${d%/}]=""; done
# Remove directories that contain a "bin" somewhere inside 'em
for d in **/bin; do unset dirs[${d%%/*}]; done
# Print the result in reproducible form
declare -p dirs
# Or print the result just as a list of words.
printf '%s\n' "${!dirs[#]}"
Note that we're storing directories in the array index, which (1) makes it easy for us to find and delete items, and (2) insures unique entries, even if one user has multiple "bin" directories under their home.
cd /home
find . -maxdepth 1 -type d ! -name . | sort > a
find . -type d -name bin | cut -d/ -f1,2 | sort > b
comm -23 a b
Here, I'm making two sorted lists. The first contains all the home directories, and the second contains the top parent of any bin subdirectory. Finally I output any items from the first list not present in the second.
I need to list the files on Directory. But only true files, not the folders.
Just couldn't find a way to test if the file is a folder or a directory....
Could some one provide a pice o script for that?
Thanks
How about using find?
To find regular files in the current directory and output a sorted list:
$ find -maxdepth 1 -type f | sort
To find anything that is not a directory (Note: there are more things than just regular files and directories in Unix-like systems):
$ find -maxdepth 1 ! -type d | sort
In bash shell test -f $file will tell you if $file is a file:
if test -f $file; then echo "File"; fi
You can use ls -l | grep ^d -v to realize what you want. I tested it in Redhat 9.0, it lists only the true files, including the Hidden Files.
If u want to get a list the folders on Directory. But only folders, not the true files. You can use ls -l | grep ^d