I have file getting generated (testfile) eachtime a script is run in working directory. I have to copy that generated file to a directory (testfolder) and have it incremented by a .ext
If the script is run for first time then copy the testfile to testfolder as "testfile.0" and when run second time copy the testfile to testfolder as "testfile.1" and so on.
My script:
#!/bin/sh
file="testfile"
n=1
ls folder/${file}* | while read i
do
if [ folder/${file}.${n} = ${i} ]
then
n=$(( $n + 1 ))
fi
done
cp testfile folder/${file}.${n}
this is only working for first increment "folder/testfile.0"
I won't correct your solution, since mklement0 does it well.
Here is another solution, without any loop:
file="testfile"
n_max=$(ls -1 "folder/${file}"* | egrep -o '[0-9]+$' | sort -rn | head -n 1)
cp "${file}" "folder/${file}.$((n_max+1))"
Here is the thing of the second line: first you list the files, then egrep extracts the digit extension, then sort -rg sorts them decreasingly, and last head catchs only the first line, i.e. the largest used index. Finally, in the third line, you add one to this max to build your new index. It is ok on my script if the list is empty.
By the way, listing a directory may be quite long, so I suggest you to store somewhere the last index you used for later use. Save it as a variable in the script, or even in a file ... it can save you also some time.
The problem is that your while loop is executed in a subshell due to use of a pipe, so your modifications of n do not work as intended.
In general, you could use process substitution with while to avoid this problem, but in the case at hand a simple for loop is the right approach:
#!/bin/sh
file="testfile"
n=1
for i in folder/${file}*
do
if [ folder/${file}.${n} = ${i} ]
then
n=$(( $n + 1 ))
fi
done
cp testfile folder/${file}.${n}
Just another suggestion - in easy to understand code:
#!/bin/bash
file="testfile"
n=1
# Check for existing files with filename and count it
n=$(ls -latr folder/ |grep -i ${file} | wc -l)
# Increment above number
n=$(( $n + 1 ))
# Copy over file with the incremented number
cp $file folder/${file}.${n}
Related
I'm trying to write a shell script that deletes duplicate commands from my zsh_history file. Having no real shell script experience and given my C background I wrote this monstrosity that seems to work (only on Mac though), but takes a couple of lifetimes to end:
#!/bin/sh
history=./.zsh_history
currentLines=$(grep -c '^' $history)
wordToBeSearched=""
currentWord=""
contrastor=0
searchdex=""
echo "Currently handling a grand total of: $currentLines lines. Please stand by..."
while (( $currentLines - $contrastor > 0 ))
do
searchdex=1
wordToBeSearched=$(awk "NR==$currentLines - $contrastor" $history | cut -d ";" -f 2)
echo "$wordToBeSearched A BUSCAR"
while (( $currentLines - $contrastor - $searchdex > 0 ))
do
currentWord=$(awk "NR==$currentLines - $contrastor - $searchdex" $history | cut -d ";" -f 2)
echo $currentWord
if test "$currentWord" == "$wordToBeSearched"
then
sed -i .bak "$((currentLines - $contrastor - $searchdex)) d" $history
currentLines=$(grep -c '^' $history)
echo "Line deleted. New number of lines: $currentLines"
let "searchdex--"
fi
let "searchdex++"
done
let "contrastor++"
done
^THIS IS HORRIBLE CODE NOONE SHOULD USE^
I'm now looking for a less life-consuming approach using more shell-like conventions, mainly sed at this point. Thing is, zsh_history stores commands in a very specific way:
: 1652789298:0;man sed
Where the command itself is always preceded by ":0;".
I'd like to find a way to delete duplicate commands while keeping the last occurrence of each command intact and in order.
Currently I'm at a point where I have a functional line that will delete strange lines that find their way into the file (newlines and such):
#sed -i '/^:/!d' $history
But that's about it. Not really sure how get the expression to look for into a sed without falling back into everlasting whiles or how to delete the duplicates while keeping the last-occurring command.
The zsh option hist_ignore_all_dups should do what you want. Just add setopt hist_ignore_all_dups to your zshrc.
I wanted something similar, but I dont care about preserving the last one as you mentioned. This is just finding duplicates and removing them.
I used this command and then removed my .zsh_history and replacing it with the .zhistory that this command outputs
So from your home folder:
cat -n .zsh_history | sort -t ';' -uk2 | sort -nk1 | cut -f2- > .zhistory
This effectively will give you the file .zhistory containing the changed list, in my case it went from 9000 lines to 3000, you can check it with wc -l .zhistory to count the number of lines it has.
Please double check and make a backup of your zsh history before doing anything with it.
The sort command might be able to be modified to sort it by numerical value and somehow archieve what you want, but you will have to investigate further about that.
I found the script here, along with some commands to avoid saving duplicates in the future
I didn't want to rename the history file.
# dedupe_lines.zsh
if [ $# -eq 0 ]; then
echo "Error: No file specified" >&2
exit 1
fi
if [ ! -f $1 ]; then
echo "Error: File not found" >&2
exit 1
fi
sort $1 | uniq >temp.txt
mv temp.txt $1
Add dedupe_lines.zsh to your home directory, then make it executable.
chmod +x dedupe_lines.zsh
Run it.
./dedupe_lines.zsh .zsh_history
there is a directory which contains folders named with numbers, i've to find the folder with largest number in that directory.
This is the script i've written to find that folder:
files='ls path/'
var=0
for file in $files
do
echo $file
tmp=$((file-"0"))
if [ $tmp -gt $var ]
then
var=$tmp
fi
done
echo $var
But it's not working. It gives below error after invoking the script using command sudo ./restore2.sh.
ls
path/
./restore2.sh: line 6: path/: syntax error: operand expected (error token is "/")
0
Try this:
#!/bin/bash
files=`ls path/`
var=0
for file in $files
do
echo $file
tmp=$((file-"0"))
if [ $tmp -gt $var ]
then
var=$tmp
fi
done
echo $var
there's a backtick here: ls path/ instead of single or double-quotes.
I've only corrected this statement and it worked. and notice to add #!/bin/bash at the top of the script. This will tell your system to run the script in a bash shell.
You're using single quotes instead of backticks files='ls path/'. It's trying to use it as a literal string instead of evaluating it.
Also, for that specific task, you can just do:
ls test | awk '{if($1 > largest){largest = $1}} END{print largest}'
To have it a bit simpler.
Use find instead:
find . -maxdepth 1 -type d -regextype "posix-extended" -regex "^.*[[:digit:]]+.*$" | sort -n | tail -1
Set the maxdepth to 1 to check for directories within this directory only and no deeper. Set the regular expression type to posix-extended and search for all directories that have one or more digits. Print the result and order through sort before taking the largest one with tail -1.
Does path/ have any files in it? It looks like it's empty.
You should be getting a completely different complaint...
You don't want the path info in the filename. Rather than strip it with ${file##*/}, just go there and use non-path'd names.
An adaptation using your own logic as its base -
cd /whatever/path/ # go where the files are
var=-1 # initialize comparator
for file in [0-9]* # each entry that starts with a digit
do [[ "$file" =~ [^0-9] ]] && continue # skip any file with nondigit contents
[[ -f "$file" ]] || continue # only process plain files
(( file > var )) && var=$file # remember largest seen
done
echo $var # report largest
If you are sure there will be no negative numbered filenames, this should do it.
If there can be valid negatives, then your initialization needs to be appropriately lower, and the exclusion of nondigits should include the minus sign, as well as the list of files to select.
Note that this doesn't parse ls and doesn't require piping through a sort or spawning any other processes -- it's all handled in the bash interpreter and should be pretty efficient.
If you are sure of your data, and know there aren't any negatives or files named just 0 or non-plain-file entries in the directory that match the [0-9]* pattern, you can simplify it to just
cd /whatever/path/ # go where the files are
for file in [0-9]*; do (( file > var )) && var=$file; done
echo $var # report largest
As an aside, if you wanted to preserve the "make a list first" logic, you should still NOT use ls. Use an array.
cd /wherever/your/files/are/
files=( [0-9]* )
for file in "${files[#]}"
do : ...
I have a directory with about 2000 files. How can I select a random sample of N files through using either a bash script or a list of piped commands?
Here's a script that uses GNU sort's random option:
ls |sort -R |tail -$N |while read file; do
# Something involving $file, or you can leave
# off the while to just get the filenames
done
You can use shuf (from the GNU coreutils package) for that. Just feed it a list of file names and ask it to return the first line from a random permutation:
ls dirname | shuf -n 1
# probably faster and more flexible:
find dirname -type f | shuf -n 1
# etc..
Adjust the -n, --head-count=COUNT value to return the number of wanted lines. For example to return 5 random filenames you would use:
find dirname -type f | shuf -n 5
Here are a few possibilities that don't parse the output of ls and that are 100% safe regarding files with spaces and funny symbols in their name. All of them will populate an array randf with a list of random files. This array is easily printed with printf '%s\n' "${randf[#]}" if needed.
This one will possibly output the same file several times, and N needs to be known in advance. Here I chose N=42.
a=( * )
randf=( "${a[RANDOM%${#a[#]}]"{1..42}"}" )
This feature is not very well documented.
If N is not known in advance, but you really liked the previous possibility, you can use eval. But it's evil, and you must really make sure that N doesn't come directly from user input without being thoroughly checked!
N=42
a=( * )
eval randf=( \"\${a[RANDOM%\${#a[#]}]\"\{1..$N\}\"}\" )
I personally dislike eval and hence this answer!
The same using a more straightforward method (a loop):
N=42
a=( * )
randf=()
for((i=0;i<N;++i)); do
randf+=( "${a[RANDOM%${#a[#]}]}" )
done
If you don't want to possibly have several times the same file:
N=42
a=( * )
randf=()
for((i=0;i<N && ${#a[#]};++i)); do
((j=RANDOM%${#a[#]}))
randf+=( "${a[j]}" )
a=( "${a[#]:0:j}" "${a[#]:j+1}" )
done
Note. This is a late answer to an old post, but the accepted answer links to an external page that shows terrible bash practice, and the other answer is not much better as it also parses the output of ls. A comment to the accepted answer points to an excellent answer by Lhunath which obviously shows good practice, but doesn't exactly answer the OP.
ls | shuf -n 10 # ten random files
A simple solution for selecting 5 random files while avoiding to parse ls. It also works with files containing spaces, newlines and other special characters:
shuf -ezn 5 * | xargs -0 -n1 echo
Replace echo with the command you want to execute for your files.
This is an even later response to #gniourf_gniourf's late answer, which I just upvoted because it's by far the best answer, twice over. (Once for avoiding eval and once for safe filename handling.)
But it took me a few minutes to untangle the "not very well documented" feature(s) this answer uses. If your Bash skills are solid enough that you saw immediately how it works, then skip this comment. But I didn't, and having untangled it I think it's worth explaining.
Feature #1 is the shell's own file globbing. a=(*) creates an array, $a, whose members are the files in the current directory. Bash understands all the weirdnesses of filenames, so that list is guaranteed correct, guaranteed escaped, etc. No need to worry about properly parsing textual file names returned by ls.
Feature #2 is Bash parameter expansions for arrays, one nested within another. This starts with ${#ARRAY[#]}, which expands to the length of $ARRAY.
That expansion is then used to subscript the array. The standard way to find a random number between 1 and N is to take the value of random number modulo N. We want a random number between 0 and the length of our array. Here's the approach, broken into two lines for clarity's sake:
LENGTH=${#ARRAY[#]}
RANDOM=${a[RANDOM%$LENGTH]}
But this solution does it in a single line, removing the unnecessary variable assignment.
Feature #3 is Bash brace expansion, although I have to confess I don't entirely understand it. Brace expansion is used, for instance, to generate a list of 25 files named filename1.txt, filename2.txt, etc: echo "filename"{1..25}".txt".
The expression inside the subshell above, "${a[RANDOM%${#a[#]}]"{1..42}"}", uses that trick to produce 42 separate expansions. The brace expansion places a single digit in between the ] and the }, which at first I thought was subscripting the array, but if so it would be preceded by a colon. (It would also have returned 42 consecutive items from a random spot in the array, which is not at all the same thing as returning 42 random items from the array.) I think it's just making the shell run the expansion 42 times, thereby returning 42 random items from the array. (But if someone can explain it more fully, I'd love to hear it.)
The reason N has to be hardcoded (to 42) is that brace expansion happens before variable expansion.
Finally, here's Feature #4, if you want to do this recursively for a directory hierarchy:
shopt -s globstar
a=( ** )
This turns on a shell option that causes ** to match recursively. Now your $a array contains every file in the entire hierarchy.
If you have Python installed (works with either Python 2 or Python 3):
To select one file (or line from an arbitrary command), use
ls -1 | python -c "import sys; import random; print(random.choice(sys.stdin.readlines()).rstrip())"
To select N files/lines, use (note N is at the end of the command, replace this by a number)
ls -1 | python -c "import sys; import random; print(''.join(random.sample(sys.stdin.readlines(), int(sys.argv[1]))).rstrip())" N
If you want to copy a sample of those files to another folder:
ls | shuf -n 100 | xargs -I % cp % ../samples/
make samples directory first obviously.
MacOS does not have the sort -R and shuf commands, so I needed a bash only solution that randomizes all files without duplicates and did not find that here. This solution is similar to gniourf_gniourf's solution #4, but hopefully adds better comments.
The script should be easy to modify to stop after N samples using a counter with if, or gniourf_gniourf's for loop with N. $RANDOM is limited to ~32000 files, but that should do for most cases.
#!/bin/bash
array=(*) # this is the array of files to shuffle
# echo ${array[#]}
for dummy in "${array[#]}"; do # do loop length(array) times; once for each file
length=${#array[#]}
randomi=$(( $RANDOM % $length )) # select a random index
filename=${array[$randomi]}
echo "Processing: '$filename'" # do something with the file
unset -v "array[$randomi]" # set the element at index $randomi to NULL
array=("${array[#]}") # remove NULL elements introduced by unset; copy array
done
If you have more files in your folder, you can use the below piped command I found in unix stackexchange.
find /some/dir/ -type f -print0 | xargs -0 shuf -e -n 8 -z | xargs -0 cp -vt /target/dir/
Here I wanted to copy the files, but if you want to move files or do something else, just change the last command where I have used cp.
This is the only script I can get to play nice with bash on MacOS. I combined and edited snippets from the following two links:
ls command: how can I get a recursive full-path listing, one line per file?
http://www.linuxquestions.org/questions/linux-general-1/is-there-a-bash-command-for-picking-a-random-file-678687/
#!/bin/bash
# Reads a given directory and picks a random file.
# The directory you want to use. You could use "$1" instead if you
# wanted to parametrize it.
DIR="/path/to/"
# DIR="$1"
# Internal Field Separator set to newline, so file names with
# spaces do not break our script.
IFS='
'
if [[ -d "${DIR}" ]]
then
# Runs ls on the given dir, and dumps the output into a matrix,
# it uses the new lines character as a field delimiter, as explained above.
# file_matrix=($(ls -LR "${DIR}"))
file_matrix=($(ls -R $DIR | awk '; /:$/&&f{s=$0;f=0}; /:$/&&!f{sub(/:$/,"");s=$0;f=1;next}; NF&&f{ print s"/"$0 }'))
num_files=${#file_matrix[*]}
# This is the command you want to run on a random file.
# Change "ls -l" by anything you want, it's just an example.
ls -l "${file_matrix[$((RANDOM%num_files))]}"
fi
exit 0
I use this: it uses temporary file but goes deeply in a directory until it find a regular file and return it.
# find for a quasi-random file in a directory tree:
# directory to start search from:
ROOT="/";
tmp=/tmp/mytempfile
TARGET="$ROOT"
FILE="";
n=
r=
while [ -e "$TARGET" ]; do
TARGET="$(readlink -f "${TARGET}/$FILE")" ;
if [ -d "$TARGET" ]; then
ls -1 "$TARGET" 2> /dev/null > $tmp || break;
n=$(cat $tmp | wc -l);
if [ $n != 0 ]; then
FILE=$(shuf -n 1 $tmp)
# or if you dont have/want to use shuf:
# r=$(($RANDOM % $n)) ;
# FILE=$(tail -n +$(( $r + 1 )) $tmp | head -n 1);
fi ;
else
if [ -f "$TARGET" ] ; then
rm -f $tmp
echo $TARGET
break;
else
# is not a regular file, restart:
TARGET="$ROOT"
FILE=""
fi
fi
done;
How about a Perl solution slightly doctored from Mr. Kang over here:
How can I shuffle the lines of a text file on the Unix command line or in a shell script?
$ ls | perl -MList::Util=shuffle -e '#lines = shuffle(<>); print
#lines[0..4]'
I have a folder with backups from a MySQL database that are created automatically. Their name consists of the date the backup was made, like so:
2010-06-12_19-45-05.mysql.gz
2010-06-14_19-45-05.mysql.gz
2010-06-18_19-45-05.mysql.gz
2010-07-01_19-45-05.mysql.gz
What is a way to get the filename of the last file in the list, i.e. of the one which in alphabetical order comes last?
In a shell script, I would like to do something like
LAST_BACKUP_FILE= ???
gunzip $LAST_BACKUP_FILE;
ls -1 | tail -n 1
If you want to assign this to a variable, use $(...) or backticks.
FILE=`ls -1 | tail -n 1`
FILE=$(ls -1 | tail -n 1)
#Sjoerd's answer is correct, I'll just pick a few nits from it:
you don't need the -1 option to enforce one path per line if you pipe the output somewhere:
ls | tail -n 1
you can use -r to get the listing in reverse order, and take the first one:
ls -r | head -n 1
gunzip some.log.gz will write uncompressed data into some.log and remove some.log.gz, which may or may not be what you want (probably isn't). if you want to keep the compressed source, pipe it into gunzip:
gunzip < some.file.gz
you might want to protect the script against situation when the dir contains no files, since
gunzip $empty_variable
expands to just
gunzip
and such invocation will wait indefinitely for data on standard input:
latest="$(ls -r /some/where/*.gz | head -1)"
if test -z "$latest"; then
# there's no logs yet, bail out
exit
fi
gunzip < $latest
ls can yield unexpected results when parsed by other commands if the filenames have unusual characters. The following always works:
for LAST_BACKUP_FILE in *; do : ; done
for LAST_BACKUP_FILE in * loops through every filename (and folder name, if there are any) in order in the current directory, storing each in $LAST_BACKUP_FILE
do : does nothing
done finishes after the last file
Now, the last file is stored in $LAST_BACKUP_FILE.
If you happen to want the first file, use this:
for FIRST_BACKUP_FILE in *; do break; done
The break statement jumps out of the loop after the first file is stored in $FIRST_BACKUPT_FILE.
(from comment below) If you want hidden files included in the search, then use the command shopt -s dotglob before running the loops.
The shell is more powerful than many think. Just let it work for you. Assuming filenames without spaces,
set -- $(ls -r *.gz)
LAST_BACKUP_FILE=$1
does the trick with a single fork, no pipes, and you can even avoid the fork if your shell supports arithmetic expansion as in
set -- *.gz
shift $(($# - 1))
LAST_BACKUP_FILE=$1
I have a directory with about 2000 files. How can I select a random sample of N files through using either a bash script or a list of piped commands?
Here's a script that uses GNU sort's random option:
ls |sort -R |tail -$N |while read file; do
# Something involving $file, or you can leave
# off the while to just get the filenames
done
You can use shuf (from the GNU coreutils package) for that. Just feed it a list of file names and ask it to return the first line from a random permutation:
ls dirname | shuf -n 1
# probably faster and more flexible:
find dirname -type f | shuf -n 1
# etc..
Adjust the -n, --head-count=COUNT value to return the number of wanted lines. For example to return 5 random filenames you would use:
find dirname -type f | shuf -n 5
Here are a few possibilities that don't parse the output of ls and that are 100% safe regarding files with spaces and funny symbols in their name. All of them will populate an array randf with a list of random files. This array is easily printed with printf '%s\n' "${randf[#]}" if needed.
This one will possibly output the same file several times, and N needs to be known in advance. Here I chose N=42.
a=( * )
randf=( "${a[RANDOM%${#a[#]}]"{1..42}"}" )
This feature is not very well documented.
If N is not known in advance, but you really liked the previous possibility, you can use eval. But it's evil, and you must really make sure that N doesn't come directly from user input without being thoroughly checked!
N=42
a=( * )
eval randf=( \"\${a[RANDOM%\${#a[#]}]\"\{1..$N\}\"}\" )
I personally dislike eval and hence this answer!
The same using a more straightforward method (a loop):
N=42
a=( * )
randf=()
for((i=0;i<N;++i)); do
randf+=( "${a[RANDOM%${#a[#]}]}" )
done
If you don't want to possibly have several times the same file:
N=42
a=( * )
randf=()
for((i=0;i<N && ${#a[#]};++i)); do
((j=RANDOM%${#a[#]}))
randf+=( "${a[j]}" )
a=( "${a[#]:0:j}" "${a[#]:j+1}" )
done
Note. This is a late answer to an old post, but the accepted answer links to an external page that shows terrible bash practice, and the other answer is not much better as it also parses the output of ls. A comment to the accepted answer points to an excellent answer by Lhunath which obviously shows good practice, but doesn't exactly answer the OP.
ls | shuf -n 10 # ten random files
A simple solution for selecting 5 random files while avoiding to parse ls. It also works with files containing spaces, newlines and other special characters:
shuf -ezn 5 * | xargs -0 -n1 echo
Replace echo with the command you want to execute for your files.
This is an even later response to #gniourf_gniourf's late answer, which I just upvoted because it's by far the best answer, twice over. (Once for avoiding eval and once for safe filename handling.)
But it took me a few minutes to untangle the "not very well documented" feature(s) this answer uses. If your Bash skills are solid enough that you saw immediately how it works, then skip this comment. But I didn't, and having untangled it I think it's worth explaining.
Feature #1 is the shell's own file globbing. a=(*) creates an array, $a, whose members are the files in the current directory. Bash understands all the weirdnesses of filenames, so that list is guaranteed correct, guaranteed escaped, etc. No need to worry about properly parsing textual file names returned by ls.
Feature #2 is Bash parameter expansions for arrays, one nested within another. This starts with ${#ARRAY[#]}, which expands to the length of $ARRAY.
That expansion is then used to subscript the array. The standard way to find a random number between 1 and N is to take the value of random number modulo N. We want a random number between 0 and the length of our array. Here's the approach, broken into two lines for clarity's sake:
LENGTH=${#ARRAY[#]}
RANDOM=${a[RANDOM%$LENGTH]}
But this solution does it in a single line, removing the unnecessary variable assignment.
Feature #3 is Bash brace expansion, although I have to confess I don't entirely understand it. Brace expansion is used, for instance, to generate a list of 25 files named filename1.txt, filename2.txt, etc: echo "filename"{1..25}".txt".
The expression inside the subshell above, "${a[RANDOM%${#a[#]}]"{1..42}"}", uses that trick to produce 42 separate expansions. The brace expansion places a single digit in between the ] and the }, which at first I thought was subscripting the array, but if so it would be preceded by a colon. (It would also have returned 42 consecutive items from a random spot in the array, which is not at all the same thing as returning 42 random items from the array.) I think it's just making the shell run the expansion 42 times, thereby returning 42 random items from the array. (But if someone can explain it more fully, I'd love to hear it.)
The reason N has to be hardcoded (to 42) is that brace expansion happens before variable expansion.
Finally, here's Feature #4, if you want to do this recursively for a directory hierarchy:
shopt -s globstar
a=( ** )
This turns on a shell option that causes ** to match recursively. Now your $a array contains every file in the entire hierarchy.
If you have Python installed (works with either Python 2 or Python 3):
To select one file (or line from an arbitrary command), use
ls -1 | python -c "import sys; import random; print(random.choice(sys.stdin.readlines()).rstrip())"
To select N files/lines, use (note N is at the end of the command, replace this by a number)
ls -1 | python -c "import sys; import random; print(''.join(random.sample(sys.stdin.readlines(), int(sys.argv[1]))).rstrip())" N
If you want to copy a sample of those files to another folder:
ls | shuf -n 100 | xargs -I % cp % ../samples/
make samples directory first obviously.
MacOS does not have the sort -R and shuf commands, so I needed a bash only solution that randomizes all files without duplicates and did not find that here. This solution is similar to gniourf_gniourf's solution #4, but hopefully adds better comments.
The script should be easy to modify to stop after N samples using a counter with if, or gniourf_gniourf's for loop with N. $RANDOM is limited to ~32000 files, but that should do for most cases.
#!/bin/bash
array=(*) # this is the array of files to shuffle
# echo ${array[#]}
for dummy in "${array[#]}"; do # do loop length(array) times; once for each file
length=${#array[#]}
randomi=$(( $RANDOM % $length )) # select a random index
filename=${array[$randomi]}
echo "Processing: '$filename'" # do something with the file
unset -v "array[$randomi]" # set the element at index $randomi to NULL
array=("${array[#]}") # remove NULL elements introduced by unset; copy array
done
If you have more files in your folder, you can use the below piped command I found in unix stackexchange.
find /some/dir/ -type f -print0 | xargs -0 shuf -e -n 8 -z | xargs -0 cp -vt /target/dir/
Here I wanted to copy the files, but if you want to move files or do something else, just change the last command where I have used cp.
This is the only script I can get to play nice with bash on MacOS. I combined and edited snippets from the following two links:
ls command: how can I get a recursive full-path listing, one line per file?
http://www.linuxquestions.org/questions/linux-general-1/is-there-a-bash-command-for-picking-a-random-file-678687/
#!/bin/bash
# Reads a given directory and picks a random file.
# The directory you want to use. You could use "$1" instead if you
# wanted to parametrize it.
DIR="/path/to/"
# DIR="$1"
# Internal Field Separator set to newline, so file names with
# spaces do not break our script.
IFS='
'
if [[ -d "${DIR}" ]]
then
# Runs ls on the given dir, and dumps the output into a matrix,
# it uses the new lines character as a field delimiter, as explained above.
# file_matrix=($(ls -LR "${DIR}"))
file_matrix=($(ls -R $DIR | awk '; /:$/&&f{s=$0;f=0}; /:$/&&!f{sub(/:$/,"");s=$0;f=1;next}; NF&&f{ print s"/"$0 }'))
num_files=${#file_matrix[*]}
# This is the command you want to run on a random file.
# Change "ls -l" by anything you want, it's just an example.
ls -l "${file_matrix[$((RANDOM%num_files))]}"
fi
exit 0
I use this: it uses temporary file but goes deeply in a directory until it find a regular file and return it.
# find for a quasi-random file in a directory tree:
# directory to start search from:
ROOT="/";
tmp=/tmp/mytempfile
TARGET="$ROOT"
FILE="";
n=
r=
while [ -e "$TARGET" ]; do
TARGET="$(readlink -f "${TARGET}/$FILE")" ;
if [ -d "$TARGET" ]; then
ls -1 "$TARGET" 2> /dev/null > $tmp || break;
n=$(cat $tmp | wc -l);
if [ $n != 0 ]; then
FILE=$(shuf -n 1 $tmp)
# or if you dont have/want to use shuf:
# r=$(($RANDOM % $n)) ;
# FILE=$(tail -n +$(( $r + 1 )) $tmp | head -n 1);
fi ;
else
if [ -f "$TARGET" ] ; then
rm -f $tmp
echo $TARGET
break;
else
# is not a regular file, restart:
TARGET="$ROOT"
FILE=""
fi
fi
done;
How about a Perl solution slightly doctored from Mr. Kang over here:
How can I shuffle the lines of a text file on the Unix command line or in a shell script?
$ ls | perl -MList::Util=shuffle -e '#lines = shuffle(<>); print
#lines[0..4]'