I'm trying to move only 100 files with a specific extensions (from the current directory to the parent directory), but the following attempt of mine does not work
for file in $(ls -U | grep *.txt | tail -100)
do
mv $file ../
done
Can you point me to the correct approach?
Since you didn't quote *.txt, the shell expanded it to all the filenames ending in .txt. So your command is something like:
ls -U | grep file1.txt file2.txt file3.txt ... | tail -100
Since grep has filename arguments, it ignores its standard input. It outputs all the lines matching file1.txt in the remaining files. There's probably no matches, so nothing is piped to tail -100. And even if there were matches, the output would be the lines from the files, not filenames, so it wouldn't be useful for the mv command.
You can loop over the filenames directly, and use a counter variable to stop after 100 files.
counter=0
for file in *.txt
do
if (( counter >= 100 ))
then break
fi
mv "$file" ../
((counter++))
done
This avoids the pitfalls of parsing the output of ls.
this will do the job:
ls -U *.txt | tail -100 | while read filename; do mv "$filename" ../; done
while read filename respect spaces in the filename.
Run this in the text file directory:
#!/bin/bash
for txt_file in ./*.txt; do
((c++==100)) && break
mv "$txt_file" ../
done
Related
My goal is copy a bulk of specific directories whose names are in a txt file as follows:
$ cat names.txt
raw1
raw2
raw3
raw4
raw5
These directories have subdirectories, hence it is important to copy all the contents. When I list in my terminal it looks like this:
$ ls -l
raw3
raw7
raw1
raw8
raw5
raw6
raw2
raw4
To perform this task, I have tried the following:
cat names.txt | while read line; do grep -l '$line' | xargs -r0 cp -t <desired_destination>; done
But, I get this mistake
cp: cannot stat No such file or directory
I suppose it's because the names in the file list (names.txt) don't match in sorting with the ones in the terminal. Notice that they are unsorted and by using while read line doesn't work. Thank you for taking the time and commitment to help me.
Having problems following the logic of the current code so in the name of K.I.S.S. I propose:
tgtdir=/my/target/directory
while read -r srcdir
do
[[ -d "${srcdir}" ]] && cp -rp "${srcdir}" "${tgtdir}"
done < <(tr -d '\r' < names.dat)
NOTES:
the < <(tr -d '\r' < names.dat) is used to remove windows/dos line endings from names.dat (per comments from OP); if names.dat is updated to remove the \r' then the tr -d with be a no-op (ie, bit of overhead to spawn the subprocess but the script should still read names.dat correctly)
assumes script is run from the directory where the source directories reside otherwise code can be modified to either cd to said directory or preface the ${srcdir} references with said directory
OP can add/modify the cp flags as needed, but I'm assuming at a minimum -r will be needed in order to recursively copy the directories
UUoC.
cat names.txt | while read line; do ...; done
is better written
while read line; do ...; done < names.txt
do grep -l '$LINE' | is eating your input.
printf "%s\n" 1 2 3 |while read line; do echo "Read: [$line]"; grep . | cat; done
Read: [1]
2
3
In your case, it is likely finding no lines that match the literal string $LINE which you have embedded in single-qote marks, which do not allow it to be parsed for content. Use "$line" (avoid capitals), and wouldn't be helpful even if it did match:
$: printf "%s\n" 1 2 3 | grep -l .
(standard input)
You didn't tell it what to read from, so -l is pointless since it's reading the same stdin stream that the read is.
I think what you want is a little simpler -
xargs cp -Rt /your/desired/target/directory/ < names.txt
Assuming you wanted to leave the originals where they were.
I have a bash command that lists a number of files and directories. I want to remove everything that is not an existing directory. Is there anyway I can do this without creating a script of my own? I.e. I want to use pre-existing programs available in linux.
E.g. Given that I have this folder:
dir1/
dir2/
file.txt
I want to be able to run something like:
echo dir1 dir2 file.txt somethingThatDoesNotExist | xargs [ theCommandIAmLookingFor]
and get
dir1
dir2
It would be better if the command generating the putative paths used a better delimeter, but you might be looking for something like:
... | xargs -n 1 sh -c 'test -d "$0" && echo $0'
You can use this command line using grep -v:
your_command | grep -vxFf <(printf '%s\n' */ | sed 's/.$//') -
This will filter out all the sub-directories in current path from your list.
If in case you want to list only existing directories then remove -v as:
your_command | grep -xFf <(printf '%s\n' */ | sed 's/.$//') -
Note that glob */ prints all sub-directories in current path with a trailing / and sed is used to remove this last /.
I have a directory that contains only .txt files. I want to print the number of lines for every file. When I write cat file.txt | wc -l the number of lines appears but when I want to make a script it's more complicated. I have this code:
for fis in `ls -R $1`
do
echo `cat $fis | wc -l`
done
I tried: wc -l $fis , with awk,grep and it doesn't work. It tells that:
cat: fis1: No such file or directory
0
How can I do to print the number of lines?
To find files recursively in subdirectories, use the find command, not ls -R, which is mainly intended for human reading.
find "$1" -type f -exec wc -l {} +
The problems with looping over the output of ls -R are:
Filenames with whitespace won't be parsed correctly.
It prints other output beside just the filenames.
Not the problem here, but the echo command is more than needed:
You can use
wc -l "${fis}"
What goes wrong?
You have a subdir called fis1. Look to the output of ls:
# ls -R fis1
fis1:
file1_in_fis1.txt
When you are parsing this output, your script will try
echo `cat fis1: | wc -l`
The cat will tell you No such file or directory and wc counts 0.
As #Barmar explained, ls prints additional output you do not want.
Do not try to patch your attempt by | grep .txt and if [ -f "${fis}"]; then .., these will fail with filename with spaces.txt. So use find or shopt (and accept the answer of #Barmar or #Cyrus).
I've a bash script like this:
for d in /home/test/*
do
echo $d
done
Which ouputs this:
/home/test/newer dir
/home/test/oldest dir
I'd like to order the folders by creation time so that the 'oldest dir' directory appears first in the list. I've tried ls and tree variations to no avail.
For example,
for d in `ls -d -c -1 $PWD/*`
Returns:
/home/test/oldest
dir
/home/test/newer
dir
Very close, but it does not respect the space in the directory name. My question, how would I have oldest dir on top and support the whitespace?
ls -d -c $PWD/* | while read line
do echo "$line"
done
Another technique, kind of a Schwartzian transform:
stat -c $'%Z\t%n' /home/test/* | sort -n | cut -f2- |
while IFS= read -r filename; do
# ...
This solution is fragile with filenames containing newlines.
Inside a directory, how can I delete files that lack any of the words specified, so that only files that contain ALL the words are left? I tried to write a simple bash shell script using grep and rm commands, but I got lost. I am totally new to Linux, any help would be appreciated
How about:
grep -L foo *.txt | xargs rm
grep -L bar *.txt | xargs rm
If a file does not contain foo, then the first line will remove it.
If a file does not contain bar, then the second line will remove it.
Only files containing both foo and bar should be left
-L, --files-without-match
Suppress normal output; instead print the name of each input
file from which no output would normally have been printed. The
scanning will stop on the first match.
See also #Mykola Golubyev's post for placing in a loop.
list=`Word1 Word2 Word3 Word4 Word5`
for word in $list
grep -L $word *.txt | xargs rm
done
Addition to the answers above: Use the newline character as delimiter to handle file names with spaces!
grep -L $word $file | xargs -d '\n' rm
grep -L word | xargs rm
To do the same matching filenames (not the contents of files as most of the solutions above) you can use the following:
for file in `ls --color=never | grep -ve "\(foo\|bar\)"`
do
rm $file
done
As per comments:
for file in `ls`
shouldn't be used. The below does the same thing without using the ls
for file in *
do
if [ x`echo $file | grep -ve "\(test1\|test3\)"` == x ]; then
rm $file
fi
done
The -ve reverses the search for the regexp pattern for either foo or bar in the filename.
Any further words to be added to the list need to be separated by \|
e.g. one\|two\|three
First, remove the file-list:
rm flist
Then, for each of the words, add the file to the filelist if it contains that word:
grep -l WORD * >>flist
Then sort, uniqify and get a count:
sort flist | uniq -c >flist_with_count
All those files in flsit_with_count that don't have the number of words should be deleted. The format will be:
2 file1
7 file2
8 file3
8 file4
If there were 8 words, then file1 and file2 should be deleted. I'll leave the writing/testing of the script to you.
Okay, you convinced me, here's my script:
#!/bin/bash
rm -rf flist
for word in fopen fclose main ; do
grep -l ${word} *.c >>flist
done
rm $(sort flist | uniq -c | awk '$1 != 3 {print $2} {}')
This removes the files in the directory that didn't have all three words:
You could try something like this but it may break
if the patterns contain shell or grep meta characters:
(in this example one two three are the patterns)
for f in *; do
unset cmd
for p in one two three; do
cmd="fgrep \"$p\" \"$f\" && $cmd"
done
eval "$cmd" >/dev/null || rm "$f"
done
This will remove all files that doesn't contain words Ping or Sent
grep -L 'Ping\|Sent' * | xargs rm