If I have the following:
ls|sort -n
How would I touch all those files in the order of the sorted files? Something like:
ls|sort -n|touch
What would be the proper syntax? Note that I need to sort touch the files in the exact order they're being sorted -- as I'm trying to sort these files for a FAT reader with minimal metadata reading.
ls -1tr | while read file; do touch "$file"; sleep 1; done
If you want to preserve distance in modification time from one file to the next then call this instead:
upmodstamps() {
oldest_elapsed=$(( $(date +%s) - $(stat -c %Y "`ls -1tr|head -1`") ))
for file in *; do
oldstamp=$(stat -c %Y "$file")
newstamp=$(( $oldstamp + $oldest_elapsed ))
newstamp_fmt=$(date --date=#${newstamp} +'%Y%m%d%H%M.%S')
touch -t ${newstamp_fmt} "$file"
done
}
Note: date usage assumes GNU
You can use this command
(ls|sort -n >> list.txt )
touch $(cat list.txt)
OR
touch $(ls /path/to/dir | sort -n)
OR if you want to copy files instead of creating empty files use this command
cp list.txt ./DirectoryWhereYouWantToCopy
Try like this
touch $(ls | sort -n)
Can you give a few file name?
if you have file names with numbers as 1file, 10file, 11file .. 20file, then you need use --general-numeric-sort
ls | sort --general-numeric-sort --output=../workingDirectory/sortedFiles.txt
cat sortedFiles.txt
1file
10file
11file
12file
20file
and move sortedFile.txt into your working directory or where ever you want.
touch $(cat ../workingDirectory/sortedFiles.txt)
this will create empty files with the exact same name
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'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
Good morning everybody,
for a website I'd like to rename files(pictures) in a folder from "1.jpg, 2.jpg, 3.jpg ..." to "yyyymmdd_hhmmss.jpg" - so I'd like to read out the creation times an set this times as names for the pics. Does anybody have an idea how to do that for example with a linux-shell or with imagemagick?
Thank you!
Naming based on file system date
In the linux shell:
for f in *.jpg
do
mv -n "$f" "$(date -r "$f" +"%Y%m%d_%H%M%S").jpg"
done
Explanation:
for f in *.jpg
do
This starts the loop over all jpeg files. A feature of this is that it will work with all file names, even ones with spaces, tabs or other difficult characters in the names.
mv -n "$f" "$(date -r "$f" +"%Y%m%d_%H%M%S").jpg"
This renames the file. It uses the -r option which tells date to display the date of the file rather than the current date. The specification +"%Y%m%d_%H%M%S" tells date to format it as you specified.
The file name, $f, is placed in double quotes where ever it is used. This assures that odd file names will not cause errors.
The -n option to mv tells move never to overwrite an existing file.
done
This completes the loop.
For interactive use, you may prefer that the command is all on one line. In that case, use:
for f in *.jpg; do mv -n "$f" "$(date -r "$f" +"%Y%m%d_%H%M%S").jpg"; done
Naming based on EXIF Create Date
To name the file based on the EXIF Create Date (instead of the file system date), we need exiftool or equivalent:
for f in *.jpg
do
mv -n "$f" "$(exiftool -d "%Y%m%d_%H%M%S" -CreateDate "$f" | awk '{print $4".jpg"}')"
done
Explanation:
The above is quite similar to the commands for the file date but with the use of exiftool and awk to extract the EXIF image Create Date.
The exiftool command provides the date in a format like:
$ exiftool -d "%Y%m%d_%H%M%S" -CreateDate sample.jpg
Create Date : 20121027_181338
The actual date that we want is the fourth field in the output.
We pass the exiftool output to awk so that it can extract the field that we want:
awk '{print $4".jpg"}'
This selects the date field and also adds on the .jpg extension.
Thanks to #John1024 !
I needed to rename files with different extensions in the same time, according to last modification date :
for f in *; do
fn=$(basename "$f")
mv "$fn" "$(date -r "$f" +"%Y-%m-%d_%H-%M-%S")_$fn"
done
"DSC_0189.JPG" ➜ "2016-02-21_18-22-15_DSC_0189.JPG"
"MOV_0131.avi" ➜ "2016-01-01_20-30-31_MOV_0131.avi"
If you don't want to keep original filename :
mv "$fn" "$(date -r "$pathAndFileName" +"%Y-%m-%d_%H-%M-%S")"
Hope it helps noobs as me !
Try this
for file in `ls -1 *.jpg`; do name=`stat -c %y $file | awk -F"." '{ print $1 }' | sed -e "s/\-//g" -e "s/\://g" -e "s/[ ]/_/g"`.jpg; mv $file $name; done
Though there might be an easier way.
I created a shell script; I think it's mac only, linux might need other arguments.
#!/bin/bash
BASEDIR=$1;
for file in `ls -1 $BASEDIR`; do
TIMESTAMP=`stat -f "%B" $BASEDIR/$file`;
DATENAME=`date -r $TIMESTAMP +'%Y%m%d-%H%M%S'`-$file
mv -v $BASEDIR/$file $BASEDIR/$DATENAME;
done
when called with a directory path, moves all files in that directory to prepend the creation date of that file, like
../camera/P1210232.JPG -> ../camera/20220121-103456-P1210232.JPG
Change filename based on file creation time:
exiftool "-filename<FileCreateDate" -d %Y%m%d_%H%M%S%z%%-c.%%le input.jpg
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 $'%Z\t%n' /home/test/* | sort -n | cut -f2- |
while IFS= read -r filename; do
# ...
This solution is fragile with filenames containing newlines.
I'm having some rather unusual problems using grep in a bash script. Below is an example of the bash script code that I'm using that exhibits the behaviour:
UNIQ_SCAN_INIT_POINT=1
cat "$FILE_BASENAME_LIST" | uniq -d >> $UNIQ_LIST
sed '/^$/d' $UNIQ_LIST >> $UNIQ_LIST_FINAL
UNIQ_LINE_COUNT=`wc -l $UNIQ_LIST_FINAL | cut -d \ -f 1`
while [ -n "`cat $UNIQ_LIST_FINAL | sed "$UNIQ_SCAN_INIT_POINT"'q;d'`" ]; do
CURRENT_LINE=`cat $UNIQ_LIST_FINAL | sed "$UNIQ_SCAN_INIT_POINT"'q;d'`
CURRENT_DUPECHK_FILE=$FILE_DUPEMATCH-$CURRENT_LINE
grep $CURRENT_LINE $FILE_LOCTN_LIST >> $CURRENT_DUPECHK_FILE
MATCH=`grep -c $CURRENT_LINE $FILE_BASENAME_LIST`
CMD_ECHO="$CURRENT_LINE matched $MATCH times," cmd_line_echo
echo "$CURRENT_DUPECHK_FILE" >> $FILE_DUPEMATCH_FILELIST
let UNIQ_SCAN_INIT_POINT=UNIQ_SCAN_INIT_POINT+1
done
On numerous occasions, when grepping for the current line in the file location list, it has put no output to the current dupechk file even though there have definitely been matches to the current line in the file location list (I ran the command in terminal with no issues).
I've rummaged around the internet to see if anyone else has had similar behaviour, and thus far all I have found is that it is something to do with buffered and unbuffered outputs from other commands operating before the grep command in the Bash script....
However no one seems to have found a solution, so basically I'm asking you guys if you have ever come across this, and any idea/tips/solutions to this problem...
Regards
Paul
The `problem' is the standard I/O library. When it is writing to a terminal
it is unbuffered, but if it is writing to a pipe then it sets up buffering.
try changing
CURRENT_LINE=`cat $UNIQ_LIST_FINAL | sed "$UNIQ_SCAN_INIT_POINT"'q;d'`
to
CURRENT LINE=`sed "$UNIQ_SCAN_INIT_POINT"'q;d' $UNIQ_LIST_FINAL`
Are there any directories with spaces in their names in $FILE_LOCTN_LIST? Because if they are, those spaces will need escaped somehow. Some combination of find and xargs can usually deal with that for you, especially xargs -0
A small bash script using md5sum and sort that detects duplicate files in the current directory:
CURRENT="" md5sum * |
sort |
while read md5sum filename;
do
[[ $CURRENT == $md5sum ]] && echo $filename is duplicate;
CURRENT=$md5sum;
done
you tagged linux, some i assume you have tools like GNU find,md5sum,uniq, sort etc. here's a simple example to find duplicate files
$ echo "hello world">file
$ md5sum file
6f5902ac237024bdd0c176cb93063dc4 file
$ cp file file1
$ md5sum file1
6f5902ac237024bdd0c176cb93063dc4 file1
$ echo "blah" > file2
$ md5sum file2
0d599f0ec05c3bda8c3b8a68c32a1b47 file2
$ find . -type f -exec md5sum "{}" \; |sort -n | uniq -w32 -D
6f5902ac237024bdd0c176cb93063dc4 ./file
6f5902ac237024bdd0c176cb93063dc4 ./file1