I have this folder in windows
if I do a simple ls , find, either in bash (cygwin) or msdos, it shows me like this.
$ ls -1
su-01-01.jpg
su-01-02-03.jpg
su-01-12-13.jpg
su-01-14.jpg
su-01-15.jpg
su-01-16.jpg
su-01-18.jpg
su-01-19.jpg
su-01-20.jpg
su-01-21.jpg
su-01-31.jpg
su-01-34.jpg
su-01-35.jpg
su-01-38.jpg
su-01-39.jpg
su-01-42-43.jpg
su-01-44.jpg
su-01-45.jpg
su-01-47.jpg
su-01-48.jpg
su01-00.jpg
su01-04.jpg
su01-05.jpg
su01-06.jpg
su01-07.jpg
su01-08.jpg
I have tried ordering and it does not take into account 0 00 1
$ ls -1 |sort -V
su01-00.jpg
su01-04.jpg
su01-05.jpg
su01-06.jpg
su01-07.jpg
su01-08.jpg
su01-09.jpg
su01-10.jpg
su01-11.jpg
su01-22-23.jpg
su01-24.jpg
su01-25.jpg
su01-26.jpg
su01-27.jpg
su01-28-29.jpg
su01-30.jpg
su01-32.jpg
su01-33.jpg
su01-40-41.jpg
su-01-01.jpg
su-01-02-03.jpg
su-01-12-13.jpg
su-01-14.jpg
su-01-15.jpg
but how do I make it ignore the (-)?
thank you very much for your help
find doesn't guaranty alphabetical ordering; ls and sort do, but the char - value is 45 while the 0 char value is 48, so su- will come ahead of the su0 in an alphabetical sorting.
While a printf '%s\n' su* | LANG=en_US.utf8 sort -n seems to display the files the way you want, the best thing to do for making your life easier would be to rename some of the files:
#!/bin/bash
for f in su0*
do
mv "$f" "su-0${f#su0}"
done
Update
renaming the files to 001.jpg 002.jpg ...
#!/bin/bash
shopt -s nullglob
n=1
while IFS='' read -r file
do
printf -v newname '%03d.%s' "$((n++))" "${file##*.}"
printf '%q %q %q\n' mv "$file" "$newname"
done < <(
printf '%s\n' su* |
sed -nE 's,su-?([^/]*)$,\1/&,p' |
LANG=C sort -nt '-' |
sed 's,[^/]*/,,'
)
The simplest way to control the sort order in Bash, both for ls and sort, so to set your LANG variable to the locale you want.
In your .bashrc or .profile, add
export LANG=en_US.utf8
and then
ls -1
or
ls -1 | sort
will output the order you're looking for.
If you want to test with different locales and see their effect, your can set LANG one command at a time. For example, compare the output of these commands:
LANG=en_US.utf8 ls -1 # what you're looking for
LANG=C ls -1 # "ASCIIbetic" order
LANG=fr_FR.utf8 ls -1 # would consider é as between e and f
I am writing a Bash Shellscript. I need to check a file for if $value1 contains $value2. $value1 is the column number (1, 4, 5 as an example) and $value2 ($value2 can be '03', '04' , '09' etc) is the String I am looking for. If the column contains the $value2 then perform a move of the file to an error directory. I was wondering what is the best approach to this. I was thinking awk or is there another way?
$value1 and $value2 are stored in a config file. I have control over what format I can use. Here's an example. The file separator is Octal \036. I just depicted with | below.
Example
$value1=5
$value2=04
Input example1.txt
example|42|udajha|llama|04
example|22|udajha|llama|02
Input example2.txt
example|22|udajha|llama|02
Result
move example1.txt to /home/user/error_directory and example2.txt stays in current directory (nothing happens)
awk can report out which files meet this condition:
awk -F"|" -v columnToSearch=$value1 -v valueToFind=$value2 '$columnToSearch==valueToFind{print FILENAME}' example1.txt example2.txt
Then you can do your mv based on that.
Example using a pipe to xargs (with smaller variable names since you get the idea by now):
awk -F"|" -v c=$value1 -v v=$value2 '$c==v{print FILENAME}' example1.txt example2.txt | xargs -I{} mv -i {} /home/user/error_directory
If you're writing a bash shell script then you can break it down by column using cut.
There are really so many options that it depends on what you want to get done.
In my experience with data I'd use a colon rather than pipe because it allows me to avoid the escape with the 'cut' command.
Changing the data files to:
cat example1.txt
example:42:udajha:llama:04
example:22:udajha:llama:02
I'd write it like this: (adding -x so that you can see the processing, but in your code you'd not need to do that.)
[root#]# cat mysript.sh
#!/bin/sh -x
one=`cat example1.txt | cut -d: -f5`
two=`cat example2.txt | cut -d: -f5`
for i in $one
do
if [ $i -eq $two ]
then
movethis=`grep $two example1.txt`
echo $movethis >> /home/me/error.txt
fi
done
cat /home/me/error.txt
[root#]# ./mysript.sh
++ cat example1.txt
++ cut -d: -f5
+ one='04
02 '
++ cat example2.txt
++ cut -d: -f5
+ two=02
+ for i in '$one'
+ '[' 04 -eq 02 ']'
+ for i in '$one'
+ '[' 02 -eq 02 ']'
++ grep 02 example1.txt
+ movethis='example:22:udajha:llama:02 '
+ echo example:22:udajha:llama:02
+ cat /home/me/error.txt
example:22:udajha:llama:02
You can use any command you live to move your content. Touch, cp, mv, what ever you want to use there.
I have a directory with image files that follow a naming scheme and are not always contiguous. e.i:
IMG_33.jpg
IMG_34.jpg
IMG_35.jpg
IMG_223.jpg
IMG_224.jpg
IMG_225.jpg
IMG_226.jpg
IMG_446.jpg
I would like to rename them so they go something like this, in the same order:
0001.jpg
0002.jpg
0003.jpg
0004.jpg
0005.jpg
0006.jpg
0007.jpg
0008.jpg
So far this is what I came up, and while it does the four-digit padding, it doesn't sort by the number values in the filenames.
#!/bin/bash
X=1;
for i in *; do
mv $i $(printf %04d.%s ${X%.*} ${i##*.})
let X="$X+1"
done
result:
IMG_1009.JPG 0009.JPG
IMG_1010.JPG 0010.JPG
IMG_101.JPG 0011.JPG
IMG_102.JPG 0012.JPG
Update:
Try this. If output is okay remove echo.
X=1; find . -maxdepth 1 -type f -name "*.jpg" -print0 | sort -z -n -t _ -k2 | while read -d $'\0' -r line; do echo mv "$line" "$(printf "%04d%s" $X .jpg)"; ((X++)); done
Using the super helpful rename. First, pads files with one digit to two digits; then pads files with two digits to three digits; etc.
rename IMG_ IMG_0 IMG_?.jpg
rename IMG_ IMG_0 IMG_??.jpg
rename IMG_ IMG_0 IMG_???.jpg
Then, your for-loop (or another similar one) that renames does the trick as the files are in both alphabetical and numerical order.
how about this :
while read f1;do
echo $f1
mv IMG_$f1 $f1
done< <(ls | cut -d '_' -f 2 | sort -n)
thanks
Michael
I am writing a bash script where i will need to check a directory for existing files and look at the last 4 digits of the first segment of the file name to set the counter when adding new files to the directory.
Naming Scructure:
yymmddHNAZXLCOM0001.835
I need to put the portion in the example 0001 into a CTR variable so the next file it puts into the directory will be
yymmddHNAZXLCOM0002.835
and so on.
what would be the easiest and shortest way to do this?
You can do this with sed:
filename="yymmddHNAZXLCOM0001.835"
first_part=$(echo $filename | sed -e 's/\(.*\)\([0-9]\{4,4\}\)\.\(.*\)/\1/')
counter=$(echo $filename | sed -e 's/\(.*\)\([0-9]\{4,4\}\)\.\(.*\)/\2/')
suffix=$(echo $filename | sed -e 's/\(.*\)\([0-9]\{4,4\}\)\.\(.*\)/\3/')
echo "$first_part$(printf "%04u" $(($counter + 1))).$suffix"
=> "yymmddHNAZXLCOM0002.835"
All three sed calls use the same regular expression. The only thing that changes is the group selected to return. There's probably a way to do all of that in one call, but my sed-fu is rusty.
Alternate version, using a Bash array:
filename="yymmddHNAZXLCOM0001.835"
ary=($(echo $filename | sed -e 's/\(.*\)\([0-9]\{4,4\}\)\.\(.*\)/\1 \2 \3/'))
echo "${ary[0]}$(printf "%04u" $((${ary[1]} + 1))).${ary[2]}"
=> "yymmddHNAZXLCOM0002.835"
Note: This version assumes that the filename does not have spaces in it.
Try this...
current=`echo yymmddHNAZXLCOM0001.835 | cut -d . -f 1 | rev | cut -c 1-4 | rev`
next=`echo $current | awk '{printf("%04i",$0+1)}'`
f() {
if [[ $1 =~ (.*)([[:digit:]]{4})(\.[^.]*)$ ]]; then
local -a ctr=("${BASH_REMATCH[#]:1}")
touch "${ctr}$((++ctr[1]))${ctr[2]}"
# ...
else
echo 'no matches'
fi
}
shopt -s nullglob
f *
What is the best way, using Bash, to rename files in the form:
(foo1, foo2, ..., foo1300, ..., fooN)
With zero-padded file names:
(foo00001, foo00002, ..., foo01300, ..., fooN)
It's not pure bash, but much easier with the Perl version of rename:
rename 's/\d+/sprintf("%05d",$&)/e' foo*
Where 's/\d+/sprintf("%05d",$&)/e' is the Perl replace regular expression.
\d+ will match the first set of numbers (at least one number)
sprintf("%05d",$&) will pass the matched numbers to Perl's sprintf, and %05d will pad to five digits
In case N is not a priori fixed:
for f in foo[0-9]*; do
mv "$f" "$(printf 'foo%05d' "${f#foo}")"
done
I had a more complex case where the file names had a postfix as well as a prefix. I also needed to perform a subtraction on the number from the filename.
For example, I wanted foo56.png to become foo00000055.png.
I hope this helps if you're doing something more complex.
#!/bin/bash
prefix="foo"
postfix=".png"
targetDir="../newframes"
paddingLength=8
for file in ${prefix}[0-9]*${postfix}; do
# strip the prefix off the file name
postfile=${file#$prefix}
# strip the postfix off the file name
number=${postfile%$postfix}
# subtract 1 from the resulting number
i=$((number-1))
# copy to a new name with padded zeros in a new folder
cp ${file} "$targetDir"/$(printf $prefix%0${paddingLength}d$postfix $i)
done
Pure Bash, no external processes other than 'mv':
for file in foo*; do
newnumber='00000'${file#foo} # get number, pack with zeros
newnumber=${newnumber:(-5)} # the last five characters
mv $file foo$newnumber # rename
done
The oneline command that I use is this:
ls * | cat -n | while read i f; do mv "$f" `printf "PATTERN" "$i"`; done
PATTERN can be for example:
rename with increment counter: %04d.${f#*.} (keep original file extension)
rename with increment counter with prefix: photo_%04d.${f#*.} (keep original extension)
rename with increment counter and change extension to jpg: %04d.jpg
rename with increment counter with prefix and file basename: photo_$(basename $f .${f#*.})_%04d.${f#*.}
...
You can filter the file to rename with for example ls *.jpg | ...
You have available the variable f that is the file name and i that is the counter.
For your question the right command is:
ls * | cat -n | while read i f; do mv "$f" `printf "foo%d05" "$i"`; done
To left-pad numbers in filenames:
$ ls -l
total 0
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:24 010
-rw-r--r-- 1 victoria victoria 0 Mar 28 18:09 050
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:23 050.zzz
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:24 10
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:23 1.zzz
$ for f in [0-9]*.[a-z]*; do tmp=`echo $f | awk -F. '{printf "%04d.%s\n", $1, $2}'`; mv "$f" "$tmp"; done;
$ ls -l
total 0
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:23 0001.zzz
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:23 0050.zzz
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:24 010
-rw-r--r-- 1 victoria victoria 0 Mar 28 18:09 050
-rw-r--r-- 1 victoria victoria 0 Mar 28 17:24 10
Explanation
for f in [0-9]*.[a-z]*; do tmp=`echo $f | \
awk -F. '{printf "%04d.%s\n", $1, $2}'`; mv "$f" "$tmp"; done;
note the backticks: `echo ... $2}\` (The backslash, \, immediately above just splits that one-liner over two lines for readability)
in a loop find files that are named as numbers with lowercase alphabet extensions: [0-9]*.[a-z]*
echo that filename ($f) to pass it to awk
-F. : awk field separator, a period (.): if matched, separates the file names as two fields ($1 = number; $2 = extension)
format with printf: print first field ($1, the number part) as 4 digits (%04d), then print the period, then print the second field ($2: the extension) as a string (%s). All of that is assigned to the $tmp variable
lastly, move the source file ($f) to the new filename ($tmp)
The following will do it:
for ((i=1; i<=N; i++)) ; do mv foo$i `printf foo%05d $i` ; done
EDIT: changed to use ((i=1,...)), thanks mweerden!
My solution replaces numbers, everywhere in a string
for f in * ; do
number=`echo $f | sed 's/[^0-9]*//g'`
padded=`printf "%04d" $number`
echo $f | sed "s/${number}/${padded}/";
done
You can easily try it, since it just prints transformed file names (no filesystem operations are performed).
Explanation:
Looping through list of files
A loop: for f in * ; do ;done, lists all files and passes each filename as $f variable to loop body.
Grabbing the number from string
With echo $f | sed we pipe variable $f to sed program.
In command sed 's/[^0-9]*//g', part [^0-9]* with modifier ^ tells to match opposite from digit 0-9 (not a number) and then remove it it with empty replacement //. Why not just remove [a-z]? Because filename can contain dots, dashes etc. So, we strip everything, that is not a number and get a number.
Next, we assign the result to number variable. Remember to not put spaces in assignment, like number = …, because you get different behavior.
We assign execution result of a command to variable, wrapping the command with backtick symbols `.
Zero padding
Command printf "%04d" $number changes format of a number to 4 digits and adds zeros if our number contains less than 4 digits.
Replacing number to zero-padded number
We use sed again with replacement command like s/substring/replacement/. To interpret our variables, we use double quotes and substitute our variables in this way ${number}.
The script above just prints transformed names, so, let's do actual renaming job:
for f in *.js ; do
number=`echo $f | sed 's/[^0-9]*//g'`
padded=`printf "%04d" $number`
new_name=`echo $f | sed "s/${number}/${padded}/"`
mv $f $new_name;
done
Hope this helps someone.
I spent several hours to figure this out.
This answer is derived from Chris Conway's accepted answer but assumes your files have an extension (unlike Chris' answer). Just paste this (rather long) one liner into your command line.
for f in foo[0-9]*; do mv "$f" "$(printf 'foo%05d' "${f#foo}" 2> /dev/null)"; done; for f in foo[0-9]*; do mv "$f" "$f.ext"; done;
OPTIONAL ADDITIONAL INFO
This script will rename
foo1.ext > foo00001.ext
foo2.ext > foo00002.ext
foo1300.ext > foo01300.ext
To test it on your machine, just paste this one liner into an EMPTY directory.
rm * 2> /dev/null; touch foo1.ext foo2.ext foo1300.ext; for f in foo[0-9]*; do mv "$f" "$(printf 'foo%05d' "${f#foo}" 2> /dev/null)"; done; for f in foo[0-9]*; do mv "$f" "$f.ext"; done;
This deletes the content of the directory, creates the files in the above example and then does the batch rename.
For those who don't need a one liner, the script indented looks like this.
for f in foo[0-9]*;
do mv "$f" "$(printf 'foo%05d' "${f#foo}" 2> /dev/null)";
done;
for f in foo[0-9]*;
do mv "$f" "$f.ext";
done;
Here's a quick solution that assumes a fixed length prefix (your "foo") and fixed length padding. If you need more flexibility, maybe this will at least be a helpful starting point.
#!/bin/bash
# some test data
files="foo1
foo2
foo100
foo200
foo9999"
for f in $files; do
prefix=`echo "$f" | cut -c 1-3` # chars 1-3 = "foo"
number=`echo "$f" | cut -c 4-` # chars 4-end = the number
printf "%s%04d\n" "$prefix" "$number"
done