I have a directory with a bunch of files with names like:
001234.jpg
001235.jpg
004729342.jpg
I want to remove the leading zeros from all file names, so I'd be left with:
1234.jpg
1235.jpg
4729342.jpg
I've been trying different configurations of sed, but I can't find the proper syntax. Is there an easy way to list all files in the directory, pipe it through sed, and either move or copy them to the new file name without the leading zeros?
for FILE in `ls`; do mv $FILE `echo $FILE | sed -e 's:^0*::'`; done
sed by itself is the wrong tool for this: you need to use some shell scripting as well.
Check Rename multiple files with Linux page for some ideas. One of the ideas suggested is to use the rename perl script:
rename 's/^0*//' *.jpg
In Bash, which is likely to be your default login shell, no external commands are necessary.
shopt -s extglob
for i in 0*[^0]; do mv "$i" "${i##*(0)}"; done
Maybe not the most elegant but it will work.
for i in 0*
do
mv "${i}" "`expr "${i}" : '0*\(.*\)'`"
done
Try using sed, e.g.:
sed -e 's:^0*::'
Complete loop:
for f in `ls`; do
mv $f $(echo $f | sed -e 's:^0*::')
done
I dont know sed at all but you can get a listing by using find:
find -type f -name *.jpg
so with the other answer it might look like
find . -type f -name *.jpg | sed -e 's:^0*::'
but i dont know if that sed command holds up or not.
Here's one that doesn't require sed:
for x in *.jpg ; do let num="10#${x%%.jpg}"; mv $x ${num}.jpg ; done
Note that this ONLY works when the filenames are all numbers. You could also remove the leading zeros using the shell:
for a in *.jpg ; do dest=${a/*(0)/} ; mv $a $dest ; done
In Bash shell you can do:
shopt -s nullglob
for file in 0*.jpg
do
echo mv "$file" "${file##*0}"
done
Related
I have a lot of files that are named as: MM-DD-YYYY.pdf. I want to rename them as YYYY-MM-DD.pdf I’m sure there is some bash magic to do this. What is it?
For files in the current directory:
for name in ./??-??-????.pdf; do
if [[ "$name" =~ (.*)/([0-9]{2})-([0-9]{2})-([0-9]{4})\.pdf ]]; then
echo mv "$name" "${BASH_REMATCH[1]}/${BASH_REMATCH[4]}-${BASH_REMATCH[3]}-${BASH_REMATCH[2]}.pdf"
fi
done
Recursively, in or under the current directory:
find . -type f -name '??-??-????.pdf' -exec bash -c '
for name do
if [[ "$name" =~ (.*)/([0-9]{2})-([0-9]{2})-([0-9]{4})\.pdf ]]; then
echo mv "$name" "${BASH_REMATCH[1]}/${BASH_REMATCH[4]}-${BASH_REMATCH[3]}-${BASH_REMATCH[2]}.pdf"
fi
done' bash {} +
Enabling the globstar shell option in bash lets us do the following (will also, like the above solution, handle all files in or below the current directory):
shopt -s globstar
for name in **/??-??-????.pdf; do
if [[ "$name" =~ (.*)/([0-9]{2})-([0-9]{2})-([0-9]{4})\.pdf ]]; then
echo mv "$name" "${BASH_REMATCH[1]}/${BASH_REMATCH[4]}-${BASH_REMATCH[3]}-${BASH_REMATCH[2]}.pdf"
fi
done
All three of these solutions uses a regular expression to pick out the relevant parts of the filenames, and then rearranges these parts into the new name. The only difference between them is how the list of pathnames is generated.
The code prefixes mv with echo for safety. To actually rename files, remove the echo (but run at least once with echo to see that it does what you want).
A direct approach example from the command line:
$ ls
10-01-2018.pdf 11-01-2018.pdf 12-01-2018.pdf
$ ls [0-9]*-[0-9]*-[0-9]*.pdf|sed -r 'p;s/([0-9]{2})-([0-9]{2})-([0-9]{4})/\3-\1-\2/'|xargs -n2 mv
$ ls
2018-10-01.pdf 2018-11-01.pdf 2018-12-01.pdf
The ls output is piped to sed , then we use the p flag to print the argument without modifications, in other words, the original name of the file, and s to perform and output the conversion.
The ls + sed result is a combined output that consist of a sequence of old_file_name and new_file_name.
Finally we pipe the resulting feed through xargs to get the effective rename of the files.
From xargs man:
-n number Execute command using as many standard input arguments as possible, up to number arguments maximum.
You can use the following command very close to the one of klashxx:
for f in *.pdf; do echo "$f"; mv "$f" "$(echo "$f" | sed 's#\(..\)-\(..\)-\(....\)#\3-\2-\1#')"; done
before:
ls *.pdf
12-01-1998.pdf 12-03-2018.pdf
after:
ls *.pdf
1998-01-12.pdf 2018-03-12.pdf
Also if you have other pdf files that does not respect this format in your folder, what you can do is to select only the files that respect the format: MM-DD-YYYY.pdf to do so use the following command:
for f in `find . -maxdepth 1 -type f -regextype sed -regex './[0-9]\{2\}-[0-9]\{2\}-[0-9]\{4\}.pdf' | xargs -n1 basename`; do echo "$f"; mv "$f" "$(echo "$f" | sed 's#\(..\)-\(..\)-\(....\)#\3-\2-\1#')"; done
Explanations:
find . -maxdepth 1 -type f -regextype sed -regex './[0-9]\{2\}-[0-9]\{2\}-[0-9]\{4\}.pdf this find command will look only for files in the current working directory that respect your syntax and extract their basename (remove the ./ at the beginning, folders and other type of files that would have the same name are not taken into account, other *.pdf files are also ignored.
for each file you do a move and the resulting file name is computed using sed and back reference to the 3 groups for MM,DD and YYYY
For these simple filenames, using a more verbose pattern, you can simplify the body of the loop a bit:
twodigit=[[:digit:]][[:digit:]]
fourdigit="$twodigit$twodigit"
for f in $twodigit-$twodigit-$fourdigit.pdf; do
IFS=- read month day year <<< "${f%.pdf}"
mv "$f" "$year-$month-$day.pdf"
done
This is basically #Kusalananda's answer, but without the verbosity of regular-expression matching.
I have a bunch of files in a folder. Some of them are of the format:
IMG_YYYYMMDD_junk.ext
I would like to rename such files into
YYYY-MM-DD junk.ext
Example: IMG_20170214_3939233.jpg becomes 2017-02-14 3939233.jpg
So far I was successful in filtering files I need:
find *.jpg *.jpeg *.png | egrep '^IMG_[0-9]{1,8}'
and I know I need to use sed but I am getting no where specifying and referencing match-groups in my regex for further filename transformation. I know I may have to use xarg later on in the pipe but so far I wasn't successful in transforming each file name just to print it out.
Perhaps, sed is not the best option here.
With Perl‘s standalone rename command and bash‘s option nullglob:
shopt -s nullglob
rename -n 's/.*_(....)(..)(..)_([0-9]+.*)/$1-$2-$3 $4/' *.jpg *.jpeg *.png
If everything looks fine remove option -n.
A logic in bash with NO external tools!
You can run the below script from inside the folder containing these images.
#!/bin/bash
for file in *.{jpg,jpeg,png}; do
IFS="_" read -ra fileNameList <<<"$file"
year="${fileNameList[1]:0:4}"
month="${fileNameList[1]:4:2}"
day="${fileNameList[1]:6:2}"
targetFileName="${year}-${month}-${day} ${fileNameList[2]}"
# Remove this line and uncomment the line with 'mv' if things look OK
echo "$file" "$targetFileName"
#mv -v "$file" "$targetFileName"
done
The idea is tot split the file name on _ and store them in array. Then parse the individual digits from the number and form the final name from the combined elements.
With GNU Parallel it looks like this:
find *.jpg *.jpeg *.png |
parallel mv {} '{= s/IMG_(....)(..)(..)_/$1-$2-$3 / =}'
Or:
parallel mv {} '{= s/IMG_(....)(..)(..)_/$1-$2-$3 / =}' ::: *.jpg *.jpeg *.png
I assume you're not really interested in the find command, but in nailing down the sed regex:
find ~ -type f -maxdepth 1 -name IMG*.jpg | sed -e 's/\(IMG_\)\([0-9]\{4\}\)\([0-9]\{2\}\)\([0-9]\{2\}\)_/\2-\3-\4 /g'
Your example:
echo "IMG_20170214_3939233.jpg" | sed -e 's/\(IMG_\)\([0-9]\{4\}\)\([0-9]\{2\}\)\([0-9]\{2\}\)_/\2-\3-\4 /g'
output:
2017-02-14 3939233.jpg
A mostly-bash solution using its regex-matching operator, =~, which supports capture groups that can be accessed via the built-in "${BASH_REMATCH[#]}" array variable:
for file in *; do
[[ $file =~ ^IMG_([0-9]{4})([0-9]{2})([0-9]{2})_(.+\.(jpg|jpeg|png))$ ]] || continue
mv "$file" "${BASH_REMATCH[1]}-${BASH_REMATCH[2]}-${BASH_REMATCH[3]} ${BASH_REMATCH[4]}"
done
I have files with name of the form "NAME-xxxxxx.tedx" and I want to remove the "-xxxxxx" part. The x are all digits.
The regex "\-[0-9]{1,6}" matches the substring, but I have no idea how to remove it from the filename.
Any idea how I can do that in the shell?
If you have the perl version of the rename command installed, you could try:
rename 's/-[0-9]+//' *.tedx
Demo:
[me#home]$ ls
hello-123.tedx world-23456.tedx
[me#home]$ rename 's/-[0-9]+//' *.tedx
[me#home]$ ls
hello.tedx world.tedx
This command is smart enough to not rename files if it means overwriting an existing file:
[me#home]$ ls
hello-123.tedx world-123.tedx world-23456.tedx
[me#home]$ rename 's/-[0-9]+//' *.tedx
world-23456.tedx not renamed: world.tedx already exists
[me#home]$ ls
hello.tedx world-23456.tedx world.tedx
echo NAME-12345.tedx | sed "s/-[0-9]*//g"
will give NAME.tedx. So you can use a loop and move the files using mv command:
for file in *.tedx; do
newfile=$(echo "$file" | sed "s/-[0-9]*//g")
mv "$file" $newfile
done
If you want to use just the shell
shopt -s extglob
for f in *-+([0-9]]).tedx; do
newname=${f%-*}.tedx # strip off the dash and all following chars
[[ -f $newname ]] || mv "$f" "$newname"
done
Dead simple.
How do I rename
05_h.png
06_h.png
to
05_half.png
06_half.png
At least, I think it's simple, but it's hard to Google for this kind of thing unless you already know.
Thanks....
Just use bash, no need to call external commands.
for file in *_h.png
do
mv "$file" "${file/_h.png/_half.png}"
done
Do not add #!/bin/sh
For those that need that one-liner:
for file in *.png; do mv "$file" "${file/_h.png/_half.png}"; done
Try rename command:
rename 's/_h.png/_half.png/' *.png
Update:
example usage:
create some content
$ mkdir /tmp/foo
$ cd /tmp/foo
$ touch one_h.png two_h.png three_h.png
$ ls
one_h.png three_h.png two_h.png
test solution:
$ rename 's/_h.png/_half.png/' *.png
$ ls
one_half.png three_half.png two_half.png
for f in *.png; do
fnew=`echo $f | sed 's/_h.png/_half.png/'`
mv $f $fnew
done
Or in one-liner:
for f in *.png; do mv "$f" "$(echo $f | sed 's/_h.png$/_half.png/g')"; done
Are you looking for a pure bash solution? There are many approaches, but here's one.
for file in *_h.png ; do mv "$file" "${file%%_h.png}_half.png" ; done
This presumes that the only files in the current directory that end in _h.png are the ones you want to rename.
Much more specifically
for file in 0{5..6}_h.png ; do mv "$file" "${file/_h./_half.}" ; done
Presuming those two examples are your only. files.
For the general case, file renaming in has
been covered
before.
Use the rename utility written in perl.
Might be that it is not available by default though...
$ touch 0{5..6}_h.png
$ ls
05_h.png 06_h.png
$ rename 's/h/half/' *.png
$ ls
05_half.png 06_half.png
for i in *_h.png ; do
mv $i `echo "$i"|awk -F'.' '{print $1"alf."$2}'`
done
I had a similar question:
In the manual, it describes rename as
rename [option] expression replacement file
so you can use it in this way
rename _h _half *.png
In the code:
'_h' is the expression that you are looking for.
'_half' is the pattern that you want to replace with.
'*.png' is the range of files that you are looking for your possible target files.
Hope this can help c:
Another approach can be manually using batch rename option
Right click on the file -> File Custom Commands -> Batch Rename
and you can replace h. with half.
This will work for linux based gui using WinSCP etc
One liner:
for file in *.php ; do mv "$file" "_$file" ; done
Although the answer set is complete, I need to add another missing one.
for i in *_h.png;
do name=`echo "$i" | cut -d'_' -f1`
echo "Executing of name $name"
mv "$i" "${name}_half.png"
done
I had to rename the prefix of files and I found this answer with a solution like this:
for i in h_*; do mv ${i/#h_/half_}; done
If pattern begins with #, it must match at the beginning of the
expanded value of parameter. If pattern begins with %, it must match
at the end of the expanded value of parameter.
from man bash
Use the rename utility:
rc#bvm3:/tmp/foo $ touch 05_h.png 06_h.png
rc#bvm3:/tmp/foo $ rename 's/_h/_half/' *
rc#bvm3:/tmp/foo $ ls -l
total 0
-rw-r--r-- 1 rc rc 0 2011-09-17 00:15 05_half.png
-rw-r--r-- 1 rc rc 0 2011-09-17 00:15 06_half.png
I have a folder with files named as
input (1).txt
input (2).txt
input (3).txt
...
input (207).txt
How do I rename them to
input_1.in
input_2.in
input_3.in
...
input_207.in
I am trying this
for f in *.txt ; do mv $f `echo $f | sed -e 's/input\ (\(\d*\))\.txt/input_\1.in/'` ; done
But it gives me
mv: target `(100).txt' is not a directory
mv: target `(101).txt' is not a directory
mv: target `(102).txt' is not a directory
...
Where did I go wrong?
I have put in the quotes now, but I get this now
mv: `input (90).txt' and `input (90).txt' are the same file
It is somehow trying to rename the file to the same name. How is that happening?
That is because bash for split the element with space ' ' so you are commanding it to move 'input' to '(1)'.
The way to solve this is to tell bash to split by new line using IFS variable.
Like this:
IFS=$'\n'
Then do your command.
However, I suggest you to use find to do this instead using -exec command.
For example:
find *.txt -exec mv "{}" `echo "{}" | sed -e 's/input\ (\([0-9]*\))\.txt/input_\1.in/'` \;
NOTE: I write this from memory and I did test this so let try and adjust it.
Hope this helps.
You're forgetting to quote your arguments.
... mv "$f" "$(echo "$f" | ... )" ; done
no need to call external commands
#!/bin/bash
shopt -s nullglob
shopt -s extglob
for file in *.txt
do
newfile="${file//[)]/}"
newfile="${file// [(]/_}"
mv "$file" "${newfile%.txt}.in"
done
As you've already fixed, you need to quote the $f argument to mv.
As to your second problem, sed doesn't support \d. You could use [0-9] instead.
for f in *.txt ; do mv "$f" `echo $f | sed -e 's/input\ (\(\d*\))\.txt/input_\1.in/'` ; done
If you have GNU Parallel http://www.gnu.org/software/parallel/ installed you can do this:
seq 1 207 | parallel -q mv 'input ({}).txt' input_{}.in
Watch the intro video for GNU Parallel to learn more:
http://www.youtube.com/watch?v=OpaiGYxkSuQ