i have a directory listing like
seascaperecovered0088crop.jpg
seascaperecovered0096crop.jpg
seascaperecovered0098crop.jpg
seascaperecovered0101crop.jpg
seascaperecovered0103crop.jpg
seascaperecovered0105crop.jpg
seascaperecovered0107crop.jpg
seascaperecovered0112crop.jpg
seascaperecovered0119crop.jpg
seascaperecovered0122crop.jpg
and i want to rename all files as seen here:
seascape_0122.jpg
i have tried something like this:
for f in `ls | egrep 'seascaperecovered.*\.jpg'`;
do mv $f ${f/seascaperecovered/seascape};
done
i have read that you can do this with mv, rename, sed, awk, etc.
can someone point me to the easiest (and clearest, hopefully) way of accomplishing this in UNIX?
FWIW, I am ssh'd into a Linux machine and running a bash shell.
thanks,
jml
Very straightforward:
for i in seascaperecovered*.jpg; do A=${i/crop/}; mv $i ${A/recovered/_}; done
(Put echo before the mv first for a dry run.)
With bash regular expressions
for file in *; do
[[ "$file" =~ [0-9]+ ]] && mv "$file" seascape_${BASH_REMATCH[0]}.jpg
done
Related
I have a bunch of files which are copied by "mv --backup=t source destination". So these files are in format *.*.~[0-9]~
For example,
some_file_name.pdf.~1~
some_file_name.pdf.~2~
another_file_name.docx.~1~
another_file_name.docx.~2~
Now, I would like to rename all such files so that the backup extension number came before the actual extension. Like
some_file_name_1.pdf
some_file_name_2.pdf
another_file_name_1.docx
another_file_name_2.docx
Is there a way to do in Unix using shell?
Any thoughts would be appreciated. Thank you in advance.
Using BASH regex directives you can do this:
for f in *~; do
[[ $f =~ ^(.+)\.([^.]+)\.~([0-9]+)~$ ]] &&
echo mv "$f" "${BASH_REMATCH[1]}_${BASH_REMATCH[3]}.${BASH_REMATCH[2]}"
done
Output:
mv CR_71050_5.3.17.pdf.~1~ CR_71050_5.3.17_1.pdf
mv another_file_name.docx.~1~ another_file_name_1.docx
mv another_file_name.docx.~2~ another_file_name_2.docx
mv some_file_name.pdf.~1~ some_file_name_1.pdf
mv some_file_name.pdf.~2~ some_file_name_2.pdf
Once you're satisfied, you can remove echo before mv
Try:
awk -F~ '{ split($1,splt,".");system("mv "$0" "splt[1]"_"$2"."splt[2]) }'
We are essentially splitting the filenames on ~ and then using awks split function to further split the text and then pass it to the system function to execute the mv command.
You can use rename (it is available as a Debian package).
$ ls
file.pdf.~1~ file.pdf.~10~ file.pdf.~2~ file.pdf.~3~ file.pdf.~9~
$ rename 's/.pdf.~([0-9]+)~$/$1.pdf/' *.pdf.~*~
$ ls
file10.pdf file1.pdf file2.pdf file3.pdf file9.pdf
So I'm trying to create a bash script that does something for every file in a directory. This is my script. The problem is that this script split on each space i would rather have it split on new line. So it look just like ls print. (I'm going to put other code inside the do, the echo is just a test)
#! /bin/bash
for file in $(ls)
do
echo $file
done
So how do i solve this?
You'd better avoid working with the result of ls (Why you shouldn't parse the output of ls).
Instead, do something like:
for file in /your/dir/*
do
echo "$file"
done
To pick files only:
while read -r file; do
echo "$file"
done < <(exec find -xtype f -maxdepth 1 -mindepth 1)
Or just filter them
for file in *; do
[[ -f $file ]] || continue
echo "$file"
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'm fairly new to bash so sorry if this is kind of a basic question. I was trying to rename a bunch of mp3 files to prepend 1- to their filenames, and mv *.mp3 1-*.mp3 didn't work unfortunately. So I tried to script it, first with echo to test the commands:
for f in *.mp3 ; do echo mv \'$f\' \'1-$f\'; done
Which seems to output the commands that I like, so I removed the echo, changing the command to
for f in *.mp3 ; do mv \'$f\' \'1-$f\'; done
Which failed. Next I tried piping the commands onward like so
for f in *.mp3 ; do echo mv \'$f\' \'1-$f\'; done | /bin/sh
Which worked, but if anyone could enlighten me as to why the middle command doesn't work I would be interested to know. Or if there is an more elegant one-liner that would do what I wanted, I would be interested to see that too.
I think you have to change the command to
for f in *.mp3 ; do mv "$f" "1-$f"; done
Otherwise you would pass something like 'file1.mp3' and '1-file1.mp3' to mv (including the single quotes).
Dry run:
rename -n 's/^/1-/' *.mp3
Remove the -n if it looks correct to run the command. man rename for details.
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