Rename files in shell - shell

I've folder and file structure like
Folder/1/fileNameOne.ext
Folder/2/fileNameTwo.ext
Folder/3/fileNameThree.ext
...
How can I rename the files such that the output becomes
Folder/1_fileNameOne.ext
Folder/2_fileNameTwo.ext
Folder/3_fileNameThree.ext
...
How can this be achieved in linux shell?

How many different ways do you want to do it?
If the names contain no spaces or newlines or other problematic characters, and the intermediate directories are always single digits, and if you have the list of the files to be renamed in a file file.list with one name per line, then one of many possible ways to do the renaming is:
sed 's%\(.*\)/\([0-9]\)/\(.*\)%mv \1/\2/\3 \1/\2_\3%' file.list | sh -x
You'd avoid running the command through the shell until you're sure it will do what you want; just look at the generated script until its right.
There is also a command called rename — unfortunately, there are several implementations, not all equally powerful. If you've got the one based on Perl (using a Perl regex to map the old name to the new name) you'd be able to use:
rename 's%/(\d)/%/${1}_%' $(< file.list)

Use a loop as follows:
while IFS= read -d $'\0' -r line
do
mv "$line" "${line%/*}_${line##*/}"
done < <(find Folder -type f -print0)
This method handle spaces, newlines and other special characters in the file names and the intermediate directories don't necessarily have to be single digits.

This may work if the name is always the same, ie "file":
for i in {1..3};
do
mv $i/file ${i}_file
done
If you have more dirs on a number range, change {1..3} for {x..y}.
I use ${i}_file instead of $i_file because it would consider $i_file a variable of name i_file, while we just want i to be the variable and file and text attached to it.

This solution from AskUbuntu worked for me.
Here is a bash script that does that:
Note: This script does not work if any of the file names contain spaces.
#! /bin/bash
# Only go through the directories in the current directory.
for dir in $(find ./ -type d)
do
# Remove the first two characters.
# Initially, $dir = "./directory_name".
# After this step, $dir = "directory_name".
dir="${dir:2}"
# Skip if $dir is empty. Only happens when $dir = "./" initially.
if [ ! $dir ]
then
continue
fi
# Go through all the files in the directory.
for file in $(ls -d $dir/*)
do
# Replace / with _
# For example, if $file = "dir/filename", then $new_file = "dir_filename"
# where $dir = dir
new_file="${file/\//_}"
# Move the file.
mv $file $new_file
done
# Remove the directory.
rm -rf $dir
done
Copy-paste the script in a file.
Make it executable using
chmod +x file_name
Move the script to the destination directory. In your case this should be inside Folder/.
Run the script using ./file_name.

Related

How to recursively find & replace whole files with bash?

I have hundreds of files that I need to recursively replace as the files are currently stored like so:
/2019/01/
file1.pdf
file2.pdf
/2019/02
file3.pdf
file4.pdf
etc
I then have all of the updated files in another directory like so:
/new-files
file1.pdf
file2.pdf
file3.pdf
file4.pdf
Could someone please tell me the best way of doing this with a bash script? I'd basically like to read the new-files directory and then replace any matching file names in the other folders.
Thanks in advance for any help!
Assuming that the 'new-files' directory and all the directory trees containing PDF files are under the current directory, try this Shellcheck-clean Bash code:
#! /bin/bash -p
find . -path ./new-files -prune -o -type f -name '*.pdf' -print0 \
| while IFS= read -r -d '' pdfpath; do
pdfname=${pdfpath##*/}
new_pdfpath=new-files/$pdfname
if [[ -f $new_pdfpath ]]; then
printf "Replace '%s' with '%s'\n" "$pdfpath" "$new_pdfpath" >&2
# cp -- "$new_pdfpath" "$pdfpath"
fi
done
The -path ./new-files -prune in the find command stops the 'new-files' directory from being searched.
The -o in the find command causes the next test and actions to be tried after checking for 'new-files'.
See BashFAQ/001 (How can I read a file (data stream, variable) line-by-line (and/or field-by-field)?) for an explanation of the use of the -print0 option to find and the while IFS= read -r -d '' .... In short, the code can handle arbitrary file paths, including ones with whitespace and newline characters in them.
See Removing part of a string (BashFAQ/100 (How do I do string manipulation in bash?)) for an explanation of ${pdfpath##*/}.
It's not clear to me if you want to copy or move the new file to replace the old file, or do something else. Run the code as it is to check if it is identifying the correct replacements to be done. If you are happy with it, uncomment the cp line, and modify it to do something different if that is what you want.
The -- in the cp command protects against arguments beginning with dash characters being interpreted as options. It's unnecessary in this case, but I always use it when arguments begin with variable (or other) expansions so the code will remain safe if it is used in other contexts.
I think this calls for a bash array.
#!/usr/bin/env bash
# Make an associative array
declare -A files=()
# Populate array as $files[file.pdf]="/path/to/file.pdf"
for f in 20*/*/*.pdf; do
files[${f##*/}]="$f"
done
# Step through files and replace
for f in new-files/*.pdf; do
if [[ ! -e "${files[${f##*/}]}" ]]; then
echo "ERROR: missing $f" >&2
continue
fi
mv -v "$f" "${files[${f##*/}]}"
done
Note that associative arrays require bash version 4 or above. If you're using the native bash on a Mac, this won't work as-is.
Note also that if you remove continue in the final lines, then the mv command will NOT safely move files that do not exist in the date hash directories, since no target is known.
If you wanted further protection you might use test -nt or friends to confirm that an update is happening in the right direction.

Bash: how to copy multiple files with same name to multiple folders

I am working on Linux machine.
I have a lot of files named the same, with a directory structure like this:
P45_input_foo/result.dat
P45_input_bar/result.dat
P45_input_tar/result.dat
P45_input_cool/result.dat ...
It is difficult to copy them one by one. I want to copy them into another folder named as data with similar folder names and file names:
/data/foo/result.dat
/data/bar/result.dat
/data/tar/result.dat
/data/cool/result.dat ...
In stead of copy them one by one what I should do?
Using a for loop in bash :
# we list every files following the pattern : ./<somedirname>/<any file>
# if you want to specify a format for the folders, you could change it here
# i.e. for your case you could write 'for f in P45*/*' to only match folders starting by P45
for f in */*
do
# we strip the path of the file from its filename
# i.e. 'P45_input_foo/result.dat' will become 'P45_input_foo'
newpath="${f%/*}"
# mkdir -p /data/${newpath##*_} will create our new data structure
# - /data/${newpath##*_} extract the last chain of character after a _, in our example, 'foo'
# - mkdir -p will recursively create our structure
# - cp "$f" "$_" will copy the file to our new directory. It will not launch if mkdir returns an error
mkdir -p /data/${newpath##*_} && cp "$f" "$_"
done
the ${newpath##*_} and ${f%/*} usage are part of Bash string manipulation methods. You can read more about it here.
You will need to extract the 3rd item after "_" :
P45_input_foo --> foo
create the directory (if needed) and copy the file to it. Something like this (not tested, might need editing):
STARTING_DIR="/"
cd "$STARTING_DIR"
VAR=$(ls -1)
while read DIR; do
TARGET_DIR=$(echo "$DIR" | cut -d'_' -f3)
NEW_DIR="/data/$DIR"
if [ ! -d "$NEW_DIR" ]; then
mkdir "$NEW_DIR"
fi
cp "$DIR/result.dat" "$NEW_DIR/result.dat"
if [ $? -ne 0 ];
echo "ERROR: encountered an error while copying"
fi
done <<<"$VAR"
Explanation: assuming all the paths you've mentioned are under root / (if not change STARTING_PATH accordingly). With ls you get the list of the directories, store the output in VAR. Pass the content of VAR to the while loop.
A bit of find and with a few bash tricks, the below script could do the trick for you. Remember to run the script without the mv and see if "/data/"$folder"/" is the actual path that you want to move the file(s).
#!/bin/bash
while IFS= read -r -d '' file
do
fileNew="${file%/*}" # Everything before the last '\'
fileNew="${fileNew#*/}" # Everything after the last '\'
IFS="_" read _ _ folder <<<"$fileNew"
mv -v "$file" "/data/"$folder"/"
done < <(find . -type f -name "result.dat" -print0)

Filenames with wildcards in variables

#!/bin/bash
outbound=/home/user/outbound/
putfile=DATA_FILE_PUT_*.CSV
cd $outbound
filecnt=0
for file in $putfile; do let filecnt=filecnt+1; done
echo "Filecount: " $filecnt
So this code works well when there are files located in the outbound directory. I can place files into the outbound path and as long as they match the putfile mask then the files are incremented as expected.
Where the problem comes in is if I run this while there are no files located in $outbound.
If there are zero files there $filecnt still returns a 1 but I'm looking to have it return a 0 if there are no files there.
Am I missing something simple?
Put set -x just below the #! line to watch what your script is doing.
If there is no matching file, then the wildcard is left unexpanded, and the loop runs once, with file having the value DATA_FILE_PUT_*.CSV.
To change that, set the nullglob option. Note that this only works in bash, not in sh.
shopt -s nullglob
putfile=DATA_FILE_PUT_*.CSV
for file in $putfile; do let filecnt=filecnt+1; done
Note that the putfile variable contains the wildcard pattern, not the list of file names. It might make more sense to put the list of matches in a variable instead. This needs to be an array variable, and you need to change the current directory first. The number of matching files is then the length of the array.
#!/bin/bash
shopt -s nullglob
outbound=/home/user/outbound/
cd "$outbound"
putfiles=(DATA_FILE_PUT_*.CSV)
echo "Filecount: " ${#putfiles}
If you need to iterate over the files, take care to protect the expansion of the array with double quotes, otherwise if a file name contains whitespace then it will be split over several words (and if a filename contains wildcard characters, they will be expanded).
#!/bin/bash
shopt -s nullglob
outbound=/home/user/outbound/
cd "$outbound"
putfiles=(DATA_FILE_PUT_*.CSV)
for file in "${putfiles[#]}"; do
echo "Processing $file"
done
You could test if file exists first
for file in $putfile; do
if [ -f "$file" ] ; then
let filecnt=filecnt+1
fi
done
Or look for your files with find
for file in $(find . -type f -name="$putfile"); do
let filecnt=filecnt+1
done
or simply (fixed)
filecnt=$(find . -type f -name "$putfile" | wc -l); echo $filecnt
This is because when no matches are found, bash by default expands the wildcard DATA_FILE_PUT_*.CSV to the word DATA_FILE_PUT_*.CSV and therefore you end up with a count of 1.
To disable this behavior, use shopt -s nullglob
Not sure why you need a piece of code here. Following one liner should do your job.
ls ${outbound}/${putfile} | wc -l
Or
find ${outbound} -maxdepth 1 -type f -name "${putfile}" | wc -l

Moving files to directories with similar names, in a script

I have a directory with sub-directories and files with names that start with a string similar to the sub-directories; e.g.
bar/
foo-1/ (dir)
foo-1-001.txt
foo-1-002.txt
foo-1-003.txt
foo-2/ (dir)
foo-2-001.txt
foo-2-002.txt
foo-2-003.txt
foo-3/ (dir)
foo-3-001.txt
foo-3-002.txt
foo-3-003.txt
etc.
All files are currently at the same level. I'd like to move the corresponding .txt files into their similarly-named directories with a script (there are > 9500 in my current situation).
I've written the following, but I'm missing something, as I can't get the files to move.
#!/bin/sh
# directory basename processing for derivatives
# create directory list in a text file
find ./ -type d > directoryList.txt
# setup while loop for moving text files around
FILE="directoryList.txt"
exec 3<&0
exec 0<$FILE
while read line
do
echo "This is a directory:`basename $line`"
filemoves=`find ./ -type f -name '*.txt' \! -name 'directoryList.txt' | sed 's|-[0-9]\{3\}\.txt$||g'`
if [ "`basename $filemoves`" == "$line" ]
then
cp $filemoves $line/
echo "copied $filemoves to $line"
fi
done
exec 0<&3
Things seem to work OK until I get to the if. I'm working across a number of *nix, so I have to be careful what arguments I'm throwing around (RHEL, FreeBSD, and possibly Mac OS X, too).
Assuming your files really match the pattern above (everything before the last dash is the directory name) this should do it:
for thefile in *.txt ; do mv -v $thefile ${thefile%-*}; done
and if it tells you command line too long (expanding *.txt into 4900 files is a lot) try this:
find . -name '*.txt' | while read thefile ; do mv -v $thefile ${thefile%-*} ; done
I'm not a shell script expert but I'm aware that in many shells (and according to this page on the internet: http://www.vectorsite.net/tsshell.html this includes SH), string comparison is done with the "=" operator, not "==".
[ "$shvar" = "fox" ] String comparison, true if match.
[code block removed]
Reason 1. Used ls instead of globbing
Reason 2. Used mv $VAR1 $VAR2 style moving without quoting variables

bash script for copying files between directories

I am writing the following script to copy *.nzb files to a folder to queue them for Download.
I wrote the following script
#!/bin/bash
#This script copies NZB files from Downloads folder to HellaNZB queue folder.
${DOWN}="/home/user/Downloads/"
${QUEUE}="/home/user/.hellanzb/nzb/daemon.queue/"
for a in $(find ${DOWN} -name *.nzb)
do
cp ${a} ${QUEUE}
rm *.nzb
done
it gives me the following error saying:
HellaNZB.sh: line 5: =/home/user/Downloads/: No such file or directory
HellaNZB.sh: line 6: =/home/user/.hellanzb/nzb/daemon.queue/: No such file or directory
Thing is that those directories exsist, I do have right to access them.
Any help would be nice.
Please and thank you.
Variable names on the left side of an assignment should be bare.
foo="something"
echo "$foo"
Here are some more improvements to your script:
#!/bin/bash
#This script copies NZB files from Downloads folder to HellaNZB queue folder.
down="/home/myusuf3/Downloads/"
queue="/home/myusuf3/.hellanzb/nzb/daemon.queue/"
find "${down}" -name "*.nzb" | while read -r file
do
mv "${file}" "${queue}"
done
Using while instead of for and quoting variables that contain filenames protects against filenames that contain spaces from being interpreted as more than one filename. Removing the rm keeps it from repeatedly producing errors and failing to copy any but the first file. The file glob for -name needs to be quoted. Habitually using lowercase variable names reduces the chances of name collisions with shell variables.
If all your files are in one directory (and not in multiple subdirectories) your whole script could be reduced to the following, by the way:
mv /home/myusuf3/Downloads/*.nzb /home/myusuf3/.hellanzb/nzb/daemon.queue/
If you do have files in multiple subdirectories:
find /home/myusuf3/Downloads/ -name "*.nzb" -exec mv {} /home/myusuf3/.hellanzb/nzb/daemon.queue/ +
As you can see, there's no need for a loop.
The correct syntax is:
DOWN="/home/myusuf3/Downloads/"
QUEUE="/home/myusuf3/.hellanzb/nzb/daemon.queue/"
for a in $(find ${DOWN} -name *.nzb)
# escape the * or it will be expanded in the current directory
# let's just hope no file has blanks in its name
do
cp ${a} ${QUEUE} # ok, although I'd normally add a -p
rm *.nzb # again, this is expanded in the current directory
# when you fix that, it will remove ${a}s before they are copied
done
Why don't you just use rm $(a}?
Why use a combination of cp and rm anyway, instead of mv?
Do you realize all files will end up in the same directory, and files with the same name from different directories will overwrite each other?
What if the cp fails? You'll lose your file.

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