I have many files of the form {a}_{b}_R1.txt I would like to change the basenames of these files to {a}.{b}.txt how would I go about this? I can only find out how to do this with extensions. Thanks.
You can just a simple loop in the native bash shell under OS X from terminal, after navigating to the folder containing those text files,
# Looping for all the text files in the current
# directory
for file in *.txt; do
# Stripping off '_R1' from the file-name
temp1="${file//_R1/}"
# Replacing '_' with '.' and renaming the file to the name generated
temp2="${temp1//_/.}"
echo "$temp2" "$file"
#mv -v "$file" "$temp2"
done
Remove the line with echo and uncomment the mv once you find the names are changed accordingly.
for file in *.txt; do temp1="${file//_R1/}"; temp2="${temp1//_/.}"; mv -v "$file" "$temp2"; done
This works as well:
for f in *.txt; do mv "$f" "echo $f | sed s/_R1//"; done then for f in *.txt; do mv "$f" "echo $f | sed s/_/./";
Related
i have found this command to replace spaces with underscores:
for file in *.jpg; do mv "$file" ${file// /_}; done
but many pics have chars like ' # ñ and want to remove those automatically and also wanted to add a suffix o prefix.
ex.
pic's#nick_0001.jpg
pic's#nick_0003.jpg
pic's#nick_0003.jpg
to
vacations_pics_nick_0001.jpg
pics_nick_0001_vacations.jpg
can you help me?
for file in *.jpg
do
mv "$file" $(sed 's/[^ [:alnum:]]//g;s/ /_/g;s/^/your_prefix/' <<<"$file")
done
Should do the job
both solution works fine.
is there a way to use that command on a single line, like the example i give to use with others comand together:
for file in * do mv "$file" "${file//[^a-z0-9]/_}" done;
??'
Using bash:
for file in *
do
mv "$file" "${file//[^a-z0-9]/_}"
done
Test:
$ touch "pic's#nick_0001.jpg"
$ ls
pic's#nick_0001.jpg
$ for file in *; do mv "$file" "${file//[^a-z0-9]/_}"; done
$ ls
pic_s_nick_0001_jpg
I am new to Unix shell scripting. I have a small task of moving files into another directory based on their names.
I have a directory named Cars containing .csv files. I have to read the .csv file names, and move each file into another directory based on the filename, as follows:
BMW_c.csv must go into the BMW/c/ directory.
Mercedes_x.csv must go into the Mercedes/x/ directory.
I tried using for and if loop but couldn`t do it till sub directory.
I highly appreciate your help.
Something like this might work
for f in *.csv; do
DIR="$( echo ${f%.*} | tr '_' '/')"
mkdir -p "./$DIR"
mv "$f" "$DIR"
done
It will read all *.csv files, replace '_' by '/' in the file names (with .csv removed), create the directory and move the file. There will be errors if there are no csv files in the current directory.
I have done this and tried,it works.If any modifications or enhancements in the script you can post your answers..
cd /DIR/Cars/BMW
for f in *.csv
do
filename="${f##*/}"
echo ${filename}
if [[ "${f}" == *c* ]]
then
echo "moving files..."
mv "${f}" /BMW/c/
fi
done
cd /DIR/Cars/Mercedes
for f in *.csv
do
filename="${f##*/}"
echo ${filename}
if [[ "${f}" == *x* ]]
then
echo "moving files..."
mv "${f}" /Mercedes/x/
fi
done
Could someone please help with this script. I need to use grep to loop to through the filenames that need to be changed.
#!/bin/bash
file=
for file in $(ls $1)
do
grep "^.old" | mv "$1/$file" "$1/$file.old"
done
bash can handle regular expressions without using grep.
for f in "$1"/*; do
[[ $f =~ \.old ]] && continue
# Or a pattern instead
# [[ $f == *.old* ]] && continue
mv "$f" "$f.old"
done
You can also move the name checking into the pattern itself:
shopt -s extglob
for f in "$1/"!(*.old*); do
mv "$f" "$f.old"
done
If I understand your question correctly, you want to make rename a file (i.e. dir/file.txt ==> dir/file.old) only if the file has not been renamed before. The solution is as follow.
#!/bin/bash
for file in "$1/"*
do
backup_file="${file%.*}.old"
if [ ! -e "$backup_file" ]
then
echo mv "$file" "$backup_file"
fi
done
Discussion
The script currently does not actual make back up, it only displays the action. Run the script once and examine the output. If this is what you want, then remove the echo from the script and run it again.
Update
Here is the no if solution:
ls "$1/"* | grep -v ".old" | while read file
do
echo mv "$file" "${file}.old"
done
Discussion
The ls command displays all files.
The grep command filter out those files that has the .old extension so they won't be displayed.
The while loop reads the file names that do not have the .old extension, one by one and rename them.
I have files with name of the form "NAME-xxxxxx.tedx" and I want to remove the "-xxxxxx" part. The x are all digits.
The regex "\-[0-9]{1,6}" matches the substring, but I have no idea how to remove it from the filename.
Any idea how I can do that in the shell?
If you have the perl version of the rename command installed, you could try:
rename 's/-[0-9]+//' *.tedx
Demo:
[me#home]$ 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
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