This question already has answers here:
Rename multiple files in shell [duplicate]
(4 answers)
Closed 4 years ago.
I've got a small problem with my bash script. I try to change file name in current directory for whole files with txt extension to text extension. For exampel 1.txt to 1.text
My script looks like this now:
#!/bin/bash
FILES=`ls /home/name/*.txt`
NAME=*.txt
RENAME=*.text
for file in FILES
do
mv $NAME $RENAME
done
i try whole combination with single, double quotes and backticks and I receive errors all the time.
Do you have some ideas how to receive wildcards "*" in bash?
Thanks.
That's not at all how you do that.
#!/bin/bash
shopt -s nullglob
OLD=.txt
NEW=.text
FILES=(/home/name/*"$OLD")
for file in "${FILES[#]}"
do
mv "$file" "${file%$OLD}$NEW}"
done
There are a number of issues with your script. Firstly, you shouldn't run ls and attempt to store its output like that. If you want to iterate through those file, just do it in the loop:
for file in /home/name/*.txt
Now the shell is doing all the work for you, and as a bonus handling any kind of weird filenames that you might have.
In your example you were looping over the literal string "FILES", not the variable, but I guess that was just a typo.
The built-in way to change the filename is to use a parameter expansion to remove the old one, then concatenate with the new one:
old=txt
new=text
for file in /home/name/*"$old"
do
mv "$file" "${file%$old}$new"
done
If it is possible that there are no files matching the glob, then by default, the /home/name/*.txt will not be expanded and your loop will just run once. then you have a couple of options:
use shopt -s nullglob so that /home/name/*.txt expands to null, and the loop is skipped
add an explicit check inside the loop to ensure that $file exists before trying to mv:
for file in /home/name/*"$old"
do
[ -e "$file" ] || continue
mv "$file" "${file%$old}$new"
done
You can use rename to rename filenames.
rename .txt .text /home/name/*.txt
And if you want to do this by looping, you can
for FILE in /data/tmp/*.txt; do
mv "${FILE}" "${FILE/.txt/.text}"
done
Related
This question already has answers here:
Rename multiple files based on pattern in Unix
(24 answers)
Closed 5 years ago.
Write a simple script that will automatically rename a number of files. As an example we want the file *001.jpg renamed to user defined string + 001.jpg (ex: MyVacation20110725_001.jpg) The usage for this script is to get the digital camera photos to have file names that make some sense.
I need to write a shell script for this. Can someone suggest how to begin?
An example to help you get off the ground.
for f in *.jpg; do mv "$f" "$(echo "$f" | sed s/IMG/VACATION/)"; done
In this example, I am assuming that all your image files contain the string IMG and you want to replace IMG with VACATION.
The shell automatically evaluates *.jpg to all the matching files.
The second argument of mv (the new name of the file) is the output of the sed command that replaces IMG with VACATION.
If your filenames include whitespace pay careful attention to the "$f" notation. You need the double-quotes to preserve the whitespace.
You can use rename utility to rename multiple files by a pattern. For example following command will prepend string MyVacation2011_ to all the files with jpg extension.
rename 's/^/MyVacation2011_/g' *.jpg
or
rename <pattern> <replacement> <file-list>
this example, I am assuming that all your image files begin with "IMG" and you want to replace "IMG" with "VACATION"
solution : first identified all jpg files and then replace keyword
find . -name '*jpg' -exec bash -c 'echo mv $0 ${0/IMG/VACATION}' {} \;
for file in *.jpg ; do mv $file ${file//IMG/myVacation} ; done
Again assuming that all your image files have the string "IMG" and you want to replace "IMG" with "myVacation".
With bash you can directly convert the string with parameter expansion.
Example: if the file is IMG_327.jpg, the mv command will be executed as if you do mv IMG_327.jpg myVacation_327.jpg. And this will be done for each file found in the directory matching *.jpg.
IMG_001.jpg -> myVacation_001.jpg
IMG_002.jpg -> myVacation_002.jpg
IMG_1023.jpg -> myVacation_1023.jpg
etcetera...
find . -type f |
sed -n "s/\(.*\)factory\.py$/& \1service\.py/p" |
xargs -p -n 2 mv
eg will rename all files in the cwd with names ending in "factory.py" to be replaced with names ending in "service.py"
explanation:
In the sed cmd, the -n flag will suppress normal behavior of echoing input to output after the s/// command is applied, and the p option on s/// will force writing to output if a substitution is made. Since a sub will only be made on match, sed will only have output for files ending in "factory.py"
In the s/// replacement string, we use "& " to interpolate the entire matching string, followed by a space character, into the replacement. Because of this, it's vital that our RE matches the entire filename. after the space char, we use "\1service.py" to interpolate the string we gulped before "factory.py", followed by "service.py", replacing it. So for more complex transformations youll have to change the args to s/// (with an re still matching the entire filename)
Example output:
foo_factory.py foo_service.py
bar_factory.py bar_service.py
We use xargs with -n 2 to consume the output of sed 2 delimited strings at a time, passing these to mv (i also put the -p option in there so you can feel safe when running this). voila.
NOTE: If you are facing more complicated file and folder scenarios, this post explains find (and some alternatives) in greater detail.
Another option is:
for i in *001.jpg
do
echo "mv $i yourstring${i#*001.jpg}"
done
remove echo after you have it right.
Parameter substitution with # will keep only the last part, so you can change its name.
Can't comment on Susam Pal's answer but if you're dealing with spaces, I'd surround with quotes:
for f in *.jpg; do mv "$f" "`echo $f | sed s/\ /\-/g`"; done;
You can try this:
for file in *.jpg;
do
mv $file $somestring_${file:((-7))}
done
You can see "parameter expansion" in man bash to understand the above better.
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(28 answers)
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(6 answers)
Closed 3 years ago.
I have numerous files in a directory and want to replace one of the libraries used in all of these files with a newer library because the old library no longer works.
I have used ls > names.txt to list all filenames into a txt document. I then attempted to write a bash script which would loop through all files, catch the old library, and replace it with the new library.
for entry in names.txt
do
sed 's/<libOld>/<libNew>/g' $entry > $entry
done
I expect the loop to go through each file name, find the old library used, and replace it with the new one. Running this script however doesn't appear to do anything.
You're bumping into a few common issues; I've closed as a duplicate, but here are the collected fixes for your specific case:
Editing a file in-place with sed
With GNU sed:
sed -i 's/<libOld>/<libNew>/g' filename
with any sed:
sed 's/<libOld>/<libNew>/g' filename > filename.tmp && mv filename.tmp filename
Looping over a file line by line
for entry in names.txt only ever sets entry to names.txt, it doesn't read its contents. This is also BashFAQ/001.
while IFS= read -r entry; do
printf '%s\n' "$entry"
done < names.txt
Looping over all files in a directory
You don't need a separate file, and you shouldn't use ls but globs:
for fname in ./*; do
printf '%s\n' "$fname"
done
Combined for your case
Notice the double quotes around $entry.
for entry in ./*; do
sed -i 's/<libOld>/<libNew>/g' "$entry"
done
which can be simplified to no loop at all:
sed -i 's/<libOld>/<libNew>/g' ./*
Is it possible to write a script to rename all files after the extension?
Example in the folder, there are :
hello.txt-123ahr
bye.txt-56athe
test.txt-98hg12
I want the output:
hello.txt
bye.txt
test.txt
If you just want to remove everything from the dash forwards, you can use Parameter expansion:
#!/bin/bash
for file in *.txt-* ; do
mv "$file" "${file%-*}"
done
Where ${file%-*} means "remove from $file everytning from the last dash". If you want to start from the first dash, use %%.
Note that you might overwrite some files if their leading parts are equivalent, e.g. hello.txt-123abc and hello.txt-456xyz.
This question already has answers here:
Do not show results if directory is empty using Bash
(3 answers)
Closed 4 years ago.
So I have to find all the files in the directory that start with the letter a, and list them out. This is pretty easy by doing
cd some_directory
for file in a*; do
echo "$file"
done
However I want that if there are no files present that match a*, then the for loop will not run at all. Currently, if this is the case then the shell will echo
a*
Is there a way to do this? Thank you
Your text is opposite of your title, in my answer below I've assumed the text is your intention and your title is incorrect:
globs can be made to act like this with the bash shell option "nullglob":
shopt -s nullglob
An alternative is to use find and ignore errors by piping stderr to /dev/null
for file in $(find a* 2>/dev/null); do
echo "$file"
done
I have a list of for example 100 files with the naming convention
<date>_<Time>_XYZ.xml.abc
<date>_<Time>_XYZ.xml
<date>_<Time>_XYZ.csv
for example
20140730_025373_XYZ.xml
20140730_015233_XYZ.xml.ab
20140730_015233_XYZ.csv
Now I want to write script which will remove anything between two underscores. for example in the above case
remove 015233 and change 20140730_015233_XYZ.xml.ab to 20140730_XYZ.xml.ab
remove 015233 and change 20140730_015233_XYZ.csv to 20140730_XYZ.csv
I have tried number of various options using rename, cut, mv but I am getting varied results, not the one which I expect.
You could use rename command if you want to rename files present inside the current directory,
rename 's/^([^_]*)_[^_]*(_.*)$/$1$2/g' *
You can use sed:
sed 's/\([^_]*\)_.*_\(.*\)/\1_\2/' files.list
You can also use cut command
cut -d'_' -f1,3 filename
for FILE in *; do mv "$FILE" "${FILE/_*_/_}"; done
And more specific is
for FILE in *.xml *.xml.ab *.csv; do mv "$FILE" "${FILE/_*_/_}"; done
Further:
for FILE in *_*_*.xml *_*_*.xml.ab *_*_*.csv; do mv "$FILE" "${FILE/_*_/_}"; done