I have a folder that is full of .bak files and some other files also. I need to remove the extension of all .bak files in that folder. How do I make a command which will accept a folder name and then remove the extension of all .bak files in that folder ?
Thanks.
To remove a string from the end of a BASH variable, use the ${var%ending} syntax. It's one of a number of string manipulations available to you in BASH.
Use it like this:
# Run in the same directory as the files
for FILENAME in *.bak; do mv "$FILENAME" "${FILENAME%.bak}"; done
That works nicely as a one-liner, but you could also wrap it as a script to work in an arbitrary directory:
# If we're passed a parameter, cd into that directory. Otherwise, do nothing.
if [ -n "$1" ]; then
cd "$1"
fi
for FILENAME in *.bak; do mv "$FILENAME" "${FILENAME%.bak}"; done
Note that while quoting your variables is almost always a good practice, the for FILENAME in *.bak is still dangerous if any of your filenames might contain spaces. Read David W.'s answer for a more-robust solution, and this document for alternative solutions.
There are several ways to remove file suffixes:
In BASH and Kornshell, you can use the environment variable filtering. Search for ${parameter%word} in the BASH manpage for complete information. Basically, # is a left filter and % is a right filter. You can remember this because # is to the left of %.
If you use a double filter (i.e. ## or %%, you are trying to filter on the biggest match. If you have a single filter (i.e. # or %, you are trying to filter on the smallest match.
What matches is filtered out and you get the rest of the string:
file="this/is/my/file/name.txt"
echo ${file#*/} #Matches is "this/` and will print out "is/my/file/name.txt"
echo ${file##*/} #Matches "this/is/my/file/" and will print out "name.txt"
echo ${file%/*} #Matches "/name.txt" and will print out "/this/is/my/file"
echo ${file%%/*} #Matches "/is/my/file/name.txt" and will print out "this"
Notice this is a glob match and not a regular expression match!. If you want to remove a file suffix:
file_sans_ext=${file%.*}
The .* will match on the period and all characters after it. Since it is a single %, it will match on the smallest glob on the right side of the string. If the filter can't match anything, it the same as your original string.
You can verify a file suffix with something like this:
if [ "${file}" != "${file%.bak}" ]
then
echo "$file is a type '.bak' file"
else
echo "$file is not a type '.bak' file"
fi
Or you could do this:
file_suffix=$(file##*.}
echo "My file is a file '.$file_suffix'"
Note that this will remove the period of the file extension.
Next, we will loop:
find . -name "*.bak" -print0 | while read -d $'\0' file
do
echo "mv '$file' '${file%.bak}'"
done | tee find.out
The find command finds the files you specify. The -print0 separates out the names of the files with a NUL symbol -- which is one of the few characters not allowed in a file name. The -d $\0means that your input separators are NUL symbols. See how nicely thefind -print0andread -d $'\0'` together?
You should almost never use the for file in $(*.bak) method. This will fail if the files have any white space in the name.
Notice that this command doesn't actually move any files. Instead, it produces a find.out file with a list of all the file renames. You should always do something like this when you do commands that operate on massive amounts of files just to be sure everything is fine.
Once you've determined that all the commands in find.out are correct, you can run it like a shell script:
$ bash find.out
rename .bak '' *.bak
(rename is in the util-linux package)
Caveat: there is no error checking:
#!/bin/bash
cd "$1"
for i in *.bak ; do mv -f "$i" "${i%%.bak}" ; done
You can always use the find command to get all the subdirectories
for FILENAME in `find . -name "*.bak"`; do mv --force "$FILENAME" "${FILENAME%.bak}"; done
Related
This question already has answers here:
Rename multiple files based on pattern in Unix
(24 answers)
Closed 5 years ago.
Write a simple script that will automatically rename a number of files. As an example we want the file *001.jpg renamed to user defined string + 001.jpg (ex: MyVacation20110725_001.jpg) The usage for this script is to get the digital camera photos to have file names that make some sense.
I need to write a shell script for this. Can someone suggest how to begin?
An example to help you get off the ground.
for f in *.jpg; do mv "$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.
I ran the following in a directory with no files:
for file in *.20191017.*;do echo ${file}; done
what it returned was this:
*.20191017.*
which is little awkward since this was just a pattern and not the filename itself.
Can anyone please help on this?
Found the reason for this anomaly (source: https://www.cyberciti.biz/faq/bash-loop-over-file/)
You can do filename expansion in loop such as work on all pdf files in current directory:
for f in *.pdf; do
echo "Removing password for pdf file - $f"
done
However, there is one problem with the above syntax. If there are no pdf files in current directory it will expand to *.pdf (i.e. f will be set to *.pdf”). To avoid this problem add the following statement before the for loop:
#!/bin/bash
# Usage: remove all utility bills pdf file password
shopt -s nullglob # expands the glob to empty string when there are no matching files in the directory.
for f in *.pdf; do
echo "Removing password for pdf file - $f"
pdftk "$f" output "output.$f" user_pw "YOURPASSWORD-HERE"
done
The for loop simply iterates over the words between in and ; (possibly expanded by bash). Here, file is just the variable name. If you want to iterate between all files that are actually present, you can, for example, add a if to check if the ${file} really exists:
for file in *.20191017.*
do
if [ -e "${file}" ]
then
echo ${file}
fi
done
Or you can use, e.g., find
find . -name '*.20191017.*' -maxdepth 1
-maxdepth 1 is to avoid recursion.
Is it possible to write a script to rename all files after the extension?
Example in the folder, there are :
hello.txt-123ahr
bye.txt-56athe
test.txt-98hg12
I want the output:
hello.txt
bye.txt
test.txt
If you just want to remove everything from the dash forwards, you can use Parameter expansion:
#!/bin/bash
for file in *.txt-* ; do
mv "$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.
I am still a newbie in shell scripting and trying to come up with a simple code. Could anyone give me some direction here. Here is what I need.
Files in path 1: /tmp
100abcd
200efgh
300ijkl
Files in path2: /home/storage
backupfile_100abcd_str1
backupfile_100abcd_str2
backupfile_200efgh_str1
backupfile_200efgh_str2
backupfile_200efgh_str3
Now I need to delete file 300ijkl in /tmp as the corresponding backup file is not present in /home/storage. The /tmp file contains more than 300 files. I need to delete the files in /tmp for which the corresponding backup files are not present and the file names in /tmp will match file names in /home/storage or directories under /home/storage.
Appreciate your time and response.
You can also approach the deletion using grep as well. You can loop though the files in /tmp checking with ls piped to grep, and deleting if there is not a match:
#!/bin/bash
[ -z "$1" -o -z "$2" ] && { ## validate input
printf "error: insufficient input. Usage: %s tmpfiles storage\n" ${0//*\//}
exit 1
}
for i in "$1"/*; do
fn=${i##*/} ## strip path, leaving filename only
## if file in backup matches filename, skip rest of loop
ls "${2}"* | grep -q "$fn" &>/dev/null && continue
printf "removing %s\n" "$i"
# rm "$i" ## remove file
done
Note: the actual removal is commented out above, test and insure there are no unintended consequences before preforming the actual delete. Call it passing the path to tmp (without trailing /) as the first argument and with /home/storage as the second argument:
$ bash scriptname /path/to/tmp /home/storage
You can solve this by
making a list of the files in /home/storage
testing each filename in /tmp to see if it is in the list from /home/storage
Given the linux+shell tags, one might use bash:
make the list of files from /home/storage an associative array
make the subscript of the array the filename
Here is a sample script to illustrate ($1 and $2 are the parameters to pass to the script, i.e., /home/storage and /tmp):
#!/bin/bash
declare -A InTarget
while read path
do
name=${path##*/}
InTarget[$name]=$path
done < <(find $1 -type f)
while read path
do
name=${path##*/}
[[ -z ${InTarget[$name]} ]] && rm -f $path
done < <(find $2 -type f)
It uses two interesting shell features:
name=${path##*/} is a POSIX shell feature which allows the script to perform the basename function without an extra process (per filename). That makes the script faster.
done < <(find $2 -type f) is a bash feature which lets the script read the list of filenames from find without making the assignments to the array run in a subprocess. Here the reason for using the feature is that if the array is updated in a subprocess, it would have no effect on the array value in the script which is passed to the second loop.
For related discussion:
Extract File Basename Without Path and Extension in Bash
Bash Script: While-Loop Subshell Dilemma
I spent some really nice time on this today because I needed to delete files which have same name but different extensions, so if anyone is looking for a quick implementation, here you go:
#!/bin/bash
# We need some reference to files which we want to keep and not delete,
# let's assume you want to keep files in first folder with jpeg, so you
# need to map it into the desired file extension first.
FILES_TO_KEEP=`ls -1 ${2} | sed 's/\.pdf$/.jpeg/g'`
#iterate through files in first argument path
for file in ${1}/*; do
# In my case, I did not want to do anything with directories, so let's continue cycle when hitting one.
if [[ -d $file ]]; then
continue
fi
# let's omit path from the iterated file with baseline so we can compare it to the files we want to keep
NAME_WITHOUT_PATH=`basename $file`
# I use mac which is equal to having poor quality clts
# when it comes to operating with strings,
# this should be safe check to see if FILES_TO_KEEP contain NAME_WITHOUT_PATH
if [[ $FILES_TO_KEEP == *"$NAME_WITHOUT_PATH"* ]];then
echo "Not deleting: $NAME_WITHOUT_PATH"
else
# If it does not contain file from the other directory, remove it.
echo "deleting: $NAME_WITHOUT_PATH"
rm -rf $file
fi
done
Usage: sh deleteDifferentFiles.sh path/from/where path/source/of/truth
I am trying to replace the prefix of all files in the directory with another prefix (renaming).
This is my script
# Script to rename the files
#!/bin/bash
for file in $1*;
do
mv $file `echo $file | sed -e 's/^$1/$2/'`;
done
Upon executing the script with
rename.sh BIT SIT
I get the following errors
mv: `BITfile.h' and `BITFile.h' are the same file
mv: `BITDefs.cpp' and `BITDefs.cpp' are the same file
mv: `BITDefs.h' and `BITDefs.h' are the same file
Seems like sed is treating $1 and $2 as the same value, but when I print those variables on another line it shows that they are different.
As Roman Newaza says, you can use " instead of ' to tell Bash that you want variables to be expanded. However, in your case, it would be safest to write:
for file in "$1"* ; do
mv -- "$file" "$2${file#$1}"
done
so that weird characters in filenames, or in your script parameters, cannot cause any problems.
You can also use parameter expansion to replace the prefix of all files in a directory.
for file in "$1"*;
do
mv ${file} ${file/#$1/$2}
done
Use double quotes instead:
# ...
mv "$file" `echo $file | sed -e "s/^$1/$2/"`
# ...
And learn Quotes and escaping in Bash.
When you don't use double quotes, the variables won't expand.
I will rather use this
#!/bin/bash
for file in $1*;
do
mv "$file" "$1${file:${#2}}"
done
where
${file:${#2}
means substring, from length of the argument 2 to the end