This question already has answers here:
How can I remove the extension of a filename in a shell script?
(15 answers)
Closed 6 years ago.
I'm new to scripting concept.. I have a requirement to rename multiple files in a directory like filename.sh.x into filename.sh
First I tried to get the file names in a particular directory.. so i followed the below scripting code
for entry in PathToThedirectory/*sh.x
do
echo $entry
done
& the above code listed down all the file names with full path..
But my expected o/p is : to get file names alone like abc.sh.x,
so that I can proceed with the split string mechanism to perform rename
operation easily...
help me to solve this ... Thanks in advance
First approach trying to follow OP suggestions:
for i in my/path/*.py.x
do
basename=$(basename "$i")
mv my/path/"$basename" my/path/"${basename%.*}"
done
And maybe, you can simplify it:
for i in my/path/*.py.x
do
mv "$i" "${i%.*}";
done
Documentation regarding this kind of operation (parameter expansion): https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html
In particular:
${parameter%word} : The word is expanded to produce a pattern just as in filename expansion. If the pattern matches a trailing portion of the expanded value of parameter, then the result of the expansion is the value of parameter with the shortest matching pattern (the ‘%’ case) or the longest matching pattern (the ‘%%’ case) deleted
So, ${i%.*} means:
Take $i
Match .* at the end of its value (. being a literal character)
Remove the shortest matching pattern
Look into prename (installed together with the perl package on ubuntu).
Then you can just do something like:
prename 's/\.x$//' *.sh.x
In ksh you can do this:
for $file in $(ls $path)
do
new_file=$(basename $path/$file .x)
mv ${path}/${file} ${path}/${new_file}
done
This should do the trick:
for file in *.sh.x;
do
mv "$file " "${file /.sh.x}";
done
Running this rename command from the root directory should work:
rename 's/\.sh\.x$/.sh/' *.sh.x
for i in ls -la /path|grep -v ^d|awk '{print $NF}'
do
echo "basename $i"
done
it will give u the base name of all files or you can try below
find /path -type f -exec basename {} \;
Related
I have the output file day by day:
linux-202105200900-foo.direct.tar.gz
The date and time string, ex: 202105200900 will change every day.
I need to manually rename these files to
linux-202105200900x86-foo.direct.tar.gz
( insert a short string x86 after date/time )
any bash script can help to do this?
If you're always inserting the string "x86" at character #18 in the string, you may use that command:
var="linux-202105200900-foo.direct.tar.gz"
var2=${var:0:18}"x86"${var:18}
echo $var2
The 2nd line means: "assign to variable var2 the first 18 characters of var, followed by x86 followed by the rest of the variable var"
If you want to insert "x86" just before the last hyphen in the string, you may write it like this:
var="linux-202105200900-foo.direct.tar.gz"
var2=${var%-*}"x86-"${var##*-}
echo $var2
The 2nd line means: "assign to variable var2:
the content of the variable var after removing the shortest matching pattern "-*" at the end
the string "x86-"
the content of the variable var after removing the longest matching pattern "*-" at the beginning
In addition to the very good answer by #Jean-Loup Sabatier another, perhaps more general way would simply be to replace the second occurrence of '-' with x86- which you can do with sed. Let's say you have:
fname=linux-202105200900-foo.direct.tar.gz
You can update that with:
fname="$(sed 's/-/x86-/2' <<< "$fname")"
Which simply uses a command substitution with sed and a herestring to modify fname assigning the modified result back to fname.
Example Use/Output
$ fname=linux-202105200900-foo.direct.tar.gz
fname="$(sed 's/-/x86-/2' <<< "$fname")"
echo $fname
linux-202105200900x86-foo.direct.tar.gz
Do you need this?
❯ dat=$(date '+%Y%m%d%H%M%S'); echo ${dat}
20210520170336
❯ filename="linux-${dat}x86-foo.direct.tar.gz"; echo ${filename}
linux-20210520170336x86-foo.direct.tar.gz
I wanted to go as simple as possible, considering only the timestamp is going to change, this script should do it. Just run it inside the folder where files are located and you'll get all of them renamed with x86.
#!/bin/bash
for file in $(ls); do
replaced=$(echo $file | sed 's|-foo|x86-foo|g')
mv $file $replaced
done
This is my output
filip#filip-ThinkPad-T14-Gen-1:~/test$ ls
linux-202105200900-foo.direct.tar.gz linux-202105201000-foo.direct.tar.gz linux-202105201100-foo.direct.tar.gz
filip#filip-ThinkPad-T14-Gen-1:~/test$ ./../development/bash-utils/bulk-rename.sh
filip#filip-ThinkPad-T14-Gen-1:~/test$ ls
linux-202105200900x86-foo.direct.tar.gz linux-202105201000x86-foo.direct.tar.gz linux-202105201100x86-foo.direct.tar.gz
Simply iterate through all the files in current folder and pipeline result to sed to replace regex -foo with x86-foo, then rename file with mv command.
As David mentioned in comment, if you're worried that there could be multiple occurrences of -foo then you can just replace g as global to 1 as first occurrence and that's it!
There is also the rename utility (https://man7.org/linux/man-pages/man1/rename.1.html), you could use:
rename -v 0-foo.direct.tar.gz 0x86-foo.direct.tar.gz *
which results in
`linux-202105200900-foo.direct.tar.gz' -> `linux-202105200900x86-foo.direct.tar.gz'
`linux-202205200900-foo.direct.tar.gz' -> `linux-202205200900x86-foo.direct.tar.gz'
`linux-202305200900-foo.direct.tar.gz' -> `linux-202305200900x86-foo.direct.tar.gz'
In addition to the very good answer by #David C. Rankin, just adding it in a loop and renaming the files
# !/usr/bin/bash
for file in `ls linux* 2>/dev/null` # Extract all files starting with linux
do
echo $file
fname="$(sed 's/-/x86-/2' <<< "$file")"
mv "$file" "$fname" # Rename file
done
Output recieved :
linux-202105200900x86-foo.direct.tar.gz
This question already has answers here:
Rename multiple files based on pattern in Unix
(24 answers)
Closed 5 years ago.
Write a simple script that will automatically rename a number of files. As an example we want the file *001.jpg renamed to user defined string + 001.jpg (ex: MyVacation20110725_001.jpg) The usage for this script is to get the digital camera photos to have file names that make some sense.
I need to write a shell script for this. Can someone suggest how to begin?
An example to help you get off the ground.
for f in *.jpg; do mv "$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.
Sorry if this is very simple compared to usual questions but I am just starting out. I have some files all with the same start name but of different file types, e.g:
1234.x
1234.y
1234.z
1234_V2.x
1234_V2.y
1234_V2.z
I want to rename the first part of these whilst keeping any ending and file type, e.g:
4321.x
4321.y
4321.z
4321_V2.x etc
I have tried using
mv 1234* 4321*
and
rename 1234* 4321*
But no luck! I have also been through all the other SO articles and although I could use a loop, most depend on the file type being the same.
Thanks in advance
You can use bash substitution:
for file in 1234*
do mv "$file" "4321${file#1234}"
done
OR, replace the do mv with the following
do mv "$file" "${file/1234/4321}"
See more in man bash under EXPANSION section, sub-section Parameter Expansion
Assuming your filenames for 1234 and 4321 i.e constant for all files, you can try this
for fn in `find . -name 1234*`
do
newf=`echo $fn | sed s/1234/4321/`
mv $fn $newfn
done
You can use a shell script, but it's kind of ugly because it will fork a lot, and thus, if you have a lot of files to rename, it will take time.
for f in 1234*; do echo mv $f $(echo $f | sed -e 's/1234/4321/'); done
Otherwize, rename is a good way to do it:
rename 's/1234/4321/' 1234*
Rename expects a regular expression as first parameter, see online documentation
See if it works:
rename "s/1234/4321/" 1234*
command means substitute(because of s) occurances of "1234" with "4321" in files that has name of pattern 1234*
You can also look at here. It is slightly more complicated than your case.
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
I'm trying to find all files with a specific extension in a directory and its subdirectories with my bash (Latest Ubuntu LTS Release).
This is what's written in a script file:
#!/bin/bash
directory="/home/flip/Desktop"
suffix="in"
browsefolders ()
for i in "$1"/*;
do
echo "dir :$directory"
echo "filename: $i"
# echo ${i#*.}
extension=`echo "$i" | cut -d'.' -f2`
echo "Erweiterung $extension"
if [ -f "$i" ]; then
if [ $extension == $suffix ]; then
echo "$i ends with $in"
else
echo "$i does NOT end with $in"
fi
elif [ -d "$i" ]; then
browsefolders "$i"
fi
done
}
browsefolders "$directory"
Unfortunately, when I start this script in terminal, it says:
[: 29: in: unexpected operator
(with $extension instead of 'in')
What's going on here, where's the error?
But this curly brace
find "$directory" -type f -name "*.in"
is a bit shorter than that whole thing (and safer - deals with whitespace in filenames and directory names).
Your script is probably failing for entries that don't have a . in their name, making $extension empty.
find {directory} -type f -name '*.extension'
Example: To find all csv files in the current directory and its sub-directories, use:
find . -type f -name '*.csv'
The syntax I use is a bit different than what #Matt suggested:
find $directory -type f -name \*.in
(it's one less keystroke).
Without using find:
du -a $directory | awk '{print $2}' | grep '\.in$'
Though using find command can be useful here, the shell itself provides options to achieve this requirement without any third party tools. The bash shell provides an extended glob support option using which you can get the file names under recursive paths that match with the extensions you want.
The extended option is extglob which needs to be set using the shopt option as below. The options are enabled with the -s support and disabled with he -u flag. Additionally you could use couple of options more i.e. nullglob in which an unmatched glob is swept away entirely, replaced with a set of zero words. And globstar that allows to recurse through all the directories
shopt -s extglob nullglob globstar
Now all you need to do is form the glob expression to include the files of a certain extension which you can do as below. We use an array to populate the glob results because when quoted properly and expanded, the filenames with special characters would remain intact and not get broken due to word-splitting by the shell.
For example to list all the *.csv files in the recursive paths
fileList=(**/*.csv)
The option ** is to recurse through the sub-folders and *.csv is glob expansion to include any file of the extensions mentioned. Now for printing the actual files, just do
printf '%s\n' "${fileList[#]}"
Using an array and doing a proper quoted expansion is the right way when used in shell scripts, but for interactive use, you could simply use ls with the glob expression as
ls -1 -- **/*.csv
This could very well be expanded to match multiple files i.e. file ending with multiple extension (i.e. similar to adding multiple flags in find command). For example consider a case of needing to get all recursive image files i.e. of extensions *.gif, *.png and *.jpg, all you need to is
ls -1 -- **/+(*.jpg|*.gif|*.png)
This could very well be expanded to have negate results also. With the same syntax, one could use the results of the glob to exclude files of certain type. Assume you want to exclude file names with the extensions above, you could do
excludeResults=()
excludeResults=(**/!(*.jpg|*.gif|*.png))
printf '%s\n' "${excludeResults[#]}"
The construct !() is a negate operation to not include any of the file extensions listed inside and | is an alternation operator just as used in the Extended Regular Expressions library to do an OR match of the globs.
Note that these extended glob support is not available in the POSIX bourne shell and its purely specific to recent versions of bash. So if your are considering portability of the scripts running across POSIX and bash shells, this option wouldn't be right.
find "$PWD" -type f -name "*.in"
There's a { missing after browsefolders ()
All $in should be $suffix
The line with cut gets you only the middle part of front.middle.extension. You should read up your shell manual on ${varname%%pattern} and friends.
I assume you do this as an exercise in shell scripting, otherwise the find solution already proposed is the way to go.
To check for proper shell syntax, without running a script, use sh -n scriptname.
To find all the pom.xml files in your current directory and print them, you can use:
find . -name 'pom.xml' -print
find $directory -type f -name "*.in"|grep $substring
for file in "${LOCATION_VAR}"/*.zip
do
echo "$file"
done