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 $file everytning from the last dash". If you want to start from the first dash, use %%.
Note that you might overwrite some files if their leading parts are equivalent, e.g. hello.txt-123abc and hello.txt-456xyz.
Related
This question already has answers here:
Rename multiple files based on pattern in Unix
(24 answers)
Closed 5 years ago.
Write a simple script that will automatically rename a number of files. As an example we want the file *001.jpg renamed to user defined string + 001.jpg (ex: MyVacation20110725_001.jpg) The usage for this script is to get the digital camera photos to have file names that make some sense.
I need to write a shell script for this. Can someone suggest how to begin?
An example to help you get off the ground.
for f in *.jpg; do mv "$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.
My files got this structure:
mynewfile-runtime-tested-1102-19.4-alpha.zip
mysdk-sdk-tested-1102-19.4-alpha.zip
sources-tested-1102-19.4-alpha.zip
I looking for a way how to dynamically detect and drop the suffix of tested-1102-19.4-alpha and to copy the files with new names so it will look like:
mynewfile-runtime.zip
mysdk-sdk.zip
sources.zip
The suffix should be detected dynamically by delimiters ('-'), my other chunk of file have suffix like nottested-404-11.2.34-beta and the other one is final-01-1-release. The only thing remain constant is the delimiter of '-'
for file in *.zip; do
mv "$file" "${file%-*-*-*-*.zip}.zip"
done
This is fully portable POSIX shell, without forks to sed or other programs.
The ${file%pattern} bit says to remove the shortest matching string.
You can also remove the longest match with %%, or from the left with # and ##, respectively.
To only move the files that match the pattern you can do this:
#!/bin/sh
suffix='-*-*-*-*.zip'
for file in *$suffix
do
trimmed=${file%$suffix}
echo mv "$file" "$trimmed".zip
done
Remove the echo when you are confident with the result.
I have 100s of files in a directory with the following naming convention.
00XYZCD042ABCDE20141002ABCDE.XML
00XYZCC011ABCDE20141002.TXT
00XYZCB165ABCDE20141002ABCDE.TXT
00XYZCB165ABCDE20141002ABCDE.CSV
I want to rename these files using bash, awk, cut, sed, so that I get the output:
XYZCD042.XML
XYZCC011.TXT
XYZCB165.TXT
XYZCB165.CSV
So basically, remove the first two 0s always, and then keep everything until ABCDE starts and then remove everything including ABCDE and keep the file extension.
You could try the below rename command,
rename 's/ABCDE.*(\..*)/$1/;s/^00//' *
Explanation:
s/ABCDE.*(\..*)/$1/ Matches all the characters from the first ABCDE upto the last and captures only the extension part. Then all the matched chars are replaced with that captured extension.
s/^00// Then this would remove the leading two zeros.
Bash only:
for fn in *; do
A=${fn#00}
mv $fn ${A/ABCDE*./.}
done
The first line in the for loop strips the 00 prefix, and the second line strips the ABCDE suffix (up to a dot), then performs the rename.
for file in *
do
mv -- "$file" "${file:2:8}.${file#*.}"
done
It's important to always quote your variables unless you have a specific purpose in mind and understand all of the effects.
for i in *; do
mv $i $(echo $i | sed -e 's/^00//' -e 's/ABCDE2014[^.]*//');
done
I have a list of for example 100 files with the naming convention
<date>_<Time>_XYZ.xml.abc
<date>_<Time>_XYZ.xml
<date>_<Time>_XYZ.csv
for example
20140730_025373_XYZ.xml
20140730_015233_XYZ.xml.ab
20140730_015233_XYZ.csv
Now I want to write script which will remove anything between two underscores. for example in the above case
remove 015233 and change 20140730_015233_XYZ.xml.ab to 20140730_XYZ.xml.ab
remove 015233 and change 20140730_015233_XYZ.csv to 20140730_XYZ.csv
I have tried number of various options using rename, cut, mv but I am getting varied results, not the one which I expect.
You could use rename command if you want to rename files present inside the current directory,
rename 's/^([^_]*)_[^_]*(_.*)$/$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
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