I want to rename (within bash) a list of files that begin with a specific string 'foo', continue with a a random string of unspecified length 'random1' and continue with another known string 'bar'. Essentially from 'foo_random1_bar_random2' I want 'foo_bar_random2'. I know how to remove specific strings using sed, but not strings of unknown length between specific strings.
for i in foo_*_bar_*
do
j=$(echo "$i"|sed -e 's/foo_.*_bar_\(.*\)/foo_bar_\1/')
echo mv "$i" "$j"
done
This shows a list of mv commands. Once you're happy with the commands, remove the echo before mv so they will be executed.
Using BASH string replacement:
f='foo_random1_bar_random2'
echo "${f/*bar/foo_bar}"
PS: As per comments below, if there is underscore before bar then better to use:
echo "${f/*_bar/foo_bar}"
Output:
foo_bar_random2
Use mv as:
for f in foo_*bar_*; do
mv -i "$f" "${f/*_bar/foo_bar}"
done
If you have the rename tool installed you can do this
rename 's/foo_.*_bar/foo_bar/' files
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" "$fname" # Rename file
done
Output recieved :
linux-202105200900x86-foo.direct.tar.gz
Given an example that my file name is
A_BC_DEF_GH_IJ_LMNO_PQ_11111111_1111111111_111111_AB.dat.meta
i am trying to rename this with unix command but when i tried using this cmd
for f in *.meta; do mv "$f" "$(echo $f|sed s/[0-9]/?/g|sed 's/-/*/g')" ; done
my file is renamed to
A_BC_DEF_GH_IJ_LMNO_PQ_????????_????????????????????_???????_AB.dat.meta
it is expected to rename the file to
A_BC_DEF_GH_IJ_LMNO_PQ__????????_????????????????????_*_AB.dat.meta
Im quite new with unix cmd , any approach that i should try ?
Since [0-9] and ? are undergoing filename expansion, you should quote them to avoid nasty error messages. With this in mind, I did a
echo A_BC_DEF_GH_IJ_LMNO_PQ_11111111_1111111111_111111_AB.dat.meta | sed 's/[0-9]/?/g'|sed 's/-/*/g'
and got as output A_BC_DEF_GH_IJ_LMNO_PQ????????????????????????AB.dat.meta, which makes sense to me. Why would you expect an asterisk in the resulting filename? In your second sed command, you are turning the hyphens into asterisks, but there is no hyphen in the input.
Of course it is pretty unsane to use question marks and asterisks in a file name, as this is just begging for trouble, but there is no law that you must not do this.
A_BC_DEF_GH_IJ_LMNO_PQ_11111111_1111111111_111111_AB.dat.meta
Match it with a regex. Remember which characters need to be escaped in sed. Remember about proper quoting - if you write $ it should be inside ". Note that if there are no files named *.meta it will just iterate over a string *.meta unless nullglob is set.
$ touch A_BC_DEF_GH_IJ_LMNO_PQ_11111111_1111111111_111111_AB.dat.meta
$ for f in *.meta; do mv "$f" "$(echo "$f" | sed 's/[0-9]/?/g; s/_\(?*\)_\(?*\)_\(?*\)_\([^_]*\)$/__\1_\2_*_\4/')" ; done
This question already has answers here:
Rename multiple files based on pattern in Unix
(24 answers)
Closed 5 years ago.
Write a simple script that will automatically rename a number of files. As an example we want the file *001.jpg renamed to user defined string + 001.jpg (ex: MyVacation20110725_001.jpg) The usage for this script is to get the digital camera photos to have file names that make some sense.
I need to write a shell script for this. Can someone suggest how to begin?
An example to help you get off the ground.
for f in *.jpg; do mv "$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 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 have a folder with about 1,700 files. They are all named like 1.txt or 1497.txt, etc. I would like to rename all the files so that all the filenames are four digits long.
I.e., 23.txt becomes 0023.txt.
What is a shell script that will do this? Or a related question: How do I use grep to only match lines that contain \d.txt (i.e., one digit, then a period, then the letters txt)?
Here's what I have so far:
for a in [command i need help with]
do
mv $a 000$a
done
Basically, run that three times, with commands there to find one digit, two digits, and three digit filenames (with the number of initial zeros changed).
Try:
for a in [0-9]*.txt; do
mv $a `printf %04d.%s ${a%.*} ${a##*.}`
done
Change the filename pattern ([0-9]*.txt) as necessary.
A general-purpose enumerated rename that makes no assumptions about the initial set of filenames:
X=1;
for i in *.txt; do
mv $i $(printf %04d.%s ${X%.*} ${i##*.})
let X="$X+1"
done
On the same topic:
Bash script to pad file names
Extract filename and extension in bash
Using the rename (prename in some cases) script that is sometimes installed with Perl, you can use Perl expressions to do the renaming. The script skips renaming if there's a name collision.
The command below renames only files that have four or fewer digits followed by a ".txt" extension. It does not rename files that do not strictly conform to that pattern. It does not truncate names that consist of more than four digits.
rename 'unless (/0+[0-9]{4}.txt/) {s/^([0-9]{1,3}\.txt)$/000$1/g;s/0*([0-9]{4}\..*)/$1/}' *
A few examples:
Original Becomes
1.txt 0001.txt
02.txt 0002.txt
123.txt 0123.txt
00000.txt 00000.txt
1.23.txt 1.23.txt
Other answers given so far will attempt to rename files that don't conform to the pattern, produce errors for filenames that contain non-digit characters, perform renames that produce name collisions, try and fail to rename files that have spaces in their names and possibly other problems.
for a in *.txt; do
b=$(printf %04d.txt ${a%.txt})
if [ $a != $b ]; then
mv $a $b
fi
done
One-liner:
ls | awk '/^([0-9]+)\.txt$/ { printf("%s %04d.txt\n", $0, $1) }' | xargs -n2 mv
How do I use grep to only match lines that contain \d.txt (IE 1 digit, then a period, then the letters txt)?
grep -E '^[0-9]\.txt$'
Let's assume you have files with datatype .dat in your folder. Just copy this code to a file named run.sh, make it executable by running chmode +x run.sh and then execute using ./run.sh:
#!/bin/bash
num=0
for i in *.dat
do
a=`printf "%05d" $num`
mv "$i" "filename_$a.dat"
let "num = $(($num + 1))"
done
This will convert all files in your folder to filename_00000.dat, filename_00001.dat, etc.
This version also supports handling strings before(after) the number. But basically you can do any regex matching+printf as long as your awk supports it. And it supports whitespace characters (except newlines) in filenames too.
for f in *.txt ;do
mv "$f" "$(
awk -v f="$f" '{
if ( match(f, /^([a-zA-Z_-]*)([0-9]+)(\..+)/, a)) {
printf("%s%04d%s", a[1], a[2], a[3])
} else {
print(f)
}
}' <<<''
)"
done
To only match single digit text files, you can do...
$ ls | grep '[0-9]\.txt'
One-liner hint:
while [ -f ./result/result`printf "%03d" $a`.txt ]; do a=$((a+1));done
RESULT=result/result`printf "%03d" $a`.txt
To provide a solution that's cautiously written to be correct even in the presence of filenames with spaces:
#!/usr/bin/env bash
pattern='%04d' # pad with four digits: change this to taste
# enable extglob syntax: +([[:digit:]]) means "one or more digits"
# enable the nullglob flag: If no matches exist, a glob returns nothing (not itself).
shopt -s extglob nullglob
for f in [[:digit:]]*; do # iterate over filenames that start with digits
suffix=${f##+([[:digit:]])} # find the suffix (everything after the last digit)
number=${f%"$suffix"} # find the number (everything before the suffix)
printf -v new "$pattern" "$number" "$suffix" # pad the number, then append the suffix
if [[ $f != "$new" ]]; then # if the result differs from the old name
mv -- "$f" "$new" # ...then rename the file.
fi
done
There is a rename.ul command installed from util-linux package (at least in Ubuntu) by default installed.
It's use is (do a man rename.ul):
rename [options] expression replacement file...
The command will replace the first occurrence of expression with the given replacement for the provided files.
While forming the command you can use:
rename.ul -nv replace-me with-this in-all?-these-files*
for not doing any changes but reading what changes that command would make. When sure just reexecute the command without the -v (verbose) and -n (no-act) options
for your case the commands are:
rename.ul "" 000 ?.txt
rename.ul "" 00 ??.txt
rename.ul "" 0 ???.txt