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
Related
How to replace only date part to current date of all files present in diretory in unix.
Folder path: C:/shan
Sample files:
CN_Apria_837p_20180924.txt
DN_Apria_837p_20150502.txt
GN_Apria_837p_20160502.txt
CH_Apria_837p_20170502.txt
CU_Apria_837p_20180502.txt
PN_Apria_837p_20140502.txt
CN_Apria_837p_20101502.txt
Desired result should be:
CN_Apria_837p_20190502.txt
DN_Apria_837p_20190502.txt
GN_Apria_837p_20190502.txt
CH_Apria_837p_20190502.txt
CU_Apria_837p_20190502.txt
PN_Apria_837p_20190502.txt
CN_Apria_837p_20190502.txt
Edit:
I'm completely new to unix sell scripting. I tried this below, however it's not working.
#!/bin/bash
for i in ls $1 | grep -E '[0-9]{4}-[0-9]{2}-[0-9]{2}'
do
x=echo $i | grep -oE '[0-9]{4}-[0-9]{2}-[0-9]{2}'
y=echo $i | sed "s/$x/$(date +%F)/g"
mv $1/$i $1/$y 2>/dev/null #incase if old date is same as current date
done
I would use regular expressions here. From the bash man-page:
An additional binary operator, =~, is available, with the same
precedence as == and !=. When it is used, the string to the right
of the operator is considered an extended regular expression and
matched accordingly (as in regex(3)). The return value is 0 if the
string matches the pattern, and 1 otherwise. .... Substrings
matched by parenthesized subexpressions within the regular
expression are saved in the array variable BASH_REMATCH. ...
The element of BASH_REMATCH with indexn is the portion of the
string matching the nth parenthesized sub-expression.
Hence, assuming that the variable x holds the name of one of the files
in question, the code
if [[ $x =~ ^(.*_)[0-9]+([.]txt$) ]]
then
mv "$x" "$BASH_REMATCH[1]$(date +%Y%m%d)$BASH_REMATCH[2]"
fi
first tests roughly whether the file indeed follows the required naming scheme, and then modifies the name accordingly.
Of course in practice, you will tailor the regexp to match your application better. Only you can know what variations in the file name are permitted.
The below should do this
for f in $(find /path/to/files -name "*_*_*_*.txt")
do
newname=$(echo "$f" | sed -r "s/[12][0-9]{3}[01][0-9][0-3][0-9]/$(date '+%Y%m%d')/g")
mv "$f" "$newname"
done
Try this Shellcheck-clean code:
#! /bin/bash -p
readonly dir=$1
shopt -s nullglob # Make glob patterns that match nothing expand to nothing
readonly dateglob='20[0-9][0-9][0-9][0-9][0-9][0-9]'
currdate=$(date '+%Y%m%d')
# shellcheck disable=SC2231
for path in "$dir"/*_${dateglob}.* ; do
name=${path##*/}
newname=${name/_${dateglob}./_${currdate}.}
if [[ $newname != "$name" ]] ; then
newpath="$dir/$newname"
printf "%q -> %q\\n" "$path" "$newpath"
mv -i -- "$path" "$newpath"
fi
done
shopt -s nullglob stops the code trying to process a garbage path if nothing matches the glob pattern in for path in ....
The pattern assigned to dateglob assumes that you will not have to process dates before 2000 (or after 2099!). Change it if that assumption is not valid.
The # shellcheck ... line is to prevent Shellcheck warning about the use of ${dateglob} without quotes. The quotes would be wrong in this case because they would prevent the glob pattern being expanded.
The pattern used to match filenames (*_${dateglob}.*) will match many more forms of filename than the examples given (e.g. A_20180313.tar.gz). You might want to change it.
See Removing part of a string (BashFAQ/100 (How do I do string manipulation in bash?)) for information about the Bash string manipulation mechanisms used (${path##...}, ${name/...}).
I've added a printf to output details of what is being moved.
The -i option to mv prompts for confirmation if a file would be overwritten. This turns out to be an issue for the example files because both CN_Apria_837p_20180924.txt and CN_Apria_837p_20101502.txt are identical except for the date, so the code tries to rename them to the same thing.
If any of the files with dates in their names have names beginning with '.', the code will not process them. Add line shopt -s dotglob somewhere before the loop if that is an issue.
I want to group the directory name with their files in bash script.
For example if I type ls /home/maindir/*
I get home/maindir/dir1: file1 file2\n file3
home/maindir/dir2: file1 file2
The directories with files are not separated by a specified delimiter because there are cases that file1 and file2, in the same directory, have a newline beetween them, so I want to tokenize with a delimiter the directory name and its file list all on one line.
Example output with newline delimiter:
home/maindir/dir1: file1 file2 file3\n
home/maindir/dir2: file1 file2\n
home/maindir/dir3: file1 file2 file4\n
I originally used an unquoted interpolation trick.
For example, if you have strings in a file, one per line, and you want them horizontalized, you don't have to use paste -
file named foo:
a
b
c
then you can say:
echo $(<foo)
and you get
a b c
But that could cause issues with filenames, especially if they have embedded special chars or whitespace.
Thanks to Gordon Davisson for a simple upgrade!
for d in /home/maindir/* # includes full path each time
do [[ -d "$d" ]] || continue # ignore nondirectories
cd "$d" # go there to make filenames path-bare
echo "$d:" *
done
Note that this still includes subdirectories. Do you need to skip those?
If you want to be more careful -
for d in /charter/apps/*
do [[ -d "$d" ]] || continue
cd "$d"
dir="$d: "
hit=0
for f in *
do if [[ -f "$f" ]]
then hit=1
dir="$dir $f "
fi
done
(( $hit )) && printf "$dir\n"
done
This one should also work on files with embedded spaces &c.
I have thousands of mp3 files but all with unusual file names such as 1-2songone.mp3, 2songtwo.mp3, 2_2_3_songthree.mp3. I want to remove all the numbers, dashes and underscores in the beginning of these files and get the result:
songone.mp3
songtwo.mp3
songthree.mp3
This can be done using extended globbing:
$ ls
1-2songone.mp3 2_2_3_songthree.mp3 2songtwo.mp3
$ shopt -s extglob
$ for fname in *.mp3; do mv -- "$fname" "${fname##*([-_[:digit:]])}"; done
$ ls
songone.mp3 songthree.mp3 songtwo.mp3
This uses parameter expansion: ${fname##pattern} removes the longest possible match from the beginning of fname. As the pattern, we use *([-_[:digit:]]), where *(pattern) stands for "zero or more matches of pattern", and the actual pattern is a bracket expression for hyhpens, underscores and digits.
Remarks:
The -- after mv indicates the end of options for move and makes sure that filenames starting with - aren't interpreted as options.
The *() expression requires the extglob shell option. As pointed out, if you don't want extended globs later, you have to unset it again with shopt -u extglob.
As per Gordon Davisson's comment: this will clobber files if you have, for example, something like 1file.mp3 and 2file.mp3. To avoid that, you can either use mv -i (or --interactive), which will prompt you before overwriting a file, or mv -n (or --noclobber), which will just not overwrite any files.
triplee points out that this needlessly moves files onto themselves if they don't start with slash, underscore or digit. To avoid that, we can iterate only over matching files with
for fname in [-_[:digit:]]*.mp3; do mv -- "$fname" "${fname##*([-_[:digit:]])}"; done
which makes sure that there is something to rename.
Benjamin W.'s answer is helpful and efficient, but has two drawbacks:
It requires setting global shell option extglob, which should be restored to its previous value afterward (the alternative, at the cost of creating an extra process, is to use a subshell: (shopt -s extglob; for fname ...)).
The extglob syntax, an extension to regular glob syntax, is familiar to few people and still less powerful than true regular expressions.
Using Bash's regex-matching operator, =~:
for f in *.mp3; do [[ $f =~ ^[0-9_-]+(.+)$ ]] && echo mv "$f" "${BASH_REMATCH[1]}"; done
Remove the echo to perform actual renaming.
$f =~ ^[0-9_-]+(.+)$ matches the longest nonempty sequence of digits, hyphens, and underscores at the start of the filename, followed by any nonempty sequence of characters captured in a parenthesized subexpression (capture group).
If the match succeeds (&&), the mv command is invoked, with the captured subexpression - accessible via element 1 of special BASH array variable ${BASH_REMATCH[#]} - forming the target filename.
You may do it this way too :
find . -type f -name "*.mp3" -print0 | while read -r -d '' line
do
mv "$line" "$( sed -E 's!(.*)/[^[:alpha:]]*([[:alpha:]].*mp3)$!\1/\2!' <<<"$line")" 2>/dev/null
done
Using sed gives you more control over the regex, I guess. Also, the 2>/dev/null is for ignoring the mv error for already converted/correct filenames.
Note:
This will recursively change the filenames across subfolders too.
I am looking for a way to make a simple loop in bash over everything my directory contains, i.e. files, directories and links including hidden ones.
I will prefer if it could be specifically in bash but it has to be the most general. Of course, file names (and directory names) can have white space, break line, symbols. Everything but "/" and ASCII NULL (0×0), even at the first character. Also, the result should exclude the '.' and '..' directories.
Here is a generator of files on which the loop has to deal with :
#!/bin/bash
mkdir -p test
cd test
touch A 1 ! "hello world" \$\"sym.dat .hidden " start with space" $'\n start with a newline'
mkdir -p ". hidden with space" $'My Personal\nDirectory'
So my loop should look like (but has to deal with the tricky stuff above):
for i in * ;
echo ">$i<"
done
My closest try was the use of ls and bash array, but it is not working with, is:
IFS=$(echo -en "\n\b")
l=( $(ls -A .) )
for i in ${l[#]} ; do
echo ">$i<"
done
unset IFS
Or using bash arrays but the ".." directory is not exclude:
IFS=$(echo -en "\n\b")
l=( [[:print:]]* .[[:print:]]* )
for i in ${l[#]} ; do
echo ">$i<"
done
unset IFS
* doesn't match files beginning with ., so you just need to be explicit:
for i in * .[^.]*; do
echo ">$i<"
done
.[^.]* will match all files and directories starting with ., followed by a non-. character, followed by zero or more characters. In other words, it's like the simpler .*, but excludes . and ... If you need to match something like ..foo, then you might add ..?* to the list of patterns.
As chepner noted in the comments below, this solution assumes you're running GNU bash along with GNU find GNU sort...
GNU find can be prevented from recursing into subdirectories with the -maxdepth option. Then use -print0 to end every filename with a 0x00 byte instead of the newline you'd usually get from -print.
The sort -z sorts the filenames between the 0x00 bytes.
Then, you can use sed to get rid of the dot and dot-dot directory entries (although GNU find seems to exclude the .. already).
I also used sed to get read of the ./ in front of every filename. basename could do that too, but older systems didn't have basename, and you might not trust it to handle the funky characters right.
(These sed commands each required two cases: one for a pattern at the start of the string, and one for the pattern between 0x00 bytes. These were so ugly I split them out into separate functions.)
The read command doesn't have a -z or -0 option like some commands, but you can fake it with -d "" and blanking the IFS environment variable.
The additional -r option prevents a backslash-newline combo from being interpreted as a line continuation. (A file called backslash\\nnewline would otherwise be mangled to backslashnewline.) It might be worth seeing if other backslash-combos get interpreted as escape sequences.
remove_dot_and_dotdot_dirs()
{
sed \
-e 's/^[.]\{1,2\}\x00//' \
-e 's/\x00[.]\{1,2\}\x00/\x00/g'
}
remove_leading_dotslash()
{
sed \
-e 's/^[.]\///' \
-e 's/\x00[.]\//\x00/g'
}
IFS=""
find . -maxdepth 1 -print0 |
sort -z |
remove_dot_and_dotdot_dirs |
remove_leading_dotslash |
while read -r -d "" filename
do
echo "Doing something with file '${filename}'..."
done
It may not be the most favorable way but I tried bellow thing
while read line ; do echo $line; done <<< $(ls -a | grep -v -w ".")
check the below trail which I did
Try the find command, something like:
find .
That will list all the files in all recursive directories.
To output only files excluding the leading . or .. try:
find . -type f -printf %P\\n
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