I want to rename all files in a directory to be sequential numbers:
1.txt
2.txt
3.txt
and so on...
Here's the code I'm currently using:
ls | cat -n | while read n f; do mv "$f" "$n.txt"; done
The code does work, but I need to start with a specific number. For example, I may want to start with the number 49 instead of the number 1.
Is there any way to do this in terminal (on a Mac)?
You could use something like nl with the -v option to set a starting line number other than 1, but instead, you can just use Bash features:
i=1
for f in *; do
[[ -f $f ]] && mv "$f" $((i++)).txt
done
where i is set to the initial value you want.
This also avoids parsing the output of ls, which is recommended to avoid. Instead, I use a glob (*) and a test (-f) to make sure that I'm actually manipulating files and not directories.
Related
How do i print out all the files of the current directory that start with the letter "k" ?Also needs to count this files.
I tried some methods but i only got errors or wrong outputs. Really stuck on this as a newbie in bash.
Try this Shellcheck-clean pure POSIX shell code:
count=0
for file in k*; do
if [ -f "$file" ]; then
printf '%s\n' "$file"
count=$((count+1))
fi
done
printf 'count=%d\n' "$count"
It works correctly (just prints count=0) when run in a directory that contains nothing starting with 'k'.
It doesn't count directories or other non-files (e.g. fifos).
It counts symlinks to files, but not broken symlinks or symlinks to non-files.
It works with 'bash' and 'dash', and should work with any POSIX-compliant shell.
Here is a pure Bash solution.
files=(k*)
printf "%s\n" "${files[#]}"
echo "${#files[#]} files total"
The shell expands the wildcard k* into the array, thus populating it with a list of matching files. We then print out the array's elements, and their count.
The use of an array avoids the various problems with metacharacters in file names (see e.g. https://mywiki.wooledge.org/BashFAQ/020), though the syntax is slightly hard on the eyes.
As remarked by pjh, this will include any matching directories in the count, and fail in odd ways if there are no matches (unless you set nullglob to true). If avoiding directories is important, you basically have to get the directories into a separate array and exclude those.
To repeat what Dominique also said, avoid parsing ls output.
Demo of this and various other candidate solutions:
https://ideone.com/XxwTxB
To start with: never parse the output of the ls command, but use find instead.
As find basically goes through all subdirectories, you might need to limit that, using the -maxdepth switch, use value 1.
In order to count a number of results, you just count the number of lines in your output (in case your output is shown as one piece of output per line, which is the case of the find command). Counting a number of lines is done using the wc -l command.
So, this comes down to the following command:
find ./ -maxdepth 1 -type f -name "k*" | wc -l
Have fun!
This should work as well:
VAR="k"
COUNT=$(ls -p ${VAR}* | grep -v ":" | wc -w)
echo -e "Total number of files: ${COUNT}\n" 1>&2
echo -e "Files,that begin with ${VAR} are:\n$(ls -p ${VAR}* | grep -v ":" )" 1>&2
I have 20 files like:
01a_AAA_qwe.sh
01b_AAA_asd.sh
01c_AAA_zxc.sh
01d_AAA_rty.sh
...
Files have a similar format in their names. They begin with 01, and they have 01*AAA*.sh format.
I wish to copy and rename files in the same directory, changing the number 01 to 02, 03, 04, and 05:
02a_AAA_qwe.sh
02b_AAA_asd.sh
02c_AAA_zxc.sh
02d_AAA_rty.sh
...
03a_AAA_qwe.sh
03b_AAA_asd.sh
03c_AAA_zxc.sh
03d_AAA_rty.sh
...
04a_AAA_qwe.sh
04b_AAA_asd.sh
04c_AAA_zxc.sh
04d_AAA_rty.sh
...
05a_AAA_qwe.sh
05b_AAA_asd.sh
05c_AAA_zxc.sh
05d_AAA_rty.sh
...
I wish to copy 20 of 01*.sh files to 02*.sh, 03*.sh, and 04*.sh. This will make the total number of files to 100 in the folder.
I'm really not sure how can I achieve this. I was trying to use for loop in the bash script. But not even sure what should I need to select as a for loop index.
for i in {1..4}; do
cp 0${i}*.sh 0${i+1}*.sh
done
does not work.
There are going to be a lot of ways to slice-n-dice this one ...
One idea using a for loop, printf + brace expansion, and xargs:
for f in 01*.sh
do
printf "%s\n" {02..05} | xargs -r -I PFX cp ${f} PFX${f:2}
done
The same thing but saving the printf in a variable up front:
printf -v prefixes "%s\n" {02..05}
for f in 01*.sh
do
<<< "${prefixes}" xargs -r -I PFX cp ${f} PFX${f:2}
done
Another idea using a pair of for loops:
for f in 01*.sh
do
for i in {02..05}
do
cp "${f}" "${i}${f:2}"
done
done
Starting with:
$ ls -1 0*.sh
01a_AAA_qwe.sh
01b_AAA_asd.sh
01c_AAA_zxc.sh
01d_AAA_rty.sh
All of the proposed code snippets leave us with:
$ ls -1 0*.sh
01a_AAA_qwe.sh
01b_AAA_asd.sh
01c_AAA_zxc.sh
01d_AAA_rty.sh
02a_AAA_qwe.sh
02b_AAA_asd.sh
02c_AAA_zxc.sh
02d_AAA_rty.sh
03a_AAA_qwe.sh
03b_AAA_asd.sh
03c_AAA_zxc.sh
03d_AAA_rty.sh
04a_AAA_qwe.sh
04b_AAA_asd.sh
04c_AAA_zxc.sh
04d_AAA_rty.sh
05a_AAA_qwe.sh
05b_AAA_asd.sh
05c_AAA_zxc.sh
05d_AAA_rty.sh
NOTE: blank lines added for readability
You can't do multiple copies in a single cp command, except when copying a bunch of files to a single target directory. cp will not do the name mapping automatically. Wildcards are expanded by the shell, they're not seen by the commands themselves, so it's not possible for them to do pattern matching like this.
To add 1 to a variable, use $((i+1)).
You can use the shell substring expansion operator to get the part of the filename after the first two characters.
for i in {1..4}; do
for file in 0${i}*.sh; do
fileend=${file:2}
cp "$file" "0$((i+1))$fileend"
done
done
I've a certain amount of files always containing same name but different extensions, for example sample.dat, sample.txt, etc.
I would like to create a script that looks where sample.dat is present and than moves all files with name sample*.* into another directory.
I know how to identify them with ls *.dat | sed 's/\(.*\)\..*/\1/', however I would like to concatenate with something like || mv (the result of the first part) *.* /otherdirectory/
You can use this bash one-liner:
for f in `ls | grep YOUR_PATTERN`; do mv ${f} NEW_DESTINATION_DIRECTORY/${f}; done
It iterates through the result of the operation ls | grep, which is the list of your files you wish to move, and then it moves each file to the new destination.
Something simple like this?
dat_roots=$(ls *.dat | sed 's/\.dat$//')
for i in $dat_roots; do
echo mv ${i}*.* other-directory
done
This will break for file names containing spaces, so be careful.
Or if spaces are an issue, this will do the job, but is less readable.
ls *.dat | sed 's/\.dat$//' | while read root; do
mv "${root}"*.* other-directory
done
Not tested, but this should do the job:
shopt -s nullglob
for f in *.dat
do
mv ${f%.dat}.* other-directory
done
Setting the nullglob option ensures that the loob is not executed, if no dat-file exists. If you use this code as part of a larger script, you might want to unset it afterwards (shopt -u nullglob).
I have duplicated files in directory on Linux machine which are listed like that:
ltulikowski#lukasz-pc:~$ ls -1
abcd
abcd.1
abcd.2
abdc
abdc.1
acbd
I want to remove all files witch aren't single so as a result I should have:
ltulikowski#lukasz-pc:~$ ls -1
acbd
The function uses extglob, so before execution, set extglob: shopt -s extglob
rm_if_dup_exist(){
arr=()
for file in *.+([0-9]);do
base=${file%.*};
if [[ -e $base ]]; then
arr+=("$base" "$file")
fi
done
rm -f -- "${arr[#]}"
}
This will also support file names with several digits after the . e.g. abcd.250 is also acceptable.
Usage example with your input:
$ touch abcd abcd.1 abcd.2 abdc abdc.1 acbd
$ rm_if_dup_exist
$ ls
acbd
Please notice that if, for example, abcd.1 exist but abcd does not exist, it won't delete abcd.1.
here is one way to do it
for f in *.[0-9]; do rm ${f%.*}*; done
may get exceptions since some files will be deleted more than once (abcd in your example). If versions always start with .1 you can restrict to match to that.
You can use:
while read -r f; do
rm "$f"*
done < <(printf "%s\n" * | cut -d. -f1 | uniq -d)
printf, cut and uniq are used to get duplicate entries (part before dot) in current directory.
The command
rm *.*
Should do the trick if I understand you correctly
Use ls to confirm first
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