In my folder, I want to delete the files that ends with 2 digits (10, 11, 12, etc.)
What I've tried is
rm [0-9]*
but it seems like it doesn't work.
What is the right syntax of doing this?
Converting comments into an answer.
Your requirement is a bit ambiguous. However, you can use:
rm -i *[0-9][0-9]
as long as you don't mind files ending with three digits being removed. If you do mind the three-digit files being removed, use:
rm -i *[!0-9][0-9][0-9]
(assuming Bash history expansion doesn't get in the way). Note that if you have file names consisting of just 2 digits, those will not be removed; that would require:
rm -i [0-9][0-9]
Caution!
The -i option is for interactive. It is generally a bad idea to experiment with globbing and rm commands because you can do a lot of damage if you get it wrong. However, you can use other techniques to neutralize the danger, such as:
echo *[!0-9][0-9]
which echoes all the file names, or:
printf '%s\n' *[!0-9][0-9]
which lists the file names one per line. Basically, be cautious when experimenting with file deletion — don't risk making a mistake unless you know you have good backups readily available. Even then, it is better not to need to use them
See also the GNU Bash manual on:
Pattern matching — which notes you might be able to use ^ in place of !.
The shopt built-in
The command
rm [0-9]*
means remove all the files that start with a digit. The range within [] is expanded single time.
Whereas you intend to remove files ending with double digit, so command should be
rm *[0-9][0-9]
If you have some file extension, the command should be modified as
rm *[0-9][0-9]* or
rm *[0-9][0-9].ext
where ext is the extension like txt
Related
I have found some similar questions here but not this specific one and I do not want to break all my files. I have a list of files and I simply need to replace all spaces with underscores. I know this is a sed command but I am not sure how to generically apply this to every file.
I do not want to rename the files, just modify them in place.
Edit: To clarify, just in case it's not clear, I only want to replace whitespace within the files, file names should not be changed.
find . -type f -exec sed -i -e 's/ /_/g' {} \;
find grabs all items in the directory (and subdirectories) that are files, and passes those filenames as arguments to the sed command using the {} \; notation. The sed command it appears you already understand.
if you only want to search the current directory, and ignore subdirectories, you can use
find . -maxdepth 1 -type f -exec sed -i -e 's/ /_/g' {} \;
This is a 2 part problem. Step 1 is providing the proper sed command, 2 is providing the proper command to replace all files in a given directory.
Substitution in sed commands follows the form s/ItemToReplace/ItemToReplaceWith/pattern, where s stands for the substitution and pattern stands for how the operation should take place. According to this super user post, in order to match whitespace characters you must use either \s or [[:space:]] in your sed command. The difference being the later is for POSIX compliance. Lastly you need to specify a global operation which is simply /g at the end. This simply replaces all spaces in a file with underscores.
Substitution in sed commands follows the form s/ItemToReplace/ItemToReplaceWith/pattern, where s stands for the substitution and pattern stands for how the operation should take place. According to this super user post, in order to match whitespace characters you must use either just a space in your sed command, \s, or [[:space:]]. The difference being the last 2 are for whitespace catching (tabs and spaces), with the last needed for POSIX compliance. Lastly you need to specify a global operation which is simply /g at the end.
Therefore, your sed command is
sed s/ /_/g FileNameHere
However this only accomplishes half of your task. You also need to be able to do this for every file within a directory. Unfortunately, wildcards won't save us in the sed command, as * > * would be ambiguous. Your only solution is to iterate through each file and overwrite them individually. For loops by default should come equipped with file iteration syntax, and when used with wildcards expands out to all files in a directory. However sed's used in this manner appear to completely lose output when redirecting to a file. To correct this, you must specify sed with the -i flag so it will edit its files. Whatever item you pass after the -i flag will be used to create a backup of the old files. If no extension is passed (-i '' for instance), no backup will be created.
Therefore the final command should simply be
for i in *;do sed -i '' 's/ /_/g' $i;done
Which looks for all files in your current directory and echos the sed output to all files (Directories do get listed but no action occurs with them).
Well... since I was trying to get something running I found a method that worked for me:
for file in `ls`; do sed -i 's/ /_/g' $file; done
I want to copy all files from /usr/lib which ends with .X.0.0 where X is an even number. Is there a better way than the following one to select all the files?
ls /usr/lib | grep "[02468].0.0$"
My problem with this solutions is that in case there are files with names like "xy.800.0.0" I need to use the bracket three times etc.
Just use a glob expansion to match the files:
cp /usr/lib/*.*[02468].0.0 /path/to/destination
The shell expands this pattern to the list of files before passing them as arguments to cp.
Since you tagged Bash, you can make the match more strict by using an extended glob:
shopt -s extglob failglob
cp /usr/lib/*.*([0-9])[02468].0.0 /path/to/destination
This matches 0 or more other digits followed by an even digit, and doesn't run the command at all if no files match.
You could use extended grep regular expressions to only match even numbers:
ls -1q /usr/lib | grep -E "\.[0-9]*[02468].0.0$"
However, as Tom suggested, there are better options than parsing the output of ls. It's generally safer and faster to use glob expansion, and more maintainable to just put everything in a python script.
I have a directory with many files with really long, repetitive names and I would like to remove the first 16 characters from each file name.
So I would like to rename files like this:
0123456789012345file1.fits
0123456789012345file2.fits
to this:
file1.fits
file2.fits
I would like to be able to do this from the command line in the terminal.
In bash, you can run
for f in *; do mv "$f" "${f:16}"; done
to rename all files stripping off the first 16 characters of the name.
You can change the * to a more restrictive pattern such as *.fits if you don't want to rename all files in the current directory. The quotes around the parameters to mv are necessary if any filenames contain whitespace.
bash's ${var:pos:len} syntax also supports more advanced usage than the above. You can take only the first five characters with ${f::5}, or the first five characters after removing the first 16 characters with ${f:16:5}. Many other variable substitution expressions are available in bash; see a reference such as TLDP's Bash Parameter Substitution for more information.
In a few words a wrote this little script to clean up some directories where I had consolidated directories/files from multiple sources where I used the cp command with the --backup=numbered feature so that files with identical names would have a suffix like .~1~ appended to avoid overwriting. I then ran fdupes to remove duplicate files, in some cases fdupes removed the file which did not have the suffix appended from the cp command (the original file) so I wanted to scan the directories looking for files with the suffix appended by the cp command and if the file does not exist with the suffix removed I would move mv the file otherwise I would leave it to avoid deleting anything as fdupes did not think it was a duplicate.
The issues is the test condition if [ -f ... ] part of the code below returns inverted results than what it should and I cannot understand why. For example, when the file exists it would return false and when the file did not exist it would return true. I fixed it by reversing the actions that I wanted to do based on the inverted return code and verified it was working as intended and it was so I ran it as such but would like to know if anyone knows why it would behave the way it did. I am not a bash script expert by any means so its possible that I missed something simple.
#!/bin/bash
logfile=$$.log
exec > $logfile 2>&1
IFS='
'
#set -f
for FILE in $(find . -type f -regextype posix-extended -regex '^.*(\.~[0-9]+~)+$')
do
FILE2=${FILE%%.~[0-9]*} # remove the suffix
if [ -f "${FILE2}" ]
then
echo ERROR: "${FILE2}" already exists!
else
echo "${FILE}" renamed "${FILE2}"
mv "${FILE}" "${FILE2}"
fi
done
You might be able to see the problem by modifying your script to show both FILE and FILE2 in the error message. There are a few minor problems with the script which could cause some confusion (but not the "inverted" logic):
find output is not sorted. If you had more than one backup file, a randomly chosen one would replace the original file;
you could sort the output using an expression like |sort -t~ -n -k2 on the end of the find-command.
the regular expression allows multiple matches of the ~[0-9]~ pattern. Conceivably you could have some odd file which ends with ~1~~2~.
the part where the suffix is removed assumes a single ~[0-9]~ is on the end of the filename. An embedded ~0, e.g., foo~0bar~1~ would reduce FILE to foo. The workaround for that would be more cumbersome (since the suffix-stripping uses globbing), but could be done with a case statement which matched an explicit number of digits (likely three digits would be enough).
I have this list of files on a Linux server:
abc.log.2012-03-14
abc.log.2012-03-27
abc.log.2012-03-28
abc.log.2012-03-29
abc.log.2012-03-30
abc.log.2012-04-02
abc.log.2012-04-04
abc.log.2012-04-05
abc.log.2012-04-09
abc.log.2012-04-10
I've been deleting selected log files one by one, using the command rm -rf see below:
rm -rf abc.log.2012-03-14
rm -rf abc.log.2012-03-27
rm -rf abc.log.2012-03-28
Is there another way, so that I can delete the selected files at once?
Bash supports all sorts of wildcards and expansions.
Your exact case would be handled by brace expansion, like so:
$ rm -rf abc.log.2012-03-{14,27,28}
The above would expand to a single command with all three arguments, and be equivalent to typing:
$ rm -rf abc.log.2012-03-14 abc.log.2012-03-27 abc.log.2012-03-28
It's important to note that this expansion is done by the shell, before rm is even loaded.
Use a wildcard (*) to match multiple files.
For example, the command below will delete all files with names beginning with abc.log.2012-03-.
rm -f abc.log.2012-03-*
I'd recommend running ls abc.log.2012-03-* to list the files so that you can see what you are going to delete before running the rm command.
For more details see the Bash man page on filename expansion.
If you want to delete all files whose names match a particular form, a wildcard (glob pattern) is the most straightforward solution. Some examples:
$ rm -f abc.log.* # Remove them all
$ rm -f abc.log.2012* # Remove all logs from 2012
$ rm -f abc.log.2012-0[123]* # Remove all files from the first quarter of 2012
Regular expressions are more powerful than wildcards; you can feed the output of grep to rm -f. For example, if some of the file names start with "abc.log" and some with "ABC.log", grep lets you do a case-insensitive match:
$ rm -f $(ls | grep -i '^abc\.log\.')
This will cause problems if any of the file names contain funny characters, including spaces. Be careful.
When I do this, I run the ls | grep ... command first and check that it produces the output I want -- especially if I'm using rm -f:
$ ls | grep -i '^abc\.log\.'
(check that the list is correct)
$ rm -f $(!!)
where !! expands to the previous command. Or I can type up-arrow or Ctrl-P and edit the previous line to add the rm -f command.
This assumes you're using the bash shell. Some other shells, particularly csh and tcsh and some older sh-derived shells, may not support the $(...) syntax. You can use the equivalent backtick syntax:
$ rm -f `ls | grep -i '^abc\.log\.'`
The $(...) syntax is easier to read, and if you're really ambitious it can be nested.
Finally, if the subset of files you want to delete can't be easily expressed with a regular expression, a trick I often use is to list the files to a temporary text file, then edit it:
$ ls > list
$ vi list # Use your favorite text editor
I can then edit the list file manually, leaving only the files I want to remove, and then:
$ rm -f $(<list)
or
$ rm -f `cat list`
(Again, this assumes none of the file names contain funny characters, particularly spaces.)
Or, when editing the list file, I can add rm -f to the beginning of each line and then:
$ . ./list
or
$ source ./list
Editing the file is also an opportunity to add quotes where necessary, for example changing rm -f foo bar to rm -f 'foo bar' .
Just use multiline selection in sublime to combine all of the files into a single line and add a space between each file name and then add rm at the beginning of the list. This is mostly useful when there isn't a pattern in the filenames you want to delete.
[$]> rm abc.log.2012-03-14 abc.log.2012-03-27 abc.log.2012-03-28 abc.log.2012-03-29 abc.log.2012-03-30 abc.log.2012-04-02 abc.log.2012-04-04 abc.log.2012-04-05 abc.log.2012-04-09 abc.log.2012-04-10
A wild card would work nicely for this, although to be safe it would be best to make the use of the wild card as minimal as possible, so something along the lines of this:
rm -rf abc.log.2012-*
Although from the looks of it, are those just single files? The recursive option should not be necessary if none of those items are directories, so best to not use that, just for safety.
I am not a linux guru, but I believe you want to pipe your list of output files to xargs rm -rf. I have used something like this in the past with good results. Test on a sample directory first!
EDIT - I might have misunderstood, based on the other answers that are appearing. If you can use wildcards, great. I assumed that your original list that you displayed was generated by a program to give you your "selection", so I thought piping to xargs would be the way to go.
if you want to delete all files that belong to a directory at once.
For example:
your Directory name is "log" and "log" directory include abc.log.2012-03-14, abc.log.2012-03-15,... etc files. You have to be above the log directory and:
rm -rf /log/*