I'm writing a deployment script that saves timestamped backup files to a backups directory. I'd like to do a rollback implementation that would roll back to the most recent file.
My backups directory:
$:ls
. 1341094065_public_html_bu 1341094788_public_html_bu
.. 1341094390_public_html_bu
1341093920_public_html_bu 1341094555_public_html_bu
I want to identify the most recent file (by timestamp in the filename) in the backup directory, and save its name to a variable, then cp it to ../public_html, and so on...
ls -t will sort files by mtime. ls -t | head -n1 will select the newest file. This is independent of any naming scheme you have, which may or may not be a plus.
...and a more "correct" way, which won't break when filenames contain newlines, and also not when there are no matching files (unexpanded glob results)
for newestfile in ./* ; do : ; done
if test -e "$newestfile"; then do something with "$newestfile" ; fi
The latest-timestamped filename should sort last alphabetically. So you can then use tail -n1 to extract it.
For files that don't have newlines in their names:
shopt -s nullglob
printf '%s\n' "$buDir"/* | tail -n 1
Related
I have 600,000+ images in a directory. The filenames look like this:
1000000-0.jpeg
1000000-1.jpeg
1000000-2.jpeg
1000001-0.jpeg
1000002-0.jpeg
1000003-0.jpeg
The first number is a unique ID and the second number is an index.
{unique-id}-{index}.jpeg
How would I load the unique-id's in from a .CSV file and remove each file whose Unique ID matches the Unique ID's in the .CSV file?
The CSV file looks like this:
1000000
1000001
1000002
... or I can have it separated by semicolons like so (if necessary):
1000000;1000001;1000002
You can set the IFS variable to ; and loop over the values read into an array:
#! /bin/bash
while IFS=';' read -a ids ; do
for id in "${ids[#]}" ; do
rm $id-*.jpg
done
done < file.csv
Try running the script with echo rm ... first to verify it does what you want.
If there's exactly one ID per line, this will show you all matching file names:
ls | grep -f unique-ids.csv
If that list looks correct, you can delete the files with:
ls | grep -f unique-ids.csv | xargs rm
Caveat: This is a quick and dirty solution. It'll work if the file names are all named the way you say. Beware it could easily be tricked into deleting the wrong things by a clever attacker or a particularly hapless user.
You could use find and sed:
find dir -regextype posix-egrep \
-regex ".*($(sed 's/\;/|/g' ids.csv))-[0-9][0-9]*\.jpeg"
replace dir with your search directory, and ids.csv with your CVS file. To delete the files you could include -delete option.
I have a few files that I want to copy and rename with the new file names generated by adding a fixed string to each of them.
E.g:
ls -ltr | tail -3
games.txt
files.sh
system.pl
Output should be:
games_my.txt
files_my.sh
system_my.pl
I am able to append at the end of file names but not before *.txt.
for i in `ls -ltr | tail -10`; do cp $i `echo $i\_my`;done
I am thinking if I am able to save the extension of each file by a simple cut as follows,
ext=cut -d'.' -f2
then I can append the same in the above for loop.
do cp $i `echo $i$ext\_my`;done
How do I achieve this?
You can use the following:
for file in *
do
name="${file%.*}"
extension="${file##*.}"
cp $file ${name}_my${extension}
done
Note that ${file%.*} returns the file name without extension, so that from hello.txt you get hello. By doing ${file%.*}_my.txt you then get from hello.txt -> hello_my.txt.
Regarding the extension, extension="${file##*.}" gets it. It is based on the question Extract filename and extension in bash.
If the shell variable expansion mechanisms provided by fedorqui's answer look too unreadable to you, you also can use the unix tool basename with a second argument to strip off the suffix:
for file in *.txt
do
cp -i "$file" "$(basename "$file" .txt)_my.txt"
done
Btw, in such cases I always propose to apply the -i option for cp to prevent any unwanted overwrites due to typing errors or similar.
It's also possible to use a direct replacement with shell methods:
cp -i "$file" "${file/.txt/_my.txt}"
The ways are numerous :)
I have a bunch of files in Unix Directory :
test_XXXXX.txt
best_YYY.txt
nest_ZZZZZZZZZ.txt
I need to rename these files as
test.txt
best.txt
nest.txt
I am using Ksh on AIX .Please let me know how i can accomplish the above using a Single command .
Thanks,
In this case, it seems you have an _ to start every section you want to remove. If that's the case, then this ought to work:
for f in *.txt
do
g="${f%%_*}.txt"
echo mv "${f}" "${g}"
done
Remove the echo if the output seems correct, or replace the last line with done | ksh.
If the files aren't all .txt files, this is a little more general:
for f in *
do
ext="${f##*.}"
g="${f%%_*}.${ext}"
echo mv "${f}" "${g}"
done
If this is a one time (or not very often) occasion, I would create a script with
$ ls > rename.sh
$ vi rename.sh
:%s/\(.*\)/mv \1 \1/
(edit manually to remove all the XXXXX from the second file names)
:x
$ source rename.sh
If this need occurs frequently, I would need more insight into what XXXXX, YYY, and ZZZZZZZZZZZ are.
Addendum
Modify this to your liking:
ls | sed "{s/\(.*\)\(............\)\.txt$/mv \1\2.txt \1.txt/}" | sh
It transforms filenames by omitting 12 characters before .txt and passing the resulting mv command to a shell.
Beware: If there are non-matching filenames, it executes the filename—and not a mv command. I omitted a way to select only matching filenames.
Hey, guys, I used zip command, but I only want to archive all the files except *.txt. For example, if two dirs file1, file2; both of them have some *.txt files. I want archive only the non-text ones from file1 and file2.
tl;dr: How to tell linux to give me all the files that don't match *.txt
$ zip -r zipfile -x'*.txt' folder1 folder2 ...
Move to you desired directory and run:
ls | grep -P '\.(?!txt$)' | zip -# zipname
This will create a zipname.zip file containing everything but .txt files. In short, what it does is:
List all files in the directory, one per line (this can be achieved by using the -1 option, however it is not needed here as it's the default when output is not the terminal, it is a pipe in this case).
Extract from that all lines that do not end in .txt. Note it's grep using a Perl regular expression (option -P) so the negative lookahead can be used.
Zip the list from stdin (-#) into zipname file.
Update
The first method I posted fails with files with two ., like I described in the comments. For some reason though, I forgot about the -v option for grep which prints only what doesn't match the regex. Plus, go ahead and include a case insensitive option.
ls | grep -vi '\.txt$' | zip -# zipname
Simple, use bash's Extended Glob option like so:
#!/bin/bash
shopt -s extglob
zip -some -options !(*.txt)
Edit
This isn't as good as the -x builtin option to zip but my solution is generic across any command that may not have this nice feature.
I have a directory where our deployments go. A deployment (which is itself a directory) is named in the format:
<application-name>_<date>
e.g. trader-gui_20091102
There are multiple applications deployed to this same parent directory, so the contents of the parent directory might look something like this:
trader-gui_20091106
trader-gui_20091102
trader-gui_20091010
simulator_20091106
simulator_20091102
simulator_20090910
simulator_20090820
I want to write a bash script to clean out all deployments except for the most current of each application. (The most current denoted by the date in the name of the deployment). So running the bash script on the above parent directory would leave:
trader-gui_20091106
simulator_20091106
Any help would be appreciated.
A quick one-liner:
ls | sed 's/_[0-9]\{8\}$//' | uniq |
while read name; do
rm $(ls -r ${name}* | tail -n +2)
done
List the files, chop off an underscore followed by eight digits, only keep unique names. For each name, remove everything but the most recent.
Assumptions:
the most recent will be last when sorted alphabetically. If that's not the case, add a sort that does what you want in the pipeline before tail -n +2
no other files in this directory. If there are, limit the output of the ls, or pipe it through a grep to select only what you want.
no weird characters in the filenames. If there are... instead of directly using the output of the inner ls pipeline, you'd probably want to pipe it into another while loop so you can quote the individual lines, or else capture it in an array so you can use the quoted expansion.
shopt -s exglob
ls|awk -F"_" '{a[$1]=$NF}END{for(i in a)print i"_"a[i]}'|while read -r F
do
rm !($F)
done
since your date in filename is already "sortable" , the awk command finds the latest file of each application. rm (!$F) just means remove those filename that is not latest.
You could try find:
# Example: Find and delete all directories in /tmp/ older than 7 days:
find /tmp/ -type d -mtime +7 -exec rm -rf {} \; &>/dev/null