How to batch-rename files using bash script?

Question

I have multiple files in one directory with below format, so I want to rename all like below:

From

27_07_2017file1vc001vpxd-1605.log
27_07_2017case2vc001vpxd-9169.log
27_07_2017server3vc001vpxd-4640.log
27_07_2017file24vc001vpxd-9170.log
27_07_2017files5vc001vpxd-4641.log

To

file1vc001vpxd-1605_27_07_2017_1.log
case2vc001vpxd-9169_27_07_2017_1.log
server3vc001vpxd-4640_27_07_2017_1.log
file24vc001vpxd-9170_27_07_2017_1.log
files5vc001vp-4641_27_07_2017_1.log

Need to change on format of files from datefilename.log to format filename.date_1.log

Please not to suggest using rename command, since it doesn't work in my linux server.

Thank you

Answer 1

Here is a bash script for that (revised for general case based on OP's revised question):

for file in ./*.log; do 
    echo mv "$file" "${file:12:$((${#file}-16))}_${file:2:10}_1.log";
done

Above the ${file:X:Y} is bash substring expansion syntax ${variable:offset:length} and get length of characters start from offset from its variable (or parameter) and rename with mv command. Just remove the echo once you ensure the result is expected as you want.

The result of above is as following:

mv ./27_07_2017case5.log case5_27_07_2017_1.log
mv ./27_07_2017file1vc001vpxd-1605.log file1vc001vpxd-1605_27_07_2017_1.log
mv ./27_07_2017file2vc001vpxd-9169.log file2vc001vpxd-9169_27_07_2017_1.log
mv ./27_07_2017file3vc001vpxd-4640.log file3vc001vpxd-4640_27_07_2017_1.log
mv ./27_07_2017file4vc001vpxd-9170.log file4vc001vpxd-9170_27_07_2017_1.log
mv ./27_07_2017file5.log file5_27_07_2017_1.log
mv ./27_07_2017file5vc001vpxd-4641.log file5vc001vpxd-4641_27_07_2017_1.log
mv ./27_07_2017number-blahblahblah5.log number-blahblahblah5_27_07_2017_1.log
mv ./27_07_2017number-blahblahblah5AAABBC.log number-blahblahblah5AAABBC_27_07_2017_1.log
mv ./27_07_2017number5.log number5_27_07_2017_1.log

Answer 2

In your case there are two components: filename and date

list all log files with recursive search

find "$(pwd)" -type f -iname "*.log" > loglist

list all log files without recurse to sub-directories

find "$(pwd)" -maxdepth 0 -type f -iname "*.log" > loglist

Now, using while loop you can move files to the desired format

while read f;
do
    # filename only and not complete path
    fname_only=$(basename "$f");
    
    # directory name only without filename
    dirname_only=$(dirname "$f");
    
    # use cut command to extract first 10 characters as date from filename
    date_only=$(echo "$fname_only" | cut -c 1-10);

    # use cut command to extract 11th character onward as new filename with extension
    fname_extract=$(echo "$fname_only" | cut -c 11-);

    # filename with complete path without extension 
    fname_no_extns="${fname_extract%.*}"

    # Now, move files to their desired format recursively
    mv "$f" "${dirname_only}/${fname_no_extns}.${date_only}_1.log"

done < loglist

Answer 3

I don't know all possible cases for filenames, so I'll show you a general approach you can adapt to your needs.

You can loop through your files with a for loop or use find -exec as you like. I think you mainly want to know how to cut your filename to pieces and set them together differently.

You can either pipe the name into a tool like sed and create a regular expression to identify the pieces you need:

newname=`echo "$oldname"|sed -E 's/([0-9-]{10})(.*).log/\2_\1_1.log/'`

If you place () around those parts you can refer to them as \1, \2 and so on in the replacement string. If you need to adapt it, read the man page of sed or ask here if you don't succeed.

The other possibility is to use POSIX variable expansion:

You probably know that

${oldname#*201?}

will remove the date prefix of your filename (the ? can match any character, so this works for years 2010 to 2019, if that's sufficient. Otherwise adapt.)

You can also combine them:

${oldname%${oldname#*201?}}

Will remove the part we extracted above, so this will give you the date only. This way you can separate the parts of the filename and paste them in a different order.

Answer 4

The obligatory zsh + zmv answer:

autoload -Uz zmv
zmv -n '(<1-31>_<1-12>-<1900-2999>)(*).log' '$2.${1}_1.log'

Remove -n (for dry-run) if happy.