How to write shell script to create zip file for the files that had same string in file name - linux

Question

How to write simple shell script to create zip file.
I want to create zip file by collecting files with same string pattern in their file names from a folder.
For example, there may be many files under a folder.
xxxxx_20140502_xxx.txt
xxxxx_20140502_xxx.txt
xxxxx_20140503_xxx.txt
xxxxx_20140503_xxx.txt
xxxxx_20140504_xxx.txt
xxxxx_20140504_xxx.txt
After running the shell script, the result must be following three zip files.
20140502.zip
20140503.zip
20140504.zip
Please give me right direction to create simple shell script to output the result as above.

Answer 1

#!/bin/bash
for file in *_????????_*.csv *_????????_*.txt; do
[ -f "${file}" ] || continue
date=${file#*_} # adjust this and next line depending
date=${date%_*} # on your actual prefix/suffix
echo "${date}"
done | sort -u | while read date; do
zip "${date}.zip" *${date}*
done

Answer 2

Since zip will update the archive, this will do:
shopt -s nullglob
for file in *.{txt,csv}; do [[ $file =~ _([[:digit:]]{8})_ ]] && zip "${BASH_REMATCH[1]}.zip" "$file"; done
The shopt -s nullglob is because you don't want to have unexpanded globs if there are no matching files.
Everything below this line is my old answer...
First, get all the possible dates. Heuristically, this could be the files ending in .txt and .csv that match the regex _[[:digit:]]{8}_:
#!/bin/bash
shopt -s nullglob
declare -A dates=()
for file in *.{csv,txt}; do
[[ $file =~ _([[:digit:]]{8})_ ]] && dates[${BASH_REMATCH[1]}]=
done
printf "Date found: %s\n" "${!dates[#]}"
This will output to stdout all the dates found in the files. E.g. (I called the previous snipped gorilla and I chmod +x gorilla and touched a few files for demo):
$ ls
banana_20010101_gorilla.csv gorilla_20140502_bonobo.csv
gorilla notthisone_123_lol.txt
gorilla_20140502_banana.txt
$ ./gorilla
Date found: 20140502
Date found: 20010101
Next step, for each date found, get all the files ending in .txt and .csv and zip them in the archive corresponding to the date: appending this to gorilla will do the job:
for date in "${!dates[#]}"; do
zip "$date.zip" *"_${date}_"*.{csv,txt}
done
Full script after removing the flooding part:
#!/bin/bash
shopt -s nullglob
declare -A dates=()
for file in *.{csv,txt}; do
[[ $file =~ _([[:digit:]]{8})_ ]] && dates[${BASH_REMATCH[1]}]=
done
for date in "${!dates[#]}"; do
zip "$date.zip" *"_${date}_"*.{csv,txt}
done
Edit. I overlooked your requirement with one line command. Then here's the one-liner:
shopt -s nullglob; declare -A dates=(); for file in *.{csv,txt}; do [[ $file =~ _([[:digit:]]{8})_ ]] && dates[${BASH_REMATCH[1]}]=; done; for date in "${!dates[#]}"; do zip "$date.zip" *"_${date}_"*.{csv,txt}; done
:)

Answer 3

#! /bin/bash
dates=$(ls ?????_[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]_???.{csv,txt} \
| cut -f2 -d_ | sort -u)
for date in $dates ; do
zip $date.zip ?????_"$date"_???.{csv,txt}
done