Select and zip files according to CSV input - linux

Question

I'm trying to read each column inside a csv file, and then search a directory using the data from the columns on the csv. Once all rows have been read, I'd like to create a new directory, copy the files to that directory, and finally zip the directory. Any ideas on how to do it?
Here's an example of the CSV file:
add an example here...

Answer 1

This is a draft of the script based on the informations you gave:
destdir="/path/to/destination/directory"
filelist=""
while IFS=, read col1 col2 col3 col4 ...
do
# check if the file should be included in filelist
if [ ... ]; then
filelist+="$col1/$col2/$col3.$col4 "
fi
if [ ! -z "$filelist" ]; then
mkdir -p $destdir
for goodfile in $filelist; do
cp $goodfile $destdir
done
fi
done < file.csv
To make things clearer I assumed the columns in the CSV file are just the parts of the path and name of the files you're interested in.
Hope this helps. :)

Answer 2

I don't know, if you would like to process every column in a different way, but here is a scratch of a script, if it does not matter.
separator=";" # select separator type in your csv file
csv_file="`pwd`/file.csv" # path to your csv file
search_dir="`pwd`/searchdir" # directory, where we search for files, that are in csv files
copy_to_dir="`pwd`/newdir" #directory where to save copied files
mkdir $copy_to_dir #making copy dir
for i in `cat ${csv_file} | tr $separator '\n'` ; do # reading all the entries in csv file
find $search_dir -name "$i" | xargs -i cp {} $copy_to_dir
# assuming, that all entries in csv are files to be found in a dir and copied
done
zip -j -r copied_files.zip $copy_to_dir # zip found contents
rmdir $copy_to_dir # if you need just the zip file