bash，列出文件名，数组

Question

I have a dir structure like

$ ls /comp/drive/
2009  2010  2011  2012  2013  2014

$ ls 2009
01  02  03  04  05  06  07  09  10  11  12

$ ls 2013
01  02  04  05  06  08  09  10  12

$ ls 2013/04/*.nc
file4.nc file44.nc file45.nc file49.nc

There are dirs like years and each year there are few months dirs and inside are .nc files.

What I want to do is get the array of filenames provided start and end years/months.

eg sYear=2011; eYear=2013; sMonth=03; eMonth=08 sYear=2011; eYear=2013; sMonth=03; eMonth=08

So, I want to get the array of all filenames from year 2011/03 to 2013/08 only without going inside the dirs.

Any bash trick?

Answer 1

sYear=2011; eYear=2013; sMonth=03; eMonth=08

# prevent bugs from interpreting numbers as hex
sMonth=$(( 10#$sMonth ))
eMonth=$(( 10#$eMonth ))

files=( )
for (( curYear=sYear; curYear <= eYear; curYear++ )); do
  # include only months after sMonth
  for monthDir in "$curYear"/*/; do
    [[ -e $monthDir ]] || continue # ignore years that don't exist
    curMonth=${monthDir##*/}
    (( curMonth )) || continue     # ignore non-numeric directory names
    (( curYear == sYear )) && (( 10#$curMonth < sMonth )) && continue
    (( curYear == eYear )) && (( 10#$curMonth > eMonth )) && continue
    files+=( "$monthDir"/*.nc )
  done
done

printf '%q\n' "${files[@]}"

Answer 2

Try this:

sYear=2011
sMonth=03

eYear=2013
eMonth=08

shopt -s nullglob
declare -a files

for year in *; do
    (( ${year} < ${sYear} || ${year} > ${eYear} )) && continue

    for year_month in ${year}/*; do

        month=${year_month##*/}
        (( ${year} == ${sYear} && ${month##0} < ${sMonth##0} )) && continue;
        (( ${year} == ${eYear} && ${month##0} > ${eMonth##0} )) && continue;

        files+=(${year_month}/*.nc)
    done
done

echo "${files[@]}"
# printf "$(pwd)/%q\n" "${files[@]}" # for full path