Question

I have written a simple script that checks the content of a zip file against the contents of a control file.

It works well but it fails with errors (that are not really there) when I receives files that have spaces in them. This is a snippet of my code (name is an array created to process ZIP files in bulk).

echo "`date '+%m/%d/%y %T:'` List ZIP file contents."                                          
LIST_Array=(`/usr/bin/unzip -l $name | head -n -2|tail -n +4 | sort -r | awk '{print $4}'`)
LISTlen=${#LIST_Array[*]}
#iterate array to 1) build report and 2) look for control file
echo "`date '+%m/%d/%y %T:'` Iterate array to 1) build report and 2) look for control files."  
echo -e "`date '+%m/%d/%y %T:'` Files in ZIP file: $name\n"                                    >> $name.report.out
for (( i = 0 ; i < ${#LIST_Array[@]} ; i++ ))
do
    echo -e "${LIST_Array[$i]}"      >> $name.report.out
done

The list of files in the ZIP is captured into $name.report.out and then compared against the contents of the control file itself.

How can I correctly display files with spaces? I though the echo -e would help but it seemingly has not effect.

Thanks.

Answer 1

So the zip file has files in it some of whose names have whitespace in them. In that case, when you list the files, awk ‘{ print $4 }’ will not capture the entire filename. read is nice in that its last argument captures the rest of the line:

LIST_Array=()
while read length date time filename; do
    LIST_Array+=( "$filename" )
done < <(/usr/bin/unzip -qql "$name")