Parsing filename from raw email

Question

Need to get a comma delimited list filenames in an email. Here is some input examples with 1 having lots of special characters:

Content-Disposition: attachment;
        filename="How-To_21_Monitor_Mode_Deployment_Guide.pdf"; size=3886046;
        creation-date="Fri, 27 Oct 2017 16:23:20 GMT";
        modification-date="Fri, 27 Oct 2017 16:24:30 GMT"

Content-Disposition: attachment; filename="How-To_24_Low_Impact_Mode.pdf";
        size=6714113; creation-date="Fri, 27 Oct 2017 16:23:20 GMT";
        modification-date="Fri, 27 Oct 2017 16:24:31 GMT"

Content-Disposition: attachment;
        filename="SBTGxYVWPE1wI9SAjl5b2PUfF1LCjbU3aChsoch5eXuI4GrIP9bRhfiaOuwL1U
 ;.,~!@#$%....txt"; size=3966; creation-date="Fri, 27 Oct 2017 16:23:20 GMT";
        modification-date="Fri, 27 Oct 2017 16:23:20 GMT"

This works, but concerned if it went more than 3 lines and it is not efficient:

grep --no-group-separator --line-buffered -A 2 '^Content-Disposition: ' | sed -e '/\;$/!{N;s/\n//}' -n -e 's/.*filename\=//p' | sed -e 's/ size\=.*//' | sed 's/\;$//' | sed ':a;N;$!ba;s/\n/,/g'

Tried to get it working with a single sed line:

sed -n '/^Content-Disposition: /,/\"\; size\=/{/\;$/!{x;N;s/\n//g}};s/.*filename\=//p;s/ size\=.*//;s/\;$//;:a;N;$!ba;s/\n/,/g;

Would be appreciated for a single command preferably with sed.

Answer 1

Satō Katsura is right. Here's a quick Perl script, assuming you have the full email message in a file.

cpan install Email::MIME File::Slurp::Tiny
perl -MEmail::MIME -MFile::Slurp::Tiny=read_file -wE '
    my $email = Email::MIME->new( read_file(shift @ARGV) );
    my $count = 0;
    $email->walk_parts(sub {
        my $part = shift;
        my %header = $part->header_str_pairs;
        if (exists $header{"Content-Disposition"}) {
            my ($filename) = $header{"Content-Disposition"} =~ m/(?<=filename=")([^"]+)/;
            say ++$count .":". $filename;
        }
    })
' email.eml

Answer 2

to get a comma delimited list filenames in an email

GNU awk solution:

awk -v RS='\n\n' 'BEGIN{ fn="" }
       match($0, /filename="([^"]+)";[[:space:]]+size=/, a){ 
           gsub(/[[:space:]]*/, "", a[1]); 
           fn = (fn!=""? fn", ":"")a[1] 
       }END{ print fn }' file

• fn - the resulting string containing all the filenames(after processing)

• -v RS='\n\n' - set record separator as double linebreak

• /filename="([^"]+)";[[:space:]]+size=/ - regex pattern to capture the filename in each Content-Disposition header into array of matches a

---

The output:

How-To_21_Monitor_Mode_Deployment_Guide.pdf, How-To_24_Low_Impact_Mode.pdf, SBTGxYVWPE1wI9SAjl5b2PUfF1LCjbU3aChsoch5eXuI4GrIP9bRhfiaOuwL1U;.,~!@#$%....txt

Answer 3

RomanPerekhrest got it 99%, thanks! This is the final line I used, just had to make a minor tweak to handle spaces in filenames and include the quotes around strings. Awk definitely better than sed in this case.

awk -v RS='\n\n' 'BEGIN{ fn="" } match($0, /filename=("[^"]+");[[:space:]]+size=/, a){gsub(/\n/, "", a[1]);fn = (fn!=""? fn",":"")a[1]}END{ print fn }'