Hi All,
I am unable to get my filename in save As Dialog Box of an
Application.Instead I am getting my Perl Script name.
The part of my code is as:
my $zipfile = "assembly$now_id.tar.gz";
if (($zipfile) && (-e $zipfile)) {
if (($userAgent =~ /MSIE/i) && ($current_status ne $last_status)){
$last_status = $current_status;
$initial_delay = 0;
}
elsif ( open( ZIP, "<$zipfile") ) {
# print zip file
#my $path = "/users/x_syed/";
#my $mimetype = 'application/x-gzip';
my $size = -s "$zipfile";
print "Content-type: application/x-zip-compressed;
name=\"$zipfile\"\n\n";
print "Content-length: $size\n";
print "Content-disposition: file;filename=\"$zipfile\"\n\n";
#print "Content-type: application/x-gzip;
name=\"$zipfile\"\n\n";
#print "Content-type: application/octet-stream;
name=\"$zipfile\"\n\n";
#print "Content-type: application/x-zip-compressed\n\n";
#print "Content-type: $mimetype; name=$zipfile\n\n";
#print "Content-disposition: filename=$zipfile\n\n";
#print "Content-length: $size\n\n";
binmode(STDOUT);
binmode(ZIP);
while (<ZIP>) {
print $_;
}
I tried following above ways to get my filename in the save As Dialog Box.
Please help me. Please don't mind. The question may look simple but I am
unable to the filename in the SaveAs Dialog Box.
Note: The behaviour should be as in Links. Once you click the link SaveAs
Dialog Box appears with the filename of the Link as default.
Please your help in this regard is greatly appreciated.
Thanks
Syed.
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