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Re: [TCLUG:1872] Is this hard to do?
> syntax error at /home/httpd/cgi-bin/script.pl line 14, near "printf"
> Execution of /home/httpd/cgi-bin/script.pl aborted due to compilation
> errors.
>
> Line 14 is printf "Content-type: image/gif\n\n";
Hmm.. I would try checking the line _before_ the error
Read below:
> > ---- Cut After this line. --
> > #!/usr/bin/perl
> >
> > #all images in the image directory.
> > $imagedir="/usr/local/www/data/banners";
> > $defaultimage="/usr/local/www/data/linux.gif";
> >
> > sub show_image # filename
> > {
> > my($file) = @_;
> >
> > #send the mime type, can make it intelligent to see what kind of image
> > file it is
> > #and then set the appropriate mime type.
> > printf "Content-type: image/gif\n\n";
You don't by chance still have 'file it is' on the 12th line, do you?
That text is supposed to be quoted..
> > #Read the input file and print it to output.
> > open( INP, "< $file" );
> > while( read( INP, $buf, 1024 ) ) {
> > print $buf;
> > }
> > close( INP );
> > }
> >
> > #Open the imagedir and read the list of files in there.
> > if ( !opendir( DIR, $imagedir ) )
> > {
> > &show_image( $defaultimage );
> > }
> > else
> > {
> > #Read the image files and pick up a random one, This will only get the files
> > # in the imagedir.
> > @filelist = grep { -f "$imagedir/$_" } readdir( DIR );
> > closedir( DIR );
> > if( $#filelist < 0 )
> > {
> > #didnot find any.
> > &show_image( $defaultimage );
> > }
> > else
> > {
> > $loc = rand; #this will call srand.
> > $loc *= $#filelist; #Use it to index into the list.
> > $loc = int($loc);
> > &show_image( "$imagedir/$filelist[$loc]" );
> > }
> > }
> >
> > __END__
> >
> > -- This is the end of the script. --
--
--== Mike Hicks ==--
http://umn.edu/~hick0088 mailto:hick0088@tc.umn.edu
ICQ:6883760
Linux User Since 1.2.13 Current Kernel: 2.1.125