Hi Everybody,
As a naive cgi programmer, I want to get rid of a
problem, for that, I am keen, awaiting your
suggestions. I doubt, I could not present my case in
front of you properly. Anyway, my problem spins around
the following.
I am running a CGI script in which I am running a
system command. The scripts is as follows:
#!/usr/bin/perl -w
use strict;
use CGI;
open(FH_seq1,">/var/www/cgi-bin/emboss/water1.seq");
open(FH_seq2,">/var/www/cgi-bin/emboss/water2.seq");
print "Content-type:text/html\n\n";
my $query = new CGI;
my $seq1 = $query->param("seq1");
my $seq2 = $query->param("seq2");
print FH_seq1 "$seq1\n";
print FH_seq2 "$seq2\n";
`/var/www/cgi-bin/emboss/water
/var/www/cgi-bin/emboss/water1.seq
/var/www/cgi-bin/emboss/water2.seq -gapopen 10
-gapextend 5 -outfile
/var/www/cgi-bin/emboss/water.out`;
Above I am using the system command to run the program
"water" which should write the output into file
"water.out", which is not writing anything to this
file.
I've given all permission to these files as:
-rwxr-xr-x 1 soumya soumya 2978221 Jun 15 10:26
blastall
-rwxr-xr-x 1 soumya soumya 4912 Jun 13 08:25 water
-rwxrwxrwx 1 apache apache 36 Jun 16 05:18
water1.seq
-rwxrwxrwx 1 apache apache 36 Jun 16 05:03
water2.seq
-rwxrwxrwx 1 soumya soumya 48 Jun 16 05:18
water.out
But if I use the first file in the directory i.e.
blastall in the script as
my $out = `./blastall`;
print $out;
I get the proper output.
Could you please suggest me any solution for this.
Looking forward for your suggestions,
With regards,
Soumyadeep
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