Please read how (not) to ask a question, and writeup formatting tips. As in: use <code> tags for your code.

Here is the code:

use File::Basename;
$infile_name = shift(@ARGV);

if ( ! -r $infile_name)
{
print STDERR "1.Cannot read $infile_name\n";
exit 99;
}
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a few more lines in between and in the main program I have the following:

my $ver_name = shift @ARGV;
my $mon_dir = dirname($infile_name); # Get directory

if (! -z $ver_name) {
(our $graph_dir = $mon_dir) =~ s?log/.*$?abinitio/run?;    #Build dire
+ctory as above if not equal to version.(Replace log/* with abinitio/r
+un)
}
else
{
(our $graph_dir = $mon_dir)=~ s?log/.*$?$ver_name/abinitio/run?; 
#Build directory as such if equal to verion(Replace log/* with $ver_na
+me/abinitio/run)
}
my $graph = "$graph_dir" . "/" . "$graphname";   
# Concatenate graph directory with graphname

if ( ! -r $graph)
{
print STDERR "2. Cannot read graph=$graph\n"
exit 99;
}
[download]

By including the if-then-else condition I am getting the following error that it "Cannot read graph=********(pathname)

A second error you haven't noticed yet: $? is a variable name. Perhaps you wanted \$?, although .*\$? is the same as just .*.

Please use code tags. I think you want (not valid perl):

my $var2;
if <cond1> {
  $var2 = a;
} else {
  $var2 = y;
}
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The following would also work:

my $var2 = <cond1> ? a : y;
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But be careful. Even slight complexity in its arguments can make it hard to read.

A third option is

my $var2 = do {
  if ( <cond1>  ) {
     a;
  } else {
     y;
  }
};
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All three constructs are more-or-less interchangable and which one you use should be decided based on which is most readable in any particular instance.

my $var2 = (cond 1) ? 'a' : 'y';
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gj