In this snippet exit doesnt end the perl cgi script as i expect it to do but instead it exits the if structure.

There's no such thing as exiting an if structure with exit. You can exit a loop with last, but exit exits the program.

if(1) {
    print "1 seems to be true\n";
    exit 0;
}
print "After exit..\n";
__END__
1 seems to be true
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The After exit. is not printed.

Wouldnt it be better to just somethign like this work?

No, because die throws its message out STDERR, which is not what you want. Read the docs.

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}

print(h1( {class=>'cyan'}, "Error occured $row->{username}! Script hav
+e to die" )), exit(0)
if $select->rows;
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Here? why the program exits without printing first?

That's because you are passing a list to print. For this list to be printed, its elements must be evaluated first. There's a exit call in that list, so that is done, and the print never gets executed. Consider

print "no rows\n", exit unless $rows;
print "we seem to have rows.\n";
__END__
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and

print "no rows\n" and exit unless $rows;
print "we seem to have rows.\n";
__END__
no rows
[download]

update: as ikegami point out below, you are not passing a list to print - since you treat print as a function, and are passing in its arguments in parens. With the comma operator, you are still evalutaing a list, and in that list exit gets evaluated..

--shmem

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}