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PHP function kills my code?

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# 8  
Old 12-13-2003
What is the value you are passing to REMOTE_ADDR? Have you tried hard coding a value to test the function? Also, what scope does the variable REMOTE_ADDR have within the function? Is PHP like shell where a variable within a function has global scope? or is it a more limited scope?
# 9  
Old 12-13-2003
$remote_addr is a system variable, it is already defined. I never pass anything to it, and it is always equal to the visitors IP address. Also this chunk of code works perfectly fine when its not in my function, only when it is in, does it give me a connection error from the fopensocket.
# 10  
Old 12-15-2003
google is right. REMOTE_ADDR is not
defined in your function.

echo it in your function and see.
echo $REMOTE_ADDR

To fix it, declare it in main code and
pass it into function, such as:

PHP Code:
<?php function sniff( $portnumber, $REMOTE_ADDR ) {
        if (! $sock = fsockopen($REMOTE_ADDR, $portnumber, $errno, $errstr, 7) )
            {
              if ($errno == 60)
                  {
                  echo "Timeout";
                  }
                  else 
                  {
                  echo "Closed $errno";
                }
            }
            else
            {
              echo "<font color=#ff3300>Open</font>";
              fclose($sock);
              }
      } 
?>

<?php
$REMOTE_ADDR = $HTTP_SERVER_VARS['REMOTE_ADDR'];
sniff(22, $REMOTE_ADDR);
?>
# 11  
Old 12-16-2003
I got this all figured out on another forum. I thought $REMOTE_ADDR was global because outside of functions you can access it no problem. But to use it in functions, you have to make it a global, which is what i did and now it works fine. Its strange though. Thats why it worked when it wasn't in a function but didn't when it was in one.
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