RE: [PHP] odd addition error<>?

What are you initializing these values to before using them? The only variable that is being initialized in the code sample that you sent is the $handling variable. Sam Masiello System Analyst Chek.Com (716) 853-1362 x289 smasiello@chekinc.com -----Original Message----- From: root [mailto:root] On Behalf Of callen Sent: Tuesday, August 15, 2000 2:24 PM To: PHP General List Subject: [PHP] odd addition error<>? Greetings! The following bit of code gives a strange result! At one point it comes out with the correct values, such that the addition that is performed is correct. At line 2 $fee always is what it should be (lets say 3.71), when it gets down to line 4 sometimes the value of $sh is 0. The value of $fee changes based upon a block of code that it comes from, these two different blocks of codes are based upon simple if statements , and the variable $fee always carries its value to line 2. If the script goes thru the first block of code it performs the addtion correctly, if it goes thru the second The addition fails at line 3 (line 4 echoes out SH is: 0), although the value $fee is correct in line 2. what can this possibly mean? ???? 1) $handling =2; 2) echo "Fee again is $fee <br>"; 3) $sh += $fee; 4) echo "SH is: $sh <br>"; 5) $sh += $handling; -- Christopher C.M. Allen http://design.driver8.org/ Email: callen@driver8.org -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net For additional commands, e-mail: php-general-help@lists.php.net To contact the list administrators, e-mail: php-list-admin@lists.php.net