Trouble in adding nested arguments to an IF statement

Author Name Status Posted Time Mr.Iestamyo Shakaaramutai 5/8/2008 11:48:57 PM Author is Offline Hi, I'm trying to add nested arguments to an IF statement. What I am trying to accomplish is to only run the code following the code if: 1) a session hasn't been started... 2) the first three octets of the IP address don't match the octets from my home network or from the W3C (I have a link to the validator on my footer)

Subject Author Status Date Re: Trouble in adding nested arguments to an IF st atement Mr.Iestamyo Shakaaramutai 5/9/2008 4:09:52 AM Currently, the opening of the IF looks like this... Code: if(!isset($_SESSION['USERID']) && ($firstthreeip != '192.168.1' || $firstthreeip != '128.30.52')) When my code only checked for no session and an IP that didn't match my network, it worked... Code: if(!isset($_SESSION['USERID']) && $firstthreeip != '192.168.1') Somehow I need to do the "my network or the W3 network". Any thoughts? Post Comments Re: Trouble in adding nested arguments to an IF st atement Chalkku Monteintah 5/9/2008 4:10:24 AM mebbe Im missing a trick here but can't you 'just' simplify the IF Code: if (!isset($_SESSION['USERID']) AND ($firstthreeip != '192.168.1' OR $firstthreeip != '128.30.52')) Post Comments Re: According to my lovely PHP reference book... Mr.Iestamyo Shakaaramutai 5/9/2008 4:12:06 AM According to my lovely PHP reference book, && = AND and || = OR, so unless I'm not seeing something, those to statements are the same? Post Comments Re: They may well be the same Chalkku Monteintah 5/9/2008 4:13:19 AM They may well be the same according to the manual but y'never know Id probably go one stage further.. Code: if (isset($_SESSION['USERID'])==FALSE AND NOT ($firstthreeip == '192.168.1' OR $firstthreeip == '128.30.52')) I did think that the && and || operators were explicitly for byte operations I think you are saying you want to process this block if the session Id isn't set AND the networkID is neither 192.168.1 OR 128.30.52 I take it you are checking to see what the value of $firstthreeip is what you are expecting and not something like "192.168.1." personally I don't like seeign operators that are !=. but thats just one of my (many) bugbears Post Comments Re: I used your code Mr.Iestamyo Shakaaramutai 5/9/2008 4:14:27 AM I used your code, and it returned a 500 Server Error. I changed the NOT to a ! and the error went away, and I'm getting the results I want. Quote: I take it you are checking to see what the value of $firstthreeip is what you are expecting and not something like "192.168.1." What I'm doing is increasing a counter number by 1 if 1)a session doesn't exist and 2)the IP address is from something other than my internal network or the W3C (because I'm constantly validating my code everytime I make an HTML change). If those two conditions don't exist, then the visitor " ;must" be new (at least since the last time they opened their browser). Post Comments Re: Glad you foudn a solution Chalkku Monteintah 5/9/2008 4:15:33 AM Glad you foudn a solution one of the problems I find is trhat with so many db's out there, and not using each and evryone daly/frequently its easy to forget the precise syntax for each SQL engine Post Comments Re: This is PHP Mr.Iestamyo Shakaaramutai 5/9/2008 4:17:57 AM AH! Maybe that was our confusion. This is PHP, not the SQL in the PHP. But yes, the syntax is always different from place to place. Regardless, I thank you for your help! Post Comments

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