[SOLVED] Are my conditions wrong? I thought this would be simple...

[alconebay]

I want my stat logging to ignore my work and home ip address so they don't get put into my statistics, but it's being logged anyway. What am I doing wrong? ("//begin stats logging" being my logging code which works fine)

 

$ip=$_SERVER['REMOTE_ADDR'];
if ( $ip != "11.111.11.11" || "22.222.222.222" ) //these are the IPs I want ignored
{
// begin stats logging...
}

 

I also tried this condition:

if (( ! $ip == '111.111.11.111' ) || ( ! $ip == '22.222.22.222' ))
{
// begin stats logging...
}

[dennismonsewicz]

are you using static IPs? if so then hardcoding your IP address is fine, but if you have are using dynamic IP addresses then it will be really hard to hardcode your IP in the IF statement

[waynew]

$ip=$_SERVER['REMOTE_ADDR'];
if ( $ip != "11.111.11.11" && $ip != "22.222.222.222" ) //these are the IPs I want ignored
{
// begin stats logging...
}

[kenrbnsn]

I would put all the IPs to be ignored in an array and then use the in_array() function:

<?php
$ignore_ips = array('11.22.33.44','44.33.22.11','111.222.111.222');
if (!in_array($ip,$ignore_ips)) {
//
// begin stats logging
//
}
?>

This make for easy changing of the IP addresses, instead of mucking with the if statement.

 

Ken

[alconebay]

The array method works. I also tried changing the "or" ( || ) to an "and" ( && ) but it did the same thing (nothing was being logged). kenrbnsn's suggestion of an array works perfectly though. I'm still wondering why the other method of and's and or's didn't work. It seemed structurally correct... But the array is working (and it is easier) so I'll leave well enough alone.

 

Thanks for the help.

[phpSensei]

Your other post didnt make sense, so I thought i would fix it.

 

<?php
$ip = $REMOTE_ADDR; // Bad coding practice lol
if(($ip != "111.111.11.111" ) || ($ip != "22.222.22.222")){ echo '1'; }else{ echo '2'; }?>

[alconebay]

What is different in your code? Also, what is the bad coding practice?

 

I don't think I need an else statement. If the "if" condition is met the logging starts, else do nothing. Right?

 

Thanks

[phpSensei]

What is different in your code? Also, what is the bad coding practice?

 

I don't think I need an else statement. If the "if" condition is met the logging starts, else do nothing. Right?

 

Thanks

 

u did this if(!$quote == something)

 

I just wanted to tell you that in your first post the second try was wrong. Thats all. use the in_array if it works.

[alconebay]

Oh, the "!" placement. Gotcha.

 

Do the quotes vs. apostrophes make a difference?

[phpSensei]

Oh, the "!" placement. Gotcha.

 

Do the quotes vs. apostrophes make a difference?

 

Not in this case but

If you’re familiar with PHP, the difference between double and single quotes is obvious. Double quotes evaluate variables and control characters (e.g. \n or \r), whereas single quotes do not.

 

Double quotes you can do "i can echo $var without using the period"

 

single quote 'I can echo' . $var . 'but i need those period';