if rule (and yes, im new)

[kennethamby]

Hello,

 

First of: I have not been working with PHP for 5 years, and suddently decided: But I miss it. So startet working on a new site, where some of the code has allreade been done, but i do not work like he would like it to. So I am taking over, and my first real problem is this:

I have this value:
<?php echo gethostbyname($server_data['ip']) . ":" . $server_data['port']; ?>
And it works fine. 

But! (and not in the ass kind of way)

There are a port i dont need it to show. So i need to modifide it so if the port number is 1234 its no shown at all. 

 

I was thinking something like this:

 

<?php echo gethostbyname($server_data['ip']) . ":" . $server_data['port'] if 'port' = 1234 return null; ?> 

 

The above do not work (and im sure its my fault, but i cannot find the right way to do this)!

 

Help ?

Regards, 
Kenneth

[jairathnem]

if($server_data['port'] != 1234){

echo gethostbyname($server_data['ip']) . ":" . $server_data['port'];

}

else

return null;

[kennethamby]

I cant test that for the next hour, but will that not return the total of 'null' (meaning: will that code not remove both the IP and the Port?)

 

atm is shows: serverIP and serverPort. (ex: 1.2.3.4:1000) and that is OK. 
but only one port should not be displayed. (the 1234 port)

 

So if a server has the adresse: 1.2.3.4:1000 it shoud display the ip and the port.

But if a server has the adresse: 1.2.3.4:1234 it shoud only display the ip and not the port
 

again: I cant test the script for the next hour, but thank you so mutch for your help! 

 

Regards Kenneth
 

[Barand]

echo gethostbyname($server_data['ip']);

if($server_data['port'] != 1234){

   echo ":" . $server_data['port'];

}

[kennethamby]

echo gethostbyname($server_data['ip']);
if($server_data['port'] != 1234){
   echo ":" . $server_data['port'];
}

 

Thank you ! That seems to work perfectly !