Reply to Re: newbie - variable to string

Your name:

Reply:


Posted by J.O. Aho on 10/11/67 11:52

Jarrod wrote:
> Hi there,
>
> I am trying to get the value of the varible into an if statment that will
> determine what to do.
>
> The value is from the end of an array, this works fine when i use echo to
> print it out onto the page but cant pass the value into the if statement,
> can someone please help me
>
> $testing = '10';
> $end = end($array); // this would equal to 10

in this example, the array is empty, so you need to assign a value to the array.
>
> if($testing == $end){
> print 'yes';
> }else{
> print 'no';
> }
>

Try this one instead

$testing = '10';
$array = array(1,2,3,10);
$end = end($array); // this would equal to 10


if($testing == $end){
print 'yes\n';
}else{
print 'no\n';
}


If you want to give up end(), you can use:
$testing = '10';
$array = array(1,2,3,10);
$end = $array[count($array)-1];


if($testing == $end){
print 'yes\n';
}else{
print 'no\n';
}


If I would be you, I would try to keep integers as integers and not as
strings, so I would have used '$testing = 10;' instead of what you have.


//Aho

[Back to original message]


Удаленная работа для программистов  •  Как заработать на Google AdSense  •  England, UK  •  статьи на английском  •  PHP MySQL CMS Apache Oscommerce  •  Online Business Knowledge Base  •  DVD MP3 AVI MP4 players codecs conversion help
Home  •  Search  •  Site Map  •  Set as Homepage  •  Add to Favourites

Copyright © 2005-2006 Powered by Custom PHP Programming

Сайт изготовлен в Студии Валентина Петручека
изготовление и поддержка веб-сайтов, разработка программного обеспечения, поисковая оптимизация