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Problem with an if instruction

Posted: Wed Jul 09, 2003 5:01 am
by MeHigh
I have this code:

Code: Select all

<?php
<?php 
$limbaje = array(0=>"perl", 1=>"PHP", 2=>"Python", 3=>"perl", 4=>"PHP", 5=>"Python");
$cauta = "PHP";
//it searches the $limbaje array the $cauta value
$limita = count($limbaje);
for ($i=0; $i < $limita; $i++)
{
echo "<br> determining a identity with $limbaje[$i]";
	if ($cauta == $limbaje[$i])
		{
		"<br>$cauta is an aprooved language.";
		}
}
?>
?>

and it prints
determining a identity with perl
determining a identity with PHP
determining a identity with Python
determining a identity with perl
determining a identity with PHP
determining a identity with Python
instead off
determining a identity with perl
determining a identity with PHP
PHP is an aprooved language.
determining a identity with Python
determining a identity with perl
determining a identity with PHP
PHP is an aprooved language.
determining a identity with Python
Thx a lot

Posted: Wed Jul 09, 2003 5:25 am
by Wayne

Code: Select all

if ($cauta == $limbaje&#1111;$i]) 
      &#123; 
      echo "<br>$cauta is an aprooved language."; 
      &#125;

Posted: Wed Jul 09, 2003 5:35 am
by santos_

Code: Select all

<?php 

	$limbaje = array(0=>"perl", 1=>"PHP", 2=>"Python", 3=>"perl", 4=>"PHP", 5=>"Python"); 
	$cauta = "PHP"; 

	for ($i=0; $i<sizeof($limbaje); $i++) 
	{ 
		echo "determining a identity with " . $limbaje[$i] . "<br>\n";

		if($limbaje[$i] == $cauta){
			echo $limbaje[$i] . " is an approved language.<br>\n";
		}
	} 

?>

Posted: Wed Jul 09, 2003 6:28 am
by MeHigh
thanx a lot, guys. :D