else if statement

[woodplease]

i'm trying to create an else of statement that will echo out database entries if certain conditions are met, but i cant get it to work. I'm also not sure if i've even got the correct syntax. could anyone help?

$percentage=$total/$questiontotal*100;
echo '<br/>' .$percentage;

$query="SELECT * FROM results WHERE quizref= " .$quizref;
$result=pg_query($query) or die ("Query failed");
$row=pg_fetch_array($result);
if ($percentage<=25) {
	echo "you are " .$row['poor'];
	}else{
	if (25<$percentage <=50){
		echo "you are " .$row['acceptable']
		}else
		if ($percentage>50<=75){
			echo "you are " .$row['good'];
			}else
			if ($percentage>75<=100){
				echo "you are " .$row['geek'];
}
}
}

Thanks

[johanafm]

While your code is syntactically correct, it's also nonsensical.

; ends a statement.

if ($something);
other_stuff();

This will do nothing if $something is true and then proceed to other_stuff(). Else it will go on to other_stuff() straight away.

Also keep your indentation straight so it's easy to follow the program flow

if ($something)
	do_stuff();
else if ($something_else) {
	if ($nested_if)
		echo 'stuff';
	do_other_stuff();
	do_stuff();
}
else
	and_so_on();

[woodplease]

right, i have sorted out the code, but i keep getting the following error, and i dont know why.

"Parse error: syntax error, unexpected T_IS_GREATER_OR_EQUAL in /berw/ugrad1/base/a/agd8/public_html/quizdom/results.php on line 39"

$percentage=$total/$questiontotal*100;
echo '<br/>' .$percentage;

$query="SELECT * FROM results WHERE quizref= " .$quizref;
$result=pg_query($query) or die ("Query failed");
$row=pg_fetch_array($result);
if ($percentage<=25) 
	{
	echo "you are " .$row['poor'];
	}
	elseif (25<$percentage >=50) //line 39
	{
	echo "you are " .$row['acceptable']
	}
	elseif (50<$percentage>=75)
	{
	echo "you are " .$row['good'];
	}		
	elseif (75<$percentage>=100)
	{			
	echo "you are " .$row['geek'];
	}


echo '<p><a href="admincontrol.php"> &lt;&lt;&lt;&lt;&lt;Back to Admin Control&lt;&lt;&lt;&lt;&lt; </a></p>

[Weedpacket]

(25<$percentage >=50) //line 39 

Presumably you mean

(25<$percentage and $percentage <= 50)

What you've written is (a) bad syntax (as you've found), and (b) incorrect (since you have >= instead of <=)

Besides, if you wrote those tests in the opposite order there wouldn't be an issue:

if(75 < $percentage)
{
    'geek';
}
elseif(50 < $percentage)
{
    'good';
}
//....

[sharad_impinge]

Use the following-------------

$percentage=$total/$questiontotal*100;
echo '<br/>' .$percentage;

$query="SELECT * FROM results WHERE quizref= " .$quizref;
$result=pg_query($query) or die ("Query failed");
$row=pg_fetch_array($result);
if ($percentage<=25)
{
echo "you are " .$row['poor'];
}
else if ($percentage>25 && $percentage <=50)
{
echo "you are " .$row['acceptable']
}
else if ($percentage>50 && $percentage <=75)

{
echo "you are " .$row['good'];
}

else if ($percentage<75 && $percentage <=100)
{

echo "you are " .$row['geek'];
}

echo '<p><a href="admincontrol.php"> &lt;&lt;&lt;&lt;&lt;Back to Admin Control&lt;&lt;&lt;&lt;&lt; </a></p>