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else if statement


woodplease

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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 ($percentage >25<=50);
	echo "you are " .$row['acceptable'];
	else
	if ($percentage>50<=75);
		echo "you are " .$row['good'];
		else
		if ($percentage>75<=100);
			echo "you are " .$row['geek'];

 

Thanks

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Am not entirely sure what you're doing with that query in the first part (it looks backwards to me), but this is your if/elseif/else statement fixed up:

if ($percentage<=25) {
echo "you are " .$row['poor'];
} elseif ($percentage >25<=50) {
	echo "you are " .$row['acceptable'];
} elseif ($percentage>50<=75) {
		echo "you are " .$row['good'];
       } else {
	echo "you are " .$row['geek'];
      }

 

More than that, I would recommend using switch/case statements for sanity's/readable sake.

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Your code doesn't make it totaly clear what your trying to but I can offer some help.

 


$questiontotal = 30;

$query = "SELECT * FROM results WHERE quizref= '$quizref'";
$result = pg_query($query) or die ("Query failed");

while ($row = pg_fetch_array($result)){
$total = $row['total'];
$var2 = $row['Field_Name_2'];

echo $result."<br>"
$percentage = ($total/$questiontotal)*100;

if ($percentage <= 25){
	echo "you are poor";
}
elseif ($percentage > 25 && $percentage <= 50) {
	echo "you are acceptable";
}
elseif ($percentage > 50 && $percentage <= 75) {
	echo "you are good";
}
else {
	echo "you are a geek!";
}
} 

 

You can retrieve more fields from the database by replacing the line $var2 = $row['Field_Name_2']; with the name of your database fields and a sensible variable name.

 

Hope this is of use.

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