can't echo out else statement

[contra10]

I'm trying to echo out a statement if somethin gis untrue but its not working

<?php
$queryear = "SELECT * FROM `ratings` WHERE `songtitle` = '$name'";
$resultear = mysql_query($queryear) or die(mysql_error());
while ($postedear = mysql_fetch_assoc($resultear))
{
$avsong = "{$postedear['songtitle']}";
$linksong = "{$postedear['song']}";
$ratedavg = "{$postedear['average']}";

if ($ratedavg > 1){
echo"<td>$ratedavg / 10</td></tr>";
}else{
echo "<td>No Rating</td></tr>";
}
}
?>

THe average shows for those that have been rated but the other statement is suppose to show for those that havn't but it doesn't[/code]

[jcombs_31]

Have you check the value of $ratedavg to debug your code.  For one you are setting it to a string and then checking against an int.

[contra10]

i see what your saying

in this case one of my $ratedavg = 5

 

5 > 1 so it would show the result but if there is no rating and the $ratedavg is equal to nothing shouldn't it show. Or would I have to put something like

 

<?php
if ($ratedavg == true){
echo"<td>$ratedavg / 10</td></tr>";
}else{
echo "<td>No Rating</td></tr>";
}
?>

 

[jcombs_31]

Are you allowing null values in your database?  What I was saying is that "5" != 5, but I think PHP may be forgiving and cast it as an int for you when doing the comparison.

[contra10]

in my databse the "average" is set to "not null"

[mikesta707]

I think it would cast the string as an int in that comparison, and all strings get cast to 0 if i'm not mistaken. So do this

$ratedavg = $postedear['average'];

 

and see if it helps. There is no reason to surround the data with quotes anyways, as if its a string it will be treated as such when assigned, and if its an int it will be treated as such, but forcing it to be a string always can run into problems