Multiple IF Statement Problem

[kevape]

Hello, You can probably see what i want to do, i want to display a certain image when the variable contains a certain value. and the 5 star image when 5.00 is reached, for some reason, even though the variable $StarRating has a value of E.G. 3.09 or 1.00 the picture that gets displayed is "4star.jpg". Can anybody see my problem, Thanks Guys/Gals.

if (($StarRating =1) && ($StarRating <=1.99)){

$StarPicture = "1star.jpg";

if (($StarRating =2) && ($StarRating <=2.99)){

$StarPicture = "2star.jpg";

if (($StarRating =3) && ($StarRating <=3.99)){

$StarPicture = "3star.jpg";

if (($StarRating =4) && ($StarRating <=4.99)){

$StarPicture = "4star.jpg";

}else{

$StarPicture = "5star.jpg";

}
}
}
}

[FourDegreez]

I assume those equals signs are meant to be greater than or equal, correct? Also, instead of nesting the if statements, just make every one but the first an else if.

[vincevincevince]

$StarPicture=(($StarRating>5)?("5"):(floor($StarRating)))."star.jpg";

[edited by: eelixduppy at 1:41 am (utc) on Jan. 29, 2008]
[edited: vincevincevince]<case correction>

[PHP_Chimp]

Vince's method above is the most compact way of doing what you want, however if you find that hard to read then you may want to use switch.

$star_rating = [url=http://uk2.php.net/manual/en/function.floor.php]floor[/url]($StarRating); // so 2.7 = 2. You may want [url=http://uk2.php.net/manual/en/function.ceil.php]ceil[/url] so that 2.7 = 3.
switch ($star_rating) {
case 1:
$StarPicture = "1star.jpg";
break
case 2:
$StarPicture = "2star.jpg";
break
case 3:
$StarPicture = "3star.jpg";
break
case 4:
$StarPicture = "4star.jpg";
break
case 5:
$StarPicture = "5star.jpg";
break;
default: // for testing, as this shouldnt ever be called
echo 'Something is wrong';
break;
}

Although with

$StarPicture=(($StarRating>5)?("5"):(floor($StarRating)))."star.jpg";

You will only get 5 stars if the rating is above 5. So I guess you need

 $StarPicture = (($StarRating == 5)?("5"):(floor($StarRating)))."star.jpg";

[kevape]

Thank you for the input guys, this is the best way for me to read it.
But i cant get it to work, im getting the echo "something is wrong" and no image is diplaying, i have the the &StarRating is currently 3.09 in this case, so i would expect to hav "3star.jpg" to diplay but no image is found?

Sorry, im new!

$star_rating = floor($StarRating); // so 2.7 = 2. You may want ceil so that 2.7 = 3.
switch ($star_rating) {
case 1:
$StarPicture = "1star.jpg";
break
case 2:
$StarPicture = "2star.jpg";
break
case 3:
$StarPicture = "3star.jpg";
break
case 4:
$StarPicture = "4star.jpg";
break
case 5:
$StarPicture = "5star.jpg";
break;
default: // for testing, as this shouldnt ever be called
echo 'Something is wrong';
break;
}

[PHP_Chimp]

If you echo $StarRating do you get the answer that you expect?

Actually scrap that...You need to put; at the end of the breaks, as I forgot them. Sorry.

$star_rating = floor($StarRating); // so 2.7 = 2. You may want ceil so that 2.7 = 3.
switch ($star_rating) {
case 1:
$StarPicture = "1star.jpg";
break;
case 2:
$StarPicture = "2star.jpg";
break;
case 3:
$StarPicture = "3star.jpg";
break;
case 4:
$StarPicture = "4star.jpg";
break;
case 5:
$StarPicture = "5star.jpg";
break;
default: // for testing, as this shouldnt ever be called
echo 'Something is wrong';
break;
}

That should work better. ;)

<edit>
You may be able to use $StarRating and be able to remove the $star_rating. I just put that other variable in there if you needed the original unaltered $StarRating later in your script.

[edited by: PHP_Chimp at 4:18 pm (utc) on Jan. 29, 2008]