PHP returning value of 1 [closed] - php

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i still cannot get this review system to work properly.
i know the average rating is 3.57 but as soon as i add the if() all i am getting is an output of 1?
// collect 5 lastest reviews for viewing
$sql = "SELECT AVG(rating) FROM reviews WHERE review_id = '$id'";
$query = mysqli_query($conn, $sql);
$AvgReview = mysqli_fetch_row($query);
// Here we have the toatal row count
$AvgReviews = $AvgReview[0];
if($AvgReviews = 1){
echo '1star';
}
//next else is new
else if($AvgReviews >= 1){
echo '2star';
}
//next else is new
else if($AvgReviews >= 2){
echo '3star';
}
//next else is new
else if($AvgReviews >= 3){
echo '4star';
}
//next else is new
else if($AvgReviews >= 4){
echo '5star';
}

Answer 1

You are getting '1star' because you are assigning a value of 1 to $AvgReviews.
Change your code to compare the value.
if ($AvgReviews == 1) {
Additionally, you need to reverse your code or you will end up getting 1 or 2 stars, only.
I would recommend to change it like that:
if ($AvgReviews > 4) {
echo '5star';
}
else if($AvgReviews > 3) {
echo '4star';
}
else if($AvgReviews > 2) {
echo '3star';
}
else if($AvgReviews > 1) {
echo '2star';
}
else {
echo '1star';
}

Answer 2

It's because you're assigning 1 to $AvgReviews. The assignment operator (=) takes the value on the right-hand side, assigns it to the name on the left-hand side, and then returns the value that was assigned. In this case, that's always 1, which PHP interprets as true.
Use the comparison operator instead:
if($AvgReviews == 1){
echo '1star';
}