Undefined variable if count 0 - php

Question

i have a little bit problem with my following code. I am trying to make a star raiting system and following code is showing raiting details. But i am getting
undefined variable rate_db notice
from if(count($rate_db)) What i missing here anyone can tell me please ?
full code is here:
<?php
$query = mysqli_query($db,"SELECT ryid,r_type FROM profileRaiting WHERE rp_uid='$profile_uid'");
while($data = mysqli_fetch_assoc($query)){
$rate_db[] = $data;
$sum_rates[] = $data['r_type'];
}
if(count($rate_db)){
$rate_times = count($rate_db);
$sum_rates = array_sum($sum_rates);
$rate_value = $sum_rates/$rate_times;
$rate_bg = (($rate_value)/5)*100;
} else {
$rate_times = 0;
$rate_value = 0;
$rate_bg = 0;
}
?>

Answer 1

It's better to initialize your both arrays at top as:
$rate_db = array();
$sum_rates = array();
And i don't know where you define this variable $profile_uid, if it's user input or session value, than your code is open for SQL Injection, you can use Prepared Statement to prevent SQL attack. And this post will help you to understand: How can I prevent SQL injection in PHP?
Side Note & Suggestion:
One more thing, you can also get the sum of r_type as:
$sum_rates += $data['r_type'];
In this case, you need to initialize $sum_rates as:
$sum_rates = 0;
And just remove this line from IF condition:
$sum_rates = array_sum($sum_rates);
Benefit, no need to use extra method.

Answer 2

$rate_db variable is defined in the while loop where you fetch the data from MySQL. However, if the query does not return any records, then while loop's body will not be executed, thus $rate_db variable will not be defined.
Solution: define $rate_db variable as an empty array before the while loop.

Answer 3

You can initialize the variable:
<?php
$rate_db = array();
$sum_rates = array();
$query = mysqli_query($db,"SELECT ryid,r_type FROM profileRaiting WHERE rp_uid='$profile_uid'");
while($data = mysqli_fetch_assoc($query)){
$rate_db[] = $data;
$sum_rates[] = $data['r_type'];
}
$rate_times = count($rate_db);
if($rate_times > 0){
$sum_rates = array_sum($sum_rates);
$rate_value = $sum_rates/$rate_times;
$rate_bg = (($rate_value)/5)*100;
} else {
$rate_times = 0;
$rate_value = 0;
$rate_bg = 0;
}
?>

Answer 4

before using count you need to be sure that it does have something in it so your complete code can be like this
<?php
$query = mysqli_query($db,"SELECT ryid,r_type FROM profileRaiting WHERE rp_uid='$profile_uid'");
while($data = mysqli_fetch_assoc($query)){
$rate_db[] = $data;
$sum_rates[] = $data['r_type'];
}
if(isset($rate_db) && count($rate_db)){
$rate_times = count($rate_db);
$sum_rates = array_sum($sum_rates);
$rate_value = $sum_rates/$rate_times;
$rate_bg = (($rate_value)/5)*100;
} else {
$rate_times = 0;
$rate_value = 0;
$rate_bg = 0;
}
?>

Answer 5

$query = mysqli_query($db,"SELECT ryid,r_type FROM profileRaiting WHERE rp_uid='$profile_uid'");
while($data = mysqli_fetch_assoc($query)){
$rate_db[] = $data;
$sum_rates[] = $data['r_type'];
}
Because when your query returns empty result the while loop will never become true, so $rate_db[] will never be reached and therefore will never be created.
to fix this, you should declare $rate_db[] as empty first at the top of your script. to avoid the error.