审查具有变量函数的系统

Question

so basically I am trying to create a little thing where it outputs stars, based on the database saved rating integer. The problem is it does not seem to put the number I from the database, in the variable. Here is the code I used:

 <?php
$productID = 100;
$con = mysqli_connect("localhost", "root", "", "example");
function connect()
{
    $con = mysqli_connect("localhost", "root", "", "example");
    // Check connection
    if (mysqli_connect_errno()) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    } else {
     return $con;
    }
}

function getStars($con)
{
    $productID = 100;
    $sql       = "SELECT rating
       FROM reviews
      -- JOIN stockitemstockgroups USING (StockItemID)
      -- JOIN stockgroups USING (StockGroupID)
       WHERE reviewID = '5'
       ";
    $result    = $con->query($sql);

    if ($con && ($result->num_rows > 0)) {
        // output data of each row
        while ($row = $result->fetch_assoc()) {
            echo $row["rating"];
        }
    } else {
        echo "error";
    }
}
$value = getStars($con);
echo $value;
for ($x = 1; $x <= $value; $x++) {
echo '<div class="rating"><span>&#9733;</span></div>';
}
?>

Answer 1

I'm having trouble finding a duplicate, though I'm sure this is one. You aren't returning anything from your function, so $value doesn't have a value.

function getStars($con)
{
    $productID = 100;
    $sql       = "SELECT rating FROM reviews WHERE reviewID = 5";
    $result    = $con->query($sql);

    if ($result && ($result->num_rows > 0)) {
        // output data of first row
        $row = $result->fetch_assoc();
        return $row["rating"];
    } else {
        return false;
    }
}

As a general rule, never echo from a function. Also, no need for a loop over what will presumably be a single result.