php function to execute sql doesn't return anything

Question

I am making a webshop and need to retrieve the price of an item I've been reading and searching for a few weeks now without any luck, anybody here with an answer?

-> little explanation :

• Every item has a few sizes it can come in and every size has a price for that item.
• The price I need to retrieve is the lowest price available for that item.

the problem:

the function getPrice() does not return anything as far as I can check.

It also does not seem to produce any error.

my code:

<?php 
$servername =   "localhost";
$username =     "beegeeky";
$password =     "pwd";
$dbname =       "beegeeky";
$conn = new mysqli($servername, $username, $password, $dbname);


function getPrice($id){
$sql = "SELECT MIN(price) FROM item_price WHERE size_id IN (SELECT id FROM item_size WHERE item_id =".$id.");";
$lowestprice = $conn->query($sql);
return $lowestprice;
}



$sql = "SELECT * FROM item ORDER BY id DESC;";
$qresult = $conn->query($sql);
if ($qresult->num_rows > 0) {
while($item = $qresult->fetch_assoc()) {
    ?>
    <div class="item" id="item_<?php echo $item["id"]; ?>" onclick="itemClick('item_<?php echo $item["id"]; ?>')">

        <div class="likeHitBox" onclick="like('like_<?php echo $item["id"]; ?>')"></div>
            <div class="like<?php if ($conn->query("SELECT * FROM item_like WHERE item_id = ". $item["id"] . " AND user_id = ". $userID .";")->num_rows > 0){?> liked<?php } ?>" id="like_<?php echo $item["id"]; ?>"></div>
        <div class="iconContainer"></div>

        <div class="itemImg">
            <img src="../img/<?php echo "BoneConductingHeadphones/BCH.png"; ?>">
        </div>

        <div class="price">
            <?php echo getPrice($item["id"]); ?>
        </div>
        <div class="priceContainer"></div>

    </div>
    <?php
}
}
$conn->close();
?>

if I need to add anything, just ask :)

**EDIT: ** thanks already, I made some changes to the function, what I have now is:

function getPrice($id){
    global $conn;
    $sql = "SELECT MIN(price) FROM item_price WHERE size_id IN (SELECT id FROM item_size WHERE item_id =".$id.");";
    $lowestprice = $conn->query($sql)->fetch_assoc();
    return $lowestprice[0];
}

this returns an array (yay something is returned now), but the array seems to be empty, Thats work for another SO-question

Thank you!

Answer 1

The $conn variable doesn't exist inside the getPrice function.

http://php.net/manual/en/language.variables.scope.php

Also, look into sql injection.

PHP MySQLI Prevent SQL Injection

Answer 2

There's a few lines missing between

$lowestprice = $conn->query($sql);

and

return $lowestprice;

You need to tech result from the recordset, like you did with $qresult.

Answer 3

Try this:

function getPrice($conn, $id)
{
    $sql = "SELECT MIN(price) FROM item_price WHERE size_id IN ( " .
           "     SELECT id FROM item_size WHERE item_id =" . $id .
           ")";
    $lowestprice = $conn->query($sql);

    return $lowestprice;
}

You have to include the MySQLi connection as an input into the function for it to work.