Mysqli INSERT command followed by an UPDATE - php

Question

I would like to have data inserted in one table, and data updated in another through prepared statements in mysqli. Trying the following only executes the INSERT command:
EDITED:
if($stmt=$mysqli->prepare("SELECT bids_id, bid, fruit_volume FROM basket ORDER BY bid DESC")) {
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($bids_id, $bid, $fruit_volume);
while($stmt->fetch()) {
$stack = array($bids_id, $bid, $fruit_volume);
array_push($all_fruits, $stack);
}
$stmt->free_result();
}
foreach ($all_fruits as $fruits) {
if ($_POST["offer"] == $fruits[1] && $volume < $fruits[2]) {
$stmt2 = $mysqli->prepare("INSERT INTO oranges (username, price, volume, date) VALUES (?, ?, ?, ?)");
$stmt2->bind_param('sdis', $user, $price, $volume, $today);
$stmt2->execute();
$stmt3 = $mysqli->prepare("UPDATE basket SET fruit_volume = ? WHERE bids_id = ?");
$stmt3->bind_param('ii', 800, 1);
$stmt3->execute();
}
}
$mysqli->close();

Answer 1

bind_param passes by reference not by value,so you need to have those values in variables before they can be referenced
$a=800;
$b=1;
foreach ($all_fruits as $fruits) {
if ($_POST["offer"] == $fruits[1] && $volume < $fruits[2]) {
$stmt2 = $mysqli->prepare("INSERT INTO oranges (username, price, volume, date) VALUES (?, ?, ?, ?)");
$stmt2->bind_param('sdis', $user, $price, $volume, $today);
$stmt2->execute();
$stmt3 = $mysqli->prepare("UPDATE basket SET fruit_volume = ? WHERE bids_id = ?");
$stmt3->bind_param('ii',$a, $b);
$stmt3->execute();
}
}

Answer 2

Try this instead
foreach ($all_fruits as $fruits) {
if ($_POST["offer"] == $fruits[1] && $volume < $fruits[2]) {
$stmt2 = $mysqli->prepare("INSERT INTO oranges (username, price, volume, date) VALUES (?, ?, ?, ?)");
$stmt2->bind_param('sdis', $user, $price, $volume, $today);
$stmt2->execute();
$stmt3 = $mysqli->prepare("UPDATE basket SET fruit_volume = ? WHERE bids_id = ?");
$stmt3->bind_param('ii', 800, 1);
$stmt3->execute();
}
}
$mysqli->close();