Prepared Statement Function - Scope Error - php

Question

I am trying to write a function for a prepared statement but when I run the code it's giving me an error:
mysqli_stmt_store_result() expects parameter 1 to be mysqli_stmt, null given
My function looks as follows:
function fn_preparedStatement($query, $types, $values){
global $dbconnection;
if (!$dbconnection) {
die("Function wm_dynamicForm connection failed.</br>");
}
$db = mysqli_stmt_init($dbconnection);
if (mysqli_stmt_prepare($db, $query )) {
mysqli_stmt_bind_param($db, $types, ...$values);
if (!mysqli_stmt_execute($db)) {
echo "Execute Error: " . mysqli_error($dbconnection);
}
} else {
echo "Prep Error: " . mysqli_error($dbconnection);
}
}
Then in my code I have:
$query = "SELECT * FROM Contacts WHERE First_Name = ?";
$types = "s";
$values = array("Mike");
fn_preparedStatement($query, $types, $values);
mysqli_stmt_store_result($db); //im getting the error on this line - null
So im thinking my issue is a scope issue. I am not sure what to "return" out of my function to make this work. When i write the code inline it works fine. It's when I move the prepared statement to a function and replace the inline code with the function im now getting error. Can someone please tell me where im messing up? Thank you so much.

Answer 1

You need to return the statement handle from the function
function fn_preparedStatement($query, $types, $values){
global $dbconnection;
if (!$dbconnection) {
die("Function wm_dynamicForm connection failed.</br>");
}
$db = mysqli_stmt_init($dbconnection);
if (mysqli_stmt_prepare($db, $query )) {
mysqli_stmt_bind_param($db, $types, ...$values);
if (!mysqli_stmt_execute($db)) {
echo "Execute Error: " . mysqli_error($dbconnection);
}
} else {
echo "Prep Error: " . mysqli_error($dbconnection);
}
//
return $db;
//
}
// main line code
$query = "SELECT * FROM Contacts WHERE First_Name = ?";
$types = "s";
$values = array("Mike");
// accept the stantement handle from the function
$db = fn_preparedStatement($query, $types, $values);
// so now you can use it
mysqli_stmt_store_result($db); //im getting the error on this line - null