Mysql to mysqli or PDO - php

Question

I have a project here that has a LOT of this kind of code below.
It is everything in mysql and I want to change do mysqli or PDO.
The problem is that the code make the fields of the table be variables, inside the loop. So, If I need to change to something different, I will have to re-write thousand of lines of code. Because inside the loops below I will change every call of the var. Example: $row["product_name"] instead of $product_name
This is the code:
$query = "SELECT product_name, product_price from products";
$result = mysql_query($query_sql);
while($row = mysql_fetch_array($result)) {
$j = mysql_num_fields($query);
for($i=0;$i<$j;$i++) {
$k = mysql_field_name($query,$i);
$$k = $row[$k];
}
//Here, inside the loop, I use $product_name instead of $row["product_name"]
}
It there a way to change the code to do the same with mysqli or PDO? I want to keep using $field_name instead of $row["field_name"].

Answer 1

mysqli object method shown. I don't recommend doing it this way, but it can be done. I would highly suggest you learn how to deal with either a result object or array directly. Dynamically setting variables like this will introduce a lot of unnecessary overhead.
$result = $mysqli->query($query_sql);
while($row = $mysqli->fetch_assoc($result)) {
extract($row);
//Here, inside the loop, I use $product_name instead of $row["product_name"]
}

Answer 2

Here is what you are looking for: http://www.doctrine-project.org/
I specifically do not like this specific project, because it tends to be slow with larger amounts of data, but provides what you need:
You specify domain classes (understand map database lines to php objects) and you only have to specify the exact name of the table column, if it does not match up with the class variable name.
So you set every column once for every object type and you use that definition throughout you code, you are making a request for the object, that has a mapping to the database.
so it goes like: $entitiy_manager->getReference('myobject', $objectid = 1 );
Creating a new object is just adding a new instance of the class.
If I understood you properly this should be you are looking for.

Answer 3

Thank you for every answer and opinion, but I just found the solution for my question.
I will leave here, for future reference.
This code is in mysql:
$result = mysql_query($query);
while($row = mysql_fetch_array($result))
{
$j = mysql_num_fields($query);
for($i=0;$i<$j;$i++) {
$k = mysql_field_name($query,$i);
$$k = $row[$k];
}
}
And this would be the same in mysqli:
$result = mysqli_query($link, $query);
while($row = $query->fetch_assoc())
{
$j = mysqli_num_fields($query);
for($i=0;$i<$j;$i++) {
$k = mysqli_fetch_field_direct($query, $i)->name;
$$k = $row[$k];
}
}