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Object PHP and MYSQL queries

98

I'm experiencing mysql with object-oriented php, so i created a db with only one table called users that i want to query using php classes (the insertion function properly works)

This is my php user class:

<?php
class user{
    var $name; 
    var $surname;
    var $email;
    var $age;
    function set_user_properties($pname, $psurname, $pemail, $page){
        include "include.php";
        $sql = "INSERT INTO users (name, surname, email, age)
        VALUES ('$pname', '$psurname','$pemail', '$page')";

        if ($conn->query($sql) === TRUE){
           echo "New record created successfully";
        }else{
           echo "Error: " . $sql . "<br>" . $conn->error;
        }   
         $conn->close();
    }
    function get_user_properties($name){
        include "include.php";
        $sql3 = "SELECT name, surname, email, age
        FROM users
        WHERE name='$name'";
        $result3 = $conn->query($sql3);
            if ($result3->num_rows > 0) {
            // output data of each row
                while($row3 = $result3->fetch_assoc()) {
                    $this->name=$row3["name"];
                    $this->surname=$row3["surname"];
                    $this->email=$row3["email"];
                    $this->age=$row3["age"];
                }
            }
        $conn->close();
        return $this;
    }
}
?>

And this is the page where i want to read all the query results:

<?php

include("class_lib.php");

$pname=$_POST["firstname"];

$user1=new user();
$user1->get_user_properties($pname);
echo $user1->name;
echo $user1->surname;
echo $user1->email;
echo $user1->age;

?>

My question is: when i make a query which is supposed to have more than just one result (like this one), how can i pass all the results to the destination page?

61

Answer

Solution:

The problem is that your class only stores information for one user, and that means you can't really expect it to store multiple users at once.

So the question is: "how do I get an array of users instead of just one?" And the answer to that is to create a class that performs DB queries and parses each row of the query results into a user object

referencing this part of your code:

while($row3 = $result3->fetch_assoc()) {
     $this->name=$row3["name"];
     $this->surname=$row3["surname"];
     $this->email=$row3["email"];
     $this->age=$row3["age"];
}

If you add a constructor function to your user class to instantiate a user with a name, surname, email, and age, you can end up with code in your DB class that looks more like the following

$users = array();
...
while($row3 = $result3->fetch_assoc()) {
    $users[] = new user($row3["name"],
                   $row3["surname"],
                   $row3["email"],
                   $this->age=$row3["age"]);
}
...
return $users;

I hope that makes sense.

258

Answer

Solution:

Thanks everyone for helping, I solved my problem adding a constructor and using a class method. Here is the code, in case someone else will struggle on the same problem.

This is the class_lib.php where is defined the all() method that make an array with the resutus of the query:

<?php
class User{
    var $name;
    var $surname;
    var $email;
    var $age;
   
   function __construct($pname, $psurname, $pemail, $page){
        $this->name=$pname;
        $this->surname=$psurname;
        $this->email=$pemail;
        $this->age=$page;
    }

public static function all(){
        include "include.php";
        $users=array();
        $sql3 = "SELECT name, surname, email, age FROM users";

        $result3 = $conn->query($sql3);
            if ($result3->num_rows > 0) {
                while($row3 = $result3->fetch_assoc()) {
                    $users[] = new User($row3["name"],$row3["surname"], $row3["email"],$row3["age"]);
                }
            }
        $conn->close();
        return $users;
    }
?>

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