$_POST gives incorrect HTML input field value

Question

I'am pissed at this code. Could someone explain where the problem is?

I have this HTML form:

<form action="item_add.php" method="POST" id="formadd">
    SKU: <input type="text" name="sku" id="form2"><br/>
    Name: <input type="text" name="name" id="form2"><br/>
    Price: <input type="text" name="price" id="form2"><br/>
    Type: <select name="type" id="form2">
        <option value="1">DVD-Disc</option>
        <option value="2">Book</option>
        <option value="3">Furniture</option>
    </select>
</form>

And this simple code for checking if everything works as I am new to this:

if ($_SERVER['REQUEST_METHOD'] == 'POST'){//&&( isset( $_POST['submit1'] )){
    $item = new c_item;
    $item->$sku = mysqli_real_escape_string($link, $_POST['sku']);
    $item->$name = mysqli_real_escape_string($link, $_POST['name']);
    $item->$price = floatval(mysqli_real_escape_string($link, $_POST['price']));
    $item->$type = intval(mysqli_real_escape_string($link, $_POST['type']));

    echo "sku: ".$item->$sku."<br/>";
    echo "name: ".$item->$name."<br/>";
    echo "price: ".$item->$price."<br/>";
    echo "type: ".$item->$type."<br/>";
}

Problem is that it outputs me :

sku: Name

name: Name

price: 2

type: 2

when sku and price is inserted something different.

Why are u using an object? A simple array would do the same. And id of html elements have to be unique! — C4pt4inC4nn4bis, Jun 27 '18 at 14:59
Your using variable variables - `$item->$sku` should probably be `$item->sku` (minus the $) — Nigel Ren, Jun 27 '18 at 15:01

score 1 · Accepted Answer · answered Jun 27 '18 at 15:05

Because you didn't paste whole code this will be just assumption

echo "sku: ".$item->$sku."<br/>";
            echo "name: ".$item->$name."<br/>";
            echo "price: ".$item->$price."<br/>";
            echo "type: ".$item->$type."<br/>";

there you execute dynamic name of variables so $name, can be the same as $sku and $price can be the same as $type

I don't know your c_item but probably what you want to do is declare in this class or something similar

class c_item 
{
    public $name;
    public $price, 
    public $sku;
    public $type;
}

and then in your code you need to use normal access to properties over dynamic with $

 $item = new c_item;

 $item->sku = mysqli_real_escape_string($link, $_POST['sku']);
 $item->name = mysqli_real_escape_string($link, $_POST['name']);
 $item->price = floatval(mysqli_real_escape_string($link, $_POST['price']));
 $item->type = intval(mysqli_real_escape_string($link, $_POST['type']));

 echo "sku: ".$item->sku."<br/>";
 echo "name: ".$item->name."<br/>";
 echo "price: ".$item->price."<br/>";
 echo "type: ".$item->type."<br/>";

notice lack of $ in property name of an object

Class asumption is correct, but it doesnt output any value if no $ is used — Rihards Egle, Jun 27 '18 at 15:08
It worked, but I still don't understand why $sku and $price would copy the values of their next class property. — Rihards Egle, Jun 27 '18 at 15:24
@RihardsEgle imagine following scenario '$sku='price'; $price = 'price'; $this->$sku; => is equal to $this->price; $this->$price is equal to $this->price so both pointing to the same variable. if you do it without second $ it points to the property of the object — Robert, Jun 27 '18 at 19:13

$_POST gives incorrect HTML input field value

1 Answers1