具有相同名称的php多个下拉菜单无法获取值

Question

okay, im new in this site also new in php and can't get the logic on this,

i have a product page that shows the name, quantity and an add to cart button in each row of product

i made this just cutted of some code

    while($showProducts = mysql_fetch_array($products))
            {
            $currenQuantity = $showProducts['current_quantity'];
            $prodid = $showProducts['product_id'];
    echo"<select name='quan'>";

                for ($x=0;$x<=$currenQuantity;$x++)
            {
                if($currenQuantity != 0)
                {
                echo "<option value=$x> $x </option>";
                }
            }
     echo"</select><br/>";
     }

now the problem is every time i tried to get the value by using $_POST['quan'] the value that i always get is the default value 1 even i select a different value of quantity of a certain product, and i'm blanked with ideas.

Answer 1

You can't use the same name for an input/select field in a form. You have to specify a diffrent name or create an indexed array:

<select name="quan[$prodid]">

You can acces it via

$_POST['quan'][$prodid]