Question

I’m getting the following error in display.php **Warning: Invalid argument supplied for foreach(). I am trying to loop through each of these numbers in my display function and generate an image for each from Main. When I echo display.php it outputs nums[0]=$displayid
nums[1]=$displayid and i want to pass the $displayid into my query statement.

generate_numbers()

 var rndNums = new Array(16,17,18,19,20,21,22,23,24,25,26,27,28);
        var generate_numbers = function()
        {
            for(var i=0; i < 13; i++)
            {
                var idx = Math.floor(Math.random()*rndNums.length);
                var rndNum = rndNums[idx];
                rndNums[idx] = rndNums[rndNums.length - 1];
                rndNums.pop();
                document.getElementsByName("nums[]").item(i).value=rndNum;
            }
        }

---

Display.php

$nums = isset($_POST['nums']); // array
              foreach($nums as $key => $displayid)
                        {
                            echo 'nums['.$key.']='.$displayid."<br>\n";
                        }


        $stmt->bind_param("i", $displayid);
        $stmt->execute();
        $stmt->bind_result($image);
        $stmt->fetch();
        header("Content-Type: image/jpeg");
        echo $image; 
        ?>

---

Main.html

<form id="numberForm" action="display.php" method="POST">
        <input type="hidden" name="nums[]">
        <input type="hidden" name="nums[]">

    </form>
    <button onClick="generate_numbers();document.getElementById('numberForm').submit()">submit</button>

Answer 1

$nums = isset($_POST['nums']); // array

returns a boolean (whether the variable is set or not) and not an array.

Maybe you are trying to do this:

$nums = isset($_POST['nums']) ? $_POST['nums'] : array(); // array

Also, note, you should not echo text before the image data, you are going to corrupt your response. And the $display_id in the query would be the last one from the loop