PHP undefined index but defined earlier on

Question

I am making a script which is going to add together a bunch of inputs. I am almost there, but when testing the script i seem to get the alert: Notice: Undefined index: number in C:\xampp\htdocs\Archers.php on line 25. Even though I have declared it earlier and even when using it now. The script is below. Thanks in advance.

<html>
<head>
</head>
<body>
  <form method="post">
    Enter how many values you would like to enter:
    <input type="number" name="number">
    <input type="submit" name="submit">
  </form>
  <?php
  $number = 0;
  $result = 0;
  if (isset($_POST["submit"])){
    $number = $_POST['number'];
    $x = 0;
    echo "<form method=\"post\">";
    while ($x != $number) {
      echo "Enter score: <input type=\"text\" name=\"".'a'.$x."\"><br>";
      $x = $x + 1;
    }
    echo "<input type=\"submit\" name=\"submit2\"></form>";
  }
  if (isset($_POST["submit2"])){
    $y = 0;
    $number = $_POST['number'];
    while ($y != $number){
      $value = $_POST["a".$y];
      $result = $result + $value;
      $y = $y + 1;
    }
    echo $result;
  }
  ?>
</body>
</html>

The error is not on about `$number` its on about `$_POST['number']` the second form is sending `a123` not `number` — Lawrence Cherone, Mar 21 '18 at 17:50
it's because `$_POST['number']` isnt defined in the second form submit2, you should either use a global variable `number` or include the first input in the second form — Abr001am, Mar 21 '18 at 17:52

Tafit · Answer 1 · 2018-03-21T18:18:02.953

The error you received is because the index of array is undefined - it simply does not exist. That's probably $number = $_POST['number'] line in your code (25th).

It surely means that you did not fill number input in your form, as you have no validation to a possibility that number field is not passed.

EDIT: I can see that you have created second form, which has submit2 key in it. When you POST for the first time, you will receive both submit and number keys (in lines 14 - 22). Then your code creates second form, which makes another POST with submit2 but without number field in it. Therefore, you don't have number key in lines 23 - 32.

EDIT2: You can, for example, change the first if to be like this one:

if (isset($_POST["submit"])){
    $number = $_POST['number'];
    $x = 0;
    echo "<form method=\"post\">";
    while ($x != $number) {
        echo "Enter score: <input type=\"text\" name=\"".'a'.$x."\"><br>";
        $x = $x + 1;
    }
    echo "<input type=\"hidden\" name=\"number\" value=\"" . $number . "\">";
    echo "<input type=\"submit\" name=\"submit2\"></form>";
}

Ok, I have removed the second line of getting the post. But i still can't get the variable to transfer across from the first isset... — Blatherwick, Mar 21 '18 at 18:07
nvm I fixed it using sessions. each form refreshed the page causinog the variables to be lost. — Blatherwick, Mar 25 '18 at 10:22

score 0 · Answer 2 · answered Mar 21 '18 at 17:56

Please try this code . it will remove your error

 <html>
    <head>
    </head>
    <body>
      <form method="post">
        Enter how many values you would like to enter:
        <input type="number" name="number">
        <input type="submit" name="submit">
      </form>
      <?php
      $number = 0;
      $result = 0;
      if (isset($_POST["submit"]) && isset( $_POST['number']) ){
        $number = $_POST['number'];
        $x = 0;
        echo "<form method=\"post\">";
        while ($x != $number) {
          echo "Enter score: <input type=\"text\" name=\"".'a'.$x."\"><br>";
          $x = $x + 1;
        }
        echo "<input type=\"submit\" name=\"submit2\"></form>";
      }
      if (isset($_POST["submit2"]) && isset( $_POST['number'])){
        $y = 0;

        $number = $_POST['number'];

        while ($y != $number){
          $value = $_POST["a".$y];
          $result = $result + $value;
          $y = $y + 1;

        echo $result;
          }
      }
      ?>
    </body>
    </html>

PHP undefined index but defined earlier on

2 Answers2