Return not working on empty check function PHP

Question

Thank you for help. Anyway, i'm a student and i'm stuck making a function that does a simple empty() check on a key in a array. If it is empty() -> it will fill in the error and print it on a div element.

This is my code:

 function validateData($data, $formErrors) {

    if (empty($data)) {
        $formErrors = 'name is required.';
      }
    echo $formErrors;
}

Executing the function:

  $data = $_POST;
  $formErrors = [];

  // Elementen valideren
  validateData($data['name'], $formErrors['name'] = '');

This is the thing that i want to achieve:

if (empty($data['surname'])) {
    $formErrors['surname'] = 'Surname is required.';
  }

Looks like it is not returning the value of that key or whatelse?

Thanks alot!

Barmar · Accepted Answer · 2020-04-21 20:31:49Z

1

You need to make it a reference parameter with & if you want to assign it in the function.

function validateData($data, &$formErrors) {

    if (empty($data)) {
        $formErrors = 'name is required.';
      }
    echo $formErrors;
}

And you can't use an expression when passing a reference parameter, it has to be just the variable. It should be:

$formErrors = ['name' => ''];
validateData($data['name'], $formErrors['name']);

answered Apr 21, 2020 at 20:31

Barmar

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