arrow_back Where is the error in php code?

Question

Dear Gurus!
Taking the first steps to learn PHP - write calulator.
The first task set is to check the existence of data coming from forms. I.e. if any field is left blank, display a message below the form - "something not given!"
The script is written, but the message does not appear. Where did I go wrong?

<form action='calc.php' method="post">
      <label>Число 1:</label>
      <br />
      <input name='num1' type='text' />
      <br />
      <label>Оператор: </label>
      <br />
	  <label for="operator">
		    <select name="operator" id="operator">
			    <option value="+">+</option>
			    <option value="-">-</option>
		    </select>
      <br />
      <label>Число 2: </label>
      <br />
      <input name='num2' type='text' />
      <br />
      <br />
      <input type='submit' value='Считать'>
    </form>
    <br />
    <br />
<?php

$num1 = int($_POST['num1']);
$num2 = int($_POST['num2']);

if (empty($num1 or $num2)) {
	echo "Что-то не передано!";
}

?>

Answer 1

Did the option to display the message under the field that is not filled:

<?php
if($_SERVER['REQUEST_METHOD'] == 'POST') {
    $require = ['num1', 'num2']; //поля которые надо заполнить.
    $errors[] = ''; //массив в котором сохраним ошибки
    
    foreach($require as $key) {
        if(empty($_POST[$key])) {
            $errors[$key] = 'Это поле надо заполнить!';
        }
    }
}
?>

<form action='calc.php' method="post">
      <label>Число 1:</label>
      <br />
      <input name='num1' type='text' />
      <?php
        $msg = isset($errors['num1']) ? "Введите первое число" : "";
        echo $msg;
      ?>
      <br />
      <label>Оператор: </label>
      <br />
    <label for="operator">
        <select name="operator" id="operator">
          <option value="+">+</option>
          <option value="-">-</option>
        </select>
      <br />
      <label>Число 2: </label>
      <br />
      <input name='num2' type='text' />
      <?php
        $msg = isset($errors['num2']) ? "Введите первое число" : "";
        echo $msg;
      ?>
      <br />
      <br />
      <input type='submit' value='Считать'>
    </form>
    <br />
    <br />
<?php
    if (count($errors) > 0) {
        echo 'Заполните все поля!';
    }
?>

Сначала проверим, была ли отправлена форма, а потом проверяем заполнены ли все обязательные поля из формы.

Answer 2

It is specifically for your piece of code. Ideally, you need to watch everything and rewrite from scratch.

<?php

$num1 = intval(isset($_POST['num1']) ? $_POST['num1'] : 0);
$num2 = intval(isset($_POST['num2']) ? $_POST['num2'] : 0);

if (empty($num1) || empty($num2)) {
  echo "Что-то не передано!";
}

?>

Comments

When you open the form ( without sending the data through the form) it appears a message that "something not given!", although this message should appear when I pressed the pedal sabmit, and sent only one number of the two.
Where could be the problem?

Answer 3

1. Since php 5.5 as You did, but for earlier versions, and empty must be a variable, not an expression, so depending on the php version, maybe worth a try
if (empty($num1) || empty($num2)) {
2. You are not true bring the data to integer

$num1 = int($_POST['num1']); // не верно
$num1 = (int) $_POST['num1'];  //верно

Comments

0ldn0mad the problem , as it should be, but if You don't need to withdraw, you should check for the existence of a post request.

if(isset($_POST)){
   $num1 = int() $_POST['num1'];
   // и так далее

Кстати если бы Вы включили вывод всех ошибок, что я крайне рекомендую сделать, особенно пока изучаете php. Вы бы уже увидели эту проблем в виде ошибки. И вообще при error_all меньше "почему так".

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When you open the form ( without sending the data through the form) it appears a message that "something not given!", although this message should appear when I pressed the pedal sabmit, and sent only one number of the two.
Where could be the problem?

Answer 4

if (isset($_POST['num1']) && isset($_POST['num2']) && isset($_POST['operator'])) {
  $num1 = intval($_POST['num1']);
  $num2 = intval($_POST['num2']);
  $operator = $_POST['operator'];

  switch ($operator) {
     case '+':
         echo $num1 + $num2;
         break;
     case '-':
         echo $num1 - $num2;
         break;
  }
   
} else {
  echo "Что-то не передано!";
}

Comments

0ldn0mad ,
because when we just print the form, we check the $_POST, it is empty, so a message is displayed.
Wrap the whole if else in extra

if ($_SERVER['REQUEST_METHOD'] === 'POST') {
    // …
}

Тем самым, мы будем проверять $_POST и вычислять, только если пользователь нажал на Отправить, а не всегда.

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Yan-s ,
Well, what makes You think that is broken? All very excellent! =)

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When you open the form ( without sending the data through the form) it appears a message that "something not given!", although this message should appear when I pressed the pedal sabmit, and sent only one number of the two.
Where could be the problem?

---

all the training the man broke off