CODEWITHSUNDEEP
php
user5237857
user5237857

Reputation: 118

PHP even or odd number script

I'm using this code to check a number if it's an even or an odd number. I'm learning PHP and when I run this code it gives an unusual error.

<html>
    <body>
    <head>
        <title>Judging even and odd numbers</title>
    </head>

    <form method = "post" action="EAO.php" >
       <font size = "20">Please enter a number to check if it is even or odd:</font>
       <input type = "text" name = "number" />
       <input type = "hidden" name="checker" value="true" />
      <input type = "submit" value = "submit" />
    </form>
    <?PHP
      if (isset($_POST['checker'])) {
        $number = $_POST['number'];
        if ($number % 2 == 0 ) {
          echo "The number is Even";
        }
        if($number % 2 == 1 ) {
          echo "The number is odd";
        }
        if ($number == "") {
          echo "Please enter a number";
        }
      }
    ?>
    </body>
</html>

It works good when you input a number but when you submit the form empty, Its says "Please enter a number. The number is even". Help me correct it if I'm doing something wrong. If my code isn't of standard to judge a number, please write your code that is better. Thank you.

Upvotes: 3

Views: 5113

Answers (4)

Red Acid
Red Acid

Reputation: 217

another working version is (improved security):

<?php
if (isset($_POST['checker']) && $_POST['checker'] == 'true') {

   if (empty($_POST['number'])) { // 0 or empty values will be blocked
    echo "Please enter a number";
   } else {
     if (ctype_digit($_POST['number'])) { // only number will be accepted

     $number = $_POST['number'];
     if ($number % 2 == 0 ) {
     echo "The number is Even";
     }
     if($number % 2 == 1 ) {
     echo "The number is odd";
     }

     else {
     echo 'Numbers only please!';
     }
   }

}
?>

Upvotes: 1

Happy Coding
Happy Coding

Reputation: 2525

Try this :

<html>
    <head>
        <title>Judging even and odd numbers</title>
    </head>
    <body>
    <form method = "post" action="" >
       <font size = "20">Please enter a number to check if it is even or odd:</font>
       <input type = "text" name = "number" />
       <input type = "hidden" name="checker" value="true" />
      <input type = "submit" value = "submit" />
    </form>
    <?PHP
      if (isset($_POST['checker'])) {
        $number = $_POST['number'];
    if(empty($number))
    {
        echo "Please enter a number";
    }
    elseif ($number % 2 == 0 ) {
          echo "The number is Even";
        }
        else {
          echo "The number is odd";
        }
      }
    ?>
    </body>
</html>

Place body tag after head tag. If you are posting the form to the same page then action parameter is optional. Use if...elseif...else condition http://www.w3schools.com/php/php_if_else.asp

Upvotes: 1

Dibyendu Konar
Dibyendu Konar

Reputation: 165

HI bro here we go with you code

<html>
<body>
<head><title>Judging even and odd numbers</title></head>

<form method = "post" action="" >
<font size = "20">Please enter a number to check if it is even or odd:</font>
<input type = "text" name = "number" />
<input type = "hidden" name="checker" value="true" />
<input type = "submit" value = "submit" name="submit" />
</form>
<?PHP
if (isset($_POST['submit'])) {
$number = $_POST['number'];
if ($number % 2 == 0 ) {
echo "The number is Even";
}
if($number % 2 == 1 ) {
echo "The number is odd";
}
if ($number == "") {
echo "Please enter a number";
}
}
?>
</body>
</html>

Upvotes: 0

ankur140290
ankur140290

Reputation: 640

Just reversing your logic will do the magic for you. You must always use if () {....} elseif () {....} so that only one logic runs at a given time.

<html>
    <head>
        <title>Judging even and odd numbers</title>
    </head>    
    <body>

    <form method = "post" action="EAO.php" >
       <font size = "20">Please enter a number to check if it is even or odd:</font>
       <input type = "text" name = "number" />
       <input type = "hidden" name="checker" value="true" />
      <input type = "submit" value = "submit" />
    </form>
    <?PHP
      if (isset($_POST['checker'])) {
        $number = $_POST['number'];
        if ($number == "") {
          echo "Please enter a number";
        }            
        else if ($number % 2 == 0 ) {
          echo "The number is Even";
        }
        else if($number % 2 == 1 ) {
          echo "The number is odd";
        }            
      }
    ?>
    </body>
</html>

Upvotes: 3

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