Question

我使用此代码检查数字是偶数还是奇数。我正在学习PHP，当我运行这段代码时，它会产生一个不寻常的错误。

<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>

输入数字时效果很好但是当您将表单提交为空时，它表示＆＃34;请输入一个数字。这个数字甚至是＃34;如果我做错了，请帮我纠正。如果我的代码不是判断数字的标准，请编写更好的代码。谢谢。

Answer 1

只需颠倒你的逻辑就能为你做出神奇的事。您必须始终使用if () {....} elseif () {....}，以便在给定时间只运行一个逻辑。

<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>

Answer 2

试试这个：

<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>

在head标签后放置body标签。如果要将表单发布到同一页面，则action参数是可选的。使用if ... elseif ... else condition http://www.w3schools.com/php/php_if_else.asp

Answer 3

另一个工作版本是（改进的安全性）：

<?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!';
     }
   }

}
?>

Answer 4

在这里，我们和你一起去代码

<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>

PHP偶数或奇数脚本

4 个答案: