Question

请帮助我使用代码来检查表中的数字是否为素数：这是我的代码。我使用了isPrime函数，但是我得到一个错误，当我在表中调用函数时返回单词“isPime”

<?php 
      $x="num1"; 
      $y="num2"; 

      if(isset($_POST['Submit'])) { 

          $start=$_POST['num1']; 
          $end=$_POST['num2']; 

          echo "<table width='250' cellpadding='2' cellspacing='4' border='1'>"; 
          echo "<tr><td>Number</td><td>Odd or Even?</td><td>Prime?</td></tr>"; 

          for ($start;$start<=$end;$start++) { 

              $answer=$start;  
              $number=$answer; 
              $check="text"; 
              $num=$answer; 

                  if($number%2==0) { 

                      $check="Even"; 

                  } else { 

                      $check="Odd"; 

                  } 

                { } 

           echo "<tr><td>$answer</td><td>$check</td><td>

    <?php ?>

    </td></tr>"; 

        }  echo "</table>"; 

   } ?>

Answer 1

您错过了整个主要检查，您甚至没有在第三栏中打印任何内容。

也有人似乎在第一次编辑中删除了您的代码，并在第二次编辑中更改了它。我以某种方式获得了第一个编辑的副本，可以作一个例子。

这是您代码中的一个工作示例：

<?php 

$x="num1"; $y="num2"; 
if(isset($_POST['Submit'])) 
{ 
    $start=$_POST['num1']; $end=$_POST['num2']; 
    echo "<table width='250' cellpadding='2' cellspacing='4' border='1'>"; 
    echo "<tr><td>Number</td><td>Odd or Even?</td><td>Prime?</td></tr>"; 
    for ($start;$start<=$end;$start++) 
    { 
        $answer=$start; $number=$answer; $check="text"; $num=$answer; 
        if($number%2==0) 
            { $check="Even"; } 
        else 
            { $check="Odd"; } 
        $prime = (isPrime($number)? 'Yes':'No');
        echo "<tr><td>$answer</td><td>$check</td><td>$prime</td></tr>"; 
    } 
    echo "</table>"; 
} 
function isPrime($num) {
    //1 is not prime. See: http://en.wikipedia.org/wiki/Prime_number#Primality_of_one
    if($num == 1)
        return false;

    //2 is prime (the only even number that is prime)
    if($num == 2)
        return true;

    /**
     * if the number is divisible by two, then it's not prime and it's no longer
     * needed to check other even numbers
     */
    if($num % 2 == 0) {
        return false;
    }

    /**
     * Checks the odd numbers. If any of them is a factor, then it returns false.
     * The sqrt can be an aproximation, hence just for the sake of
     * security, one rounds it to the next highest integer value.
     */
    for($i = 3; $i <= ceil(sqrt($num)); $i = $i + 2) {
        if($num % $i == 0)
            return false;
    }

    return true;
}
?>

Answer 2

试试这个

<?php 
  $x="num1"; 
  $y="num2"; 

  if(isset($_POST['Submit'])) { 

      $start=$_POST['num1']; 
      $end=$_POST['num2']; 

      echo "<table width='250' cellpadding='2' cellspacing='4' border='1'>"; 
        echo "<tr><td>Number</td><td>Odd or Even?</td><td>Prime?</td></tr>"; 

      for ($start;$start<=$end;$start++) { 

          $answer=$start;  
          $number=$answer; 
          $check="text"; 
          $num=$answer; 

          if($number%2==0) 
          { 
              $check="Even"; 
          } 
          else 
          { 
              $check="Odd"; 
          } 

        if(isPrimeNumber($number))
        {   
            $pirme_status = "Yes";  
        }
        else
        {
            $pirme_status = "No";   
        }   

       echo "<tr><td>$answer</td><td>$check</td><td>$pirme_status</td><</tr>";



    }  echo "</table>"; 

}

和PHP函数是

function isPrimeNumber($number)
{
    $flag = true;
    $max_count = ceil($number/2);
    for($i=2; $i<$max_count; $i++)
    {
        if($number%$i==0)
        {
            $flag = false;
                    break;
        }
    }
    return $flag;
}
?>

PHP函数isPrime（）不起作用

2 个答案: