Question

我是PHP新手并尝试创建以下内容，同时尽量减少所需的代码量。 PHP应该显示100的列表然后显示如果数字是/由3,5或3和5.如果没有任何显示什么。

这是我到目前为止所得到的，但任何帮助都会很棒，因为不确定/通过3位和5位，如下所示。

<?php $var = range(0, 100); ?>
  <table>
<?php foreach ($var as &$number) {
 echo " <tr>
    <td>$number</td>
    <td>";

    if($number % 3 == 0)  {
    echo "BY3";
} elseif ($number % 5 == 0) {
    echo "BY5";
} elseif ($number % 3 and 5 == 0) {
        echo "BY3 AND 5";
}
 echo "</td></tr>";
}
?>

  </table>

由于

Answer 1

没有......你应该首先检查它是否可分为15（3x5）（或3和5）以及你可以进行其他检查后：

if($number % 15 == 0)  {
    echo "BY3 AND 5";
} elseif ($number % 5 == 0) {
    echo "BY5";
} elseif ($number % 3 == 0) {
    echo "BY3";
}
 echo "</td></tr>";

?>

因为每个15分的数字也可分为3和5.所以你的最后一次检查永远不会命中

Answer 2

如果我正在阅读你的问题，那么你正在寻找：

if ($number % 3 == 0 && $number %5 == 0) {
        echo "BY3 AND 5";
} elseif ($number % 3 == 0)  {
    echo "BY3";
} elseif ($number % 5 == 0) {
    echo "BY5";
}

替代版本：

echo ($number % 3 ? ($number % 5 ? "BY3 and 5" : "BY 3") : ($number % 5 ? "BY 5" : ""));

Answer 3

$num_count = 100;
    $div_3 = "Divisible by 3";
    $div_5 = "Divisible by 5";
    $div_both = "Divisible by 3 and 5";
    $not_div = "Not Divisible by 3 or 5";

    for($i=0;$i<=$num_count;$i++)
    {
        switch($i)
        {
            case ($i%15==0):
            echo $i." (".$div_both.")</br>";
            break;
            case ($i%3==0):
            echo $i." (".$div_3.")</br>";
            break;
            case ($i%5==0):
            echo $i." (".$div_5.")</br>";
            break;
            default:
            echo $i."</br>";
            break;
        }
    }

Answer 4

无需做三个if语句：

echo "<table border='1'>";

for ($i = 1; $i <= 100; $i++) {

    echo "<tr><td>{$i}</td><td>";

    if ($i % 3 == 0) echo "BY3 ";
    if ($i % 5 == 0) echo "BY5";

    echo "</td></tr>\n";
}
echo "</table>";

Answer 5

更新下面给出的代码

<?php $var = range(0, 100); ?>
<table>
<?php foreach ($var as &$number)
{
echo " <tr>
<td>$number</td>
<td>";

if($number % 3 == 0 &&  $number % 5 == 0) 
{
   echo "BY3 AND 5";
} 
elseif ($number % 5 == 0) 
{
echo "BY5";
}
elseif ($number % 3 == 0) 
{
    echo "BY3";
}
echo "</td></tr>";
}
?>

Answer 6

<?php

if($number % 5 == 0 && $number % 3 == 0)  {
    echo "BY3 AND 5";
} elseif ($number % 5 == 0) {
    echo "BY5";
} elseif ($number % 3 == 0) {
    echo "BY3";
} else{
    echo "NOT BY3 OR 5";
}   
?>

Answer 7

if($number % 15 == 0)  
{
     echo "Divisible by 3 and 5";
} 
elseif ($number % 5 == 0) 
{
    echo "Divisible by 5";
} 
elseif ($number % 3 == 0) 
{
 echo "Divisible by 3";
}

Answer 8

这比较整洁，可以运行了：

<?php

for ($i = 1; $i <= 100; $i++) { 
                if ($i % 15 == 0)
                {
                    echo"Divisible by 3 and 5</br>";
                }
                elseif ($i%3==0)
                {
                    echo"Divisible by 3</br>";
                }
                elseif ($i%5==0)
                {
                    echo"Divisible by 5</br>";
                }
                else
                {
                    echo $i,"</br>";
                }

}
?>

Answer 9

      <?php

for ($i = 1; $i <= 100; $i++) {

    if ($i % 15 == 0) echo "This Number is Divisible by 3 and 5<br>";
    else if ($i % 3 == 0) echo "This Number is Divisible by 3 only<br>";
    else if ($i % 5 == 0) echo "This number is Divisible by 5 only<br>";
    else{
        echo "$i<br>";
    }

}

   ?>

PHP - 如果数字可以被3和5整除，那么回声

9 个答案: