PHP - 如果数字可以被 3 和 5 整除，则回显
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PHP - If number is divisible by 3 and 5 then echo
提问by user2039554
I'm new to PHP and trying to create the following whilst minimizing the amount of code needed. PHP should show a list of 100 then display if the number is / by 3, 5 or 3 and 5. If not by any then show nothing.
我是 PHP 新手，并尝试创建以下内容，同时最大限度地减少所需的代码量。PHP 应该显示一个 100 的列表，然后显示数字是 / by 3、5 还是 3 和 5。如果不是，则不显示任何内容。
This is what I've got so far, but any help would be great since not sure about the / by 3 and 5 bit as you can see below.
这是我到目前为止所得到的，但任何帮助都会很好，因为不确定 / by 3 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 ($number % 3 and 5 == 0) {
echo "BY3 AND 5";
}
echo "</td></tr>";
}
?>
</table>
Thanks
谢谢
回答by Luca Rainone
Nope... you should check first if it's divisble for 15 (3x5) (or 3 and 5) and after you can do other checks:
不......你应该首先检查它是否可以被 15 (3x5)（或 3 和 5）整除，然后你可以做其他检查：
if($number % 15 == 0) {
echo "BY3 AND 5";
} elseif ($number % 5 == 0) {
echo "BY5";
} elseif ($number % 3 == 0) {
echo "BY3";
}
echo "</td></tr>";
?>
Because every number divisble for 15 is also divisble for 3 and 5. So your last check could never hit
因为每个可以被 15 整除的数字也可以被 3 和 5 整除。所以你的最后一次检查永远不会命中
回答by Neo
if I'm reading your question correct then you are looking for :
如果我正确阅读您的问题，那么您正在寻找：
if ($number % 3 == 0 && $number %5 == 0) {
echo "BY3 AND 5";
} elseif ($number % 3 == 0) {
echo "BY3";
} elseif ($number % 5 == 0) {
echo "BY5";
}
Alternative version :
替代版本：
echo ($number % 3 ? ($number % 5 ? "BY3 and 5" : "BY 3") : ($number % 5 ? "BY 5" : ""));
回答by pooja haldankar
$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 $i."</br>";
break;
}
}
回答by pbarney
No need to do three if statements:
不需要做三个 if 语句：
echo "<table border='1'>";
for ($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>";
回答by Deep123
Update the code as given below
更新代码如下
<?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 ($number % 3 == 0)
{
echo "BY3";
}
echo "</td></tr>";
}
?>
回答by Prasanth Bendra
<?php
if($number % 5 == 0 && $number % 3 == 0) {
echo "BY3 AND 5";
} elseif ($number % 5 == 0) {
echo "BY5";
} elseif ($number % 3 == 0) {
echo "BY3";
} else{
echo "NOT BY3 OR 5";
}
?>
回答by KAS
if($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";
}
回答by Mr-Programs
This is neater and completed to be run:
这是更整洁和完整的运行：
<?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 $i,"</br>";
}
}
?>
回答by Noobie
<?php
for ($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>";
}
}
?>
