一个包装内同时3物体或更少/的foreach在PHP [英] Wrapping 3 objects or less inside a while / foreach in PHP

问题描述

简单的问题。我有21个元素的数组，并显示出＆LT内部每三个人; DIV＆GT; 块。在code是的的东西的是这样的：

Simple question. I have an array of 21 elements, and show every three of them inside a <div> block. The code is something like this:

<?php
$faces= array(
  1 => 'happy',
  2 => 'sad',
  (sic)
  21 => 'angry'
);

$i = 1;
foreach ($faces as $face) {
  echo $face;
  $i++;
}

?>

问题在于，当这阵没有21元，有时它得到24，其他时间17.我如何包装每三个人，独自包裹休息吗？我想出了用开关和情况，但仅当只有21个元素。我想我可以<一个href=\"http://stackoverflow.com/questions/665135/find-the-last-element-of-an-array-while-using-a-foreach-loop-in-php\">count他们事先并把结束了最后一个（即使它是一组一个元素的）。

The problem lies when this array doesn't have 21 elements, sometimes it gets 24, an other times 17. How I wrap every three of them, and wrap alone the rest? I came up with using switch and case, but that works only when there are 21 elements only. I think I could count them beforehand and put a closing in the last one (even if it is a group of one element).

推荐答案

您已经有大部分是在这里。所有你缺少的东西来测试，如果你已经准备好包裹。你递增所以在 $ I ，请尝试：

You already have most of it here. All you're missing is something to test if you're ready to wrap. So before you increment $i, try:

$i = 1;

foreach ($faces as $face)
{
    echo $face;

    if ($i % 3 == 0)
    {
        echo "<br />"; // or some other wrapping thing
    }
    $i++;
}

这将确保你每包3个面，留在最后的单位剩余部分。

This will ensure you're wrapping every 3 faces, leaving any remainder in the final unit.

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