Php - return count（）in words [英] Php - return count() in words

问题描述

我的代码中有以下内容：

I have the following in my code:

 $my_count = count($total_elements[$array_object]);

$ my_count 现在包含元素数在 $ total_elements [$ array_object] 。我想将此数字转换为其对应的自然数（零，一，二，...）

$my_count now contains the number of elements in $total_elements[$array_object]. I want to convert this number into its corresponding natural number (zero, one, two, ...)

在此特定情况下， ：

$numbers  = array(
    2  => 'two',
    3  => 'three',
    4  => 'four',
    5  => 'five',
    6  => 'six',
);

如何获取给定数组中的元素数量，然后 echo 相应的自然数（人类可读数）从数组？或者更好 - 有更好的方法吗？

How to retrieve the number of elements in a given array and then echo the corresponding natural number (human readable number) from the array? Or better yet - is there a better way of doing this?

（我发现了一些功能或

(I have found some functions or classes to do just that - but now those are way too bloated for the simple case I need now)

现在，当然可以用 switch（）：

switch ($my_count) {
    case 0:
        echo "zero";
        break;
    case 1:
        echo "one";
        break;
    case 2:
        echo "two";
        break;
      // etc...
}

优雅的我。

我相信有一个更优雅的方式实现，即使现在我只有5个数字，我想在其他情况下重新使用一些功能。

I am sure that there is a more elegant way of achieving that, and even though now I have only 5 numbers, I would like to have some function to re-use in other cases .

（对不起，如果这是一个愚蠢的问题，但在SE或Google上搜索与关键字 PHP - 字和

(I am sorry if this is a stupid question - but - searching here on SE or Google with keywords PHP - words and count() I only found answers related to counting words in string)

推荐答案

如果只有少数几个：

$number = count($yournameit);
$count_words = array( "zero", "one" , "two", "three", "four" );
echo $count_words[$number];

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