Question

可能重复：
  How to get first x chars from a string, without cutting off the last word?
  php trim a string

我的问题是我需要将一个字符串剪切成少于30个字符，同时还要确保字符串完成一个单词。

我一直在用这个：

$slidedescription = substr($slidedescription,0,30).'...';

问题是它可以切入字符串中间字。有没有简单的方法来确保它完成一个单词，但它长度少于30个字符？

Answer 1

我试图覆盖尽可能多的意外情况：

<?

    $string_1  = "anavaragelongword shortword";
    $string_2 = "averylongwordwhichisprobablymorethan30characters word1";
    $string_3 = "word2 word3 averylongwordwhichisprobablymorethan30characters";
    $string_4 = "three avarege words";

    $char_length = 30;
    echo slidedescription($string_1, $char_length);
    echo slidedescription($string_2, $char_length);
    echo slidedescription($string_3, $char_length);
    echo slidedescription($string_4, $char_length);

    function slidedescription($string, $char_length)
    {   
    $total_length = null;
    $slidedescription = null;

    $length = strlen($string);
    if($length>$char_length) { 

    $array = explode(" ", $string);
    foreach ($array as $key => $value) {
    $value_length[$key] = strlen($value);
    $total_length = $total_length + $value_length[$key];
    if ($total_length<=$char_length) {
    $slidedescription .= $value." "; 
    }
    if (!$slidedescription) {
    $slidedescription = substr($value,0,$char_length).'...';
    }
    }
    } else {
    $slidedescription = $string; 
    }

    $last = trim($slidedescription).'... <br />';
    return $last;
    }

Answer 2

您可以尝试使用

查找空格发生的最后位置

strrpos

你可以像这样使用它：

$spacelocation = strrpos($slidedescription, " ");

if($spacelocation != false)
{
    if($spacelocation < 30)
    {
        $slidedescription = substr($slidedescription,0,$spacelocation).'...';
    }
}
else
{
    $slidedescription = substr($slidedescription,0,30).'...';
}

PHP - 将字符串剪切为一定数量的字符

2 个答案: