也许这能帮助到一些人:

<?php

    $string = "Your line of text";
    $spl = preg_match("/([, \.\d\-''\"\"_()]*\w+[, \.\d\-''\"\"_()]*){50}/", $string, $matches);
    if (isset($matches[0])) {
        $matches[0] .= "...";
        echo "<br />" . $matches[0];
    } else {
        echo "<br />" . $string;
    }

?>

找到这个问题的完美解决方案是多么棘手，这令人惊讶。我还没有在本页上找到至少在某些情况下不会失败的答案(特别是如果字符串包含换行符或制表符，或者如果单词break不是空格，或者如果字符串有UTF-8多字节字符)。

这里有一个简单的解决方案，适用于所有情况。这里也有类似的答案，但是如果希望它处理多行输入，“s”修饰符很重要，而“u”修饰符使它正确地计算UTF-8多字节字符。

function wholeWordTruncate($s, $characterCount) 
{
    if (preg_match("/^.{1,$characterCount}\b/su", $s, $match)) return $match[0];
    return $s;
}

一个可能的边缘情况是…如果字符串的第一个$characterCount字符中没有任何空格，它将返回整个字符串。如果你喜欢在$characterCount处强制中断，即使它不是一个单词边界，你可以使用这个:

function wholeWordTruncate($s, $characterCount) 
{
    if (preg_match("/^.{1,$characterCount}\b/su", $s, $match)) return $match[0];
    return mb_substr($return, 0, $characterCount);
}

最后一个选项，如果你想让它添加省略号如果它截断字符串…

function wholeWordTruncate($s, $characterCount, $addEllipsis = ' …') 
{
    $return = $s;
    if (preg_match("/^.{1,$characterCount}\b/su", $s, $match)) 
        $return = $match[0];
    else
        $return = mb_substr($return, 0, $characterCount);
    if (strlen($s) > strlen($return)) $return .= $addEllipsis;
    return $return;
}

我相信这是最简单的方法:

$lines = explode('♦♣♠',wordwrap($string, $length, '♦♣♠'));
$newstring = $lines[0] . ' • • •';

我正在使用特殊字符分割文本并剪切它。

Keep in mind whenever you're splitting by "word" anywhere that some languages such as Chinese and Japanese do not use a space character to split words. Also, a malicious user could simply enter text without any spaces, or using some Unicode look-alike to the standard space character, in which case any solution you use may end up displaying the entire text anyway. A way around this may be to check the string length after splitting it on spaces as normal, then, if the string is still above an abnormal limit - maybe 225 characters in this case - going ahead and splitting it dumbly at that limit.

当涉及到非ascii字符时，还有一个类似的警告;包含它们的字符串可能会被PHP的标准strlen()解释为比实际更长，因为单个字符可能占用两个或更多字节，而不是一个字节。如果你只是使用strlen()/substr()函数来分割字符串，你可能会在字符中间分割字符串!如果有疑问，mb_strlen()/mb_substr()更简单一些。

这将返回单词的前200个字符:

preg_replace('/\s+?(\S+)?$/', '', substr($string, 0, 201));

Dave和AmalMurali的代码中添加了IF/ELSEIF语句，用于处理没有空格的字符串

if ((strpos($string, ' ') !== false) && (strlen($string) > 200)) { 
    $WidgetText = substr($string, 0, strrpos(substr($string, 0, 200), ' ')); 
} 
elseif (strlen($string) > 200) {
    $WidgetText = substr($string, 0, 200);
}