用这个:

下面的代码将删除'，'。如果你有任何其他字符或子字符串，你可以用它来代替'，'

substr($string, 0, strrpos(substr($string, 0, $comparingLength), ','))

//如果你有另一个字符串帐户

substr($string, 0, strrpos(substr($string, 0, $comparingLength-strlen($currentString)), ','))

我以前用过这个

<?php
    $your_desired_width = 200;
    $string = $var->content;
    if (strlen($string) > $your_desired_width) {
        $string = wordwrap($string, $your_desired_width);
        $string = substr($string, 0, strpos($string, "\n")) . " More...";
    }
    echo $string;
?>

我将使用preg_match函数来做到这一点，因为您想要的是一个相当简单的表达式。

$matches = array();
$result = preg_match("/^(.{1,199})[\s]/i", $text, $matches);

表达式的意思是“匹配从长度1-200开始以空格结尾的任何子字符串”。结果在$result中，匹配在$matches中。这就解决了你最初的问题，即以任意空格结束的问题。如果你想让它以换行符结束，将正则表达式更改为:

$result = preg_match("/^(.{1,199})[\n]/i", $text, $matches);

这是一个小修复mattmac的答案:

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

唯一的区别是在$string的末尾添加一个空格。这确保了最后一个词不会像ReX357的注释那样被切断。

我没有足够的代表点添加这作为一个评论。

我知道这很旧，但是…

function _truncate($str, $limit) {
    if(strlen($str) < $limit)
        return $str;
    $uid = uniqid();
    return array_shift(explode($uid, wordwrap($str, $limit, $uid)));
}

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。