我发现这很有效:

function abbreviate_string_to_whole_word($string, $max_length, $buffer) {
    if (strlen($string) > $max_length) {
        $string_cropped = substr($string, 0, $max_length - $buffer);
        $last_space = strrpos($string_cropped, " ");
        if ($last_space > 0) {
            $string_cropped = substr($string_cropped, 0, $last_space);
        }
        $abbreviated_string = $string_cropped . " ...";
    }
    else {
        $abbreviated_string = $string;
    }
    return $abbreviated_string;
}

缓冲区允许您调整返回字符串的长度。

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。

我发现这很有效:

function abbreviate_string_to_whole_word($string, $max_length, $buffer) {
    if (strlen($string) > $max_length) {
        $string_cropped = substr($string, 0, $max_length - $buffer);
        $last_space = strrpos($string_cropped, " ");
        if ($last_space > 0) {
            $string_cropped = substr($string_cropped, 0, $last_space);
        }
        $abbreviated_string = $string_cropped . " ...";
    }
    else {
        $abbreviated_string = $string;
    }
    return $abbreviated_string;
}

缓冲区允许您调整返回字符串的长度。

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);

你可以试试这个

substr( $str, 0, strpos($str, ' ', 200) ); 

我是这样做的:

$string = "I appreciate your service & idea to provide the branded toys at a fair rent price. This is really a wonderful to watch the kid not just playing with variety of toys but learning faster compare to the other kids who are not using the BooksandBeyond service. We wish you all the best";

print_r(substr($string, 0, strpos(wordwrap($string, 250), "\n")));