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));

使用strpos和substr:

<?php

$longString = "I have a code snippet written in PHP that pulls a block of text.";
$truncated = substr($longString,0,strpos($longString,' ',30));

echo $truncated;

这将为您提供一个在30个字符后的第一个空格处截断的字符串。

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。

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)), ','))

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);
}