帮助解析PHP中的字符串

我认为这应该完成你的目标:

$string = 'Bob is a boy. Bob is 1000 years old! <b>Bob loves you!</b> Do you love bob?'; 

// parser
$array = preg_split('/[\.|\!\?]|[\s]*<b>|<\/b>[\s]*/', $string, 0, PREG_SPLIT_NO_EMPTY | PREG_SPLIT_OFFSET_CAPTURE);
foreach ($array as $key => $element) $array[$key] = trim($element[0]).substr($string,$element[1]+strlen($element[0]),1);

print_r($array);

它产生:

Array
(
    [0] => Bob is a boy.
    [2] => Bob is 1000 years old!
    [4] => Bob loves you!
    [6] => Do you love bob?
)

如果没有人能提供更好的解决方案,这几乎是可行的:

(?:<b>|[.!?]*)((?:[^<]+?)(?:[.!?]+|</b>))\s+

只有它会回来 Bob loves you!</b> strip_tags() 结果我想。。。

分而治之?

假设$myString是您的字符串。。。

先拿你的东西:

preg_match (" /(.*?)<b>(.*?)<\/b>(.*?)/", $myString);

现在你有1美元,2美元和3美元

$firstMatches = preg_split("/[\.\!\?]/", $1);

$lastMatches = preg_split("/[\.\!\?]/", $3);

然后拿回你的标点符号:

function addPunctuation($matches, $myString)
{
    $punctuadedResults = array();
    foreach($matches as $match)
    {
       $position = strpos( $myString, $match);
       #position is the offset of the start of your match. Find the character after your match.
       $punctMark = substr($myString, $position + length($match), 1);
       $punctuadedResults[] = $match . $punctMark;

    }
    return $punctuadedResults;
}


$allMatches = addPunctuation($firstMatches, $myString);
$allMatches[] = $2;

$allMatches = array_merge($allMatches, addPunctuation($lastMatches, $myString) );