Question

我的目标是每7个单词分割一次字符串。如果第7个单词的逗号（，），请移至下一个带有句点（。）或感叹号（！）的单词

到目前为止，我已经能够将字符串除以7个单词，但无法检查它是否包含逗号（，）或使用移至下一个单词。或！

   $string = "I have a feeling about the rain, let us ride. He figured I was only joking.";
   $explode = explode(" ", $string);
   $chunk_str = array_chunk($explode, 7);

   for ($x=0; $x<count($chunk_str); $x++)
   {
       $strip = implode(" ",$chunk_str[$x]);
       echo $strip.'<br><br>';
   }

我希望

I have a feeling about the rain, let us ride.

He figured I was only joking.

但是实际输出是

I have a feeling about the rain,

let us ride. He figured I was

only joking.

Answer 1

这是做您想要的事情的一种方法。一次遍历单词列表一次，每次7次。如果第7个单词以逗号结尾，请增加列表指针，直到到达以句点或感叹号（或字符串结尾）结尾的单词为止。输出当前块。当您到达字符串的末尾时，输出所有剩余的单词。

$string = "I have a feeling about the rain, let us ride. He figured I was only joking.";
$explode = explode(' ', $string);
$num_words = count($explode);
if ($num_words < 7) {
    echo $string;
}
else {
    $k = 0;
    for ($i = 6; $i < count($explode); $i += 7) {
        if (substr($explode[$i], -1) == ',') {
            while ($i < $num_words && substr($explode[$i], -1) != '.' && substr($explode[$i], -1) != '!') $i++;
        }
        echo implode(' ', array_slice($explode, $k, $i - $k + 1)) . PHP_EOL;
        $k = $i + 1;
    }
}
echo implode(' ', array_slice($explode, $k)) . PHP_EOL;

输出：

I have a feeling about the rain, let us ride. 
He figured I was only joking.

Demo on 3v4l.org

PHP将字符串拆分到下一个句点（。）

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