PHP 字符串 - 删除数组或单词并检查其他

Question

我有一个字符串:

$data = "String contains works like apples, peaches, banana, bananashake, appletart";

我还有 2 个标准数组，如下所示，其中包含许多单词:

$profanityTextAllowedArray = array();
$profanityTextNotAllowedArray = array();

例如:

$profanityTextAllowedArray
(
    [0] => apples
    [1] => kiwi
    [2] => mango
    [3] => pineapple
)

如何获取字符串 $data 并首先从 $profanityTextAllowedArray 中删除任何单词，然后检查字符串 $data 中是否有任何单词$profanityTextNotAllowedArray 中应标记哪些单词？

Answer 1

$list = explode( ' ', $data );

foreach( $list as $key => $word ) {
  $cleanWord = str_replace( array(','), '', $word ); // Clean word from commas, etc.
  if( !in_array( $cleanWord, $profanityTextAllowedArray ) ) {
    unset($list[$key]);
  }
}

$newData = implode( ' ', $list );

如果清楚的话请告诉我。