PHP字符串 - 删除阵列或词语检查其他

Question

我有一个字符串：

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

我也有2个STD阵列如下包含多个字的：

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

例如：

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

我怎样才能把字符串$data并首先从$profanityTextAllowedArray和删除任何话然后检查字符串$data中$profanityTextNotAllowedArray中应标记的任何字词？

Answer 1

像这样的东西可以帮助你：

$data = "apples, peaches, banana, bananashake, appletart"; 

$allowedWords = array('apples', 'peaches', 'banana'); 
$notAllowedWords = array('foo', 'appletart', 'bananashake'); 

$allowedWordsFilteredString = preg_replace('/\b('.implode('|', $allowedWords).')\b/', '', $data); 

$wordsThatNeedsToBeFlagged = array_filter($notAllowedWords, function ($word) use ($allowedWordsFilteredString) { 
    return false !== strpos($allowedWordsFilteredString, $word); 
}); 

var_dump($wordsThatNeedsToBeFlagged); 

Answer 2

$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); 

让我知道，如果是明确的。