[英]PHP String - Remove Array or Words & Check for Others
我有一個字符串:
$data = "String contains works like apples, peaches, banana, bananashake, appletart";
我還有2個std數組,其中包含許多單詞:
$profanityTextAllowedArray = array();
$profanityTextNotAllowedArray = array();
例如:
$profanityTextAllowedArray
(
[0] => apples
[1] => kiwi
[2] => mango
[3] => pineapple
)
如何獲取字符串$data
並首先從$profanityTextAllowedArray
刪除任何單詞,然后在字符串$data
檢查$profanityTextNotAllowedArray
中應標記的單詞?
$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 );
讓我知道是否清楚。
這樣的事情可以幫助您:
$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);
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