在字符串中檢測混淆的粗俗名詞

Question

我想在PHP中使用一種函數或方法來檢測隱藏文本中的粗言穢語。 的東西，將檢查字符串，如：

$string = "hey you swearword!" or 
$string = "hey you swear#word!" 

，或者甚至

$string = "hey you sw3arw0rd!" 

的「臭罵」，如果它包含壞臭罵和虛假的，如果它不將返回true。 我不希望人們在我的網站上使用不好的字，請幫忙！

Answer 1

只是一個簡單的例子來說明方向：

$stopwords = ['swearword']; 

$test = ['swear#word','sw3arw0rd','goodword','swearw*rd','swe*rw*rd','swe*!**rd']; 

foreach($test as $word){ 
    foreach($stopwords as $stopword){ 
     if(levenshtein($stopword,$word)<=2){ 
      print "levenshtein: '$word' seems to mean $stopword<br/>"; 
      continue 2; 
     } 
    } 
    if(strlen(preg_replace('#[a-zA-Z]+#','',$word))!==0){#special char found 
     print "preg_replace: '$word' seems to have illegal chars<br/>"; 
     continue; 
    } 
    print "'$word' seems be NO stopword<br/>"; 
}