Question

如何使用stripos过滤掉自身存在的不需要的单词。如何扭曲下面的代码，在语法中搜索“won”将不会返回true，因为“wonderful”本身就是另一个词。

$grammar = 'it is a wonderful day';
$bad_word = 'won';

$res = stripos($grammar, $bad_word,0);

if($res === true){
       echo 'bad word present';
}else{
       echo 'no bad word'; 
}

//result  'bad word present'

Answer 1

使用preg_match

$grammar = 'it is a wonderful day';
$bad_word = 'won';
$pattern = "/ +" . $bad_word . " +/i"; 

// works with one ore more spaces around the bad word, /i means it's not case sensitive

$res = preg_match($pattern, $grammar);
// returns 1 if the pattern has been found

if($res == 1){
  echo 'bad word present';
}
else{
  echo 'no bad word'; 
}

Answer 2

 $grammar = 'it is a wonderful day';
 $bad_word = 'won';

        /*  \b   \b indicates a word boundary, so only the distinct won not wonderful is  searched   */

 if(preg_match("/\bwon\b/i","it is a wonderful day")){ 
    echo "bad word was found";} 
 else { 
    echo "bad word not found"; 
    } 

//result is : bad word not found

在逻辑过滤中使用PHP stripos（）

2 个答案: