[英]Using PHP stripos() in Logical Filtering
How can i use the stripos to filter out unwanted word existing on itself. 我如何使用Stripos过滤掉自身上存在的不需要的单词。 How do i twist the code below that a search for '
won
' in grammar will not return true, since ' wonderful
' is another word itself. 我如何扭曲之下,一个关于“代码
won
语法不会回到真正的”,因为“ wonderful
”是另一个词本身。
$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'
Use preg_match
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';
}
$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
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.