Question

如何在不影响单词的情况下过滤掉粗鲁的单词。在我的例子中，由于草中的“屁股”，草一词正在被审查。如何更改此代码，以便用 * *替换屁股，但草不受影响？

$string = 'cut the grass spread the aftercut and watch the difference after a few days';

$naughtyWords = array("ahole","anus","ash0le","ash0les","asholes","ass","Ass Monkey","Assface"); //etc

$string = str_replace($naughtyWords, "****", $string);

echo $string;

结果

切割gr * * 传播后切，并在几天后观察差异

Answer 1

您需要检查相关字符串之前和之后的空格。

示例：

<?php

$string = 'cut the grass spread the aftercut and watch the difference after a few days';

$naughtyWords = array("ahole","anus","ash0le","ash0les","asholes","ass","Ass Monkey","Assface"); //etc

foreach ($naughtyWords as &$word) {
    $word=' '.$word.' ';
}

$string = str_replace($naughtyWords, " **** ", ' '.$string.' ');

echo $string;

?>

Answer 2

我认为戴夫的说法是：

您需要在每个单词之前检查空格。

$string 'cut the grass spread the aftercut and watch the difference after a few days';

$naughtyWords = array(" ahole"," anus"," ash0le"," ash0les"," asholes"," ass"," Ass Monkey"," Assface"); //etc

$string = str_replace($naughtyWords, "****", $string);

echo $string;

或者更好的是：

$string 'cut the grass spread the aftercut and watch the difference after a few days';
$patterns = array('/ahole/','/anus/','/ash0le/','/ash0les/','/asholes/','/ass/','/Ass Monkey/','/Assface/');
$string = preg_replace($patterns, '****', $string);

审查员从字符串中粗鲁地说出来但不影响安全的话

2 个答案: