2010-01-06 26 views
1

我想做一些事情,如果我的句子包括這個數組中的單詞之一,該怎麼做?如果句子包含此數組中的單詞之一,該怎麼做?

$sentence = "I dont give a badwordtwo"; 

$values = array("badwordone","badwordtwo","badwordthree","badwordfour"); 

謝謝...

+5

您可能會考慮爲您的示例選擇不同的單詞。 – 2010-01-06 02:17:22

+3

@Mark:我想他可能會試圖從用戶字符串中篩選出不好的單詞.. [但我同意你] – 2010-01-06 02:18:35

+1

Sanitized :) ...我做了我自己的'壞話' – 2010-01-06 02:19:33

回答

3

如果要審查的話數組中一些字符串就可以使用str_ireplace

$var = "This is my phrase."; 
$var = str_ireplace(array("this", "phrase"), array("****", "*****"), $var); 

編輯:作爲chacha102筆記,你只需要使用第二陣列以改變星星的數量,

$var = str_ireplace(array("this", "phrase"), "", $var); 

同樣是VA蓋。我還應該注意,如果你使用第二個數組,它的長度必須與第一個數組完全匹配,並且替換符合索引。

+0

除非要讓恆星數與單詞相匹配,否則不需要爲第二個參數創建數組。如果你只是想要4顆星,你可以將第二個參數傳遞給一個字符串。 – 2010-01-06 02:24:07

+0

如果任務是過濾令人討厭的內容,則必須對輸入字符串進行規範化(正如Gordon在評論中已經指出的那樣)。刪除所有空格和分隔符,以及所有特殊字符,刪除變音符號並將特殊字符(如omicron)轉換爲其基本等效字符。 – 2010-01-06 11:20:31

+0

事實上,在創建良好的褻瀆過濾器的方法上有一個相當不錯的帖子(http://stackoverflow.com/questions/273516/how-do-you-implement-a-good-profanity-filter);這裏我只談到一個簡單的例子。 – 2010-01-06 12:31:33

1

單程

$sentence = "I dont give a badwordtwo"; 
$values = array("badwordone","badwordtwo","badwordthree","badwordfour"); 
$s = explode(" ",$sentence); 
foreach ($s as $a=>$b){ 
    if (in_array($b, $values)) { 
     echo "Got $b"; 
    } 
} 

輸出

$ php test.php 
Got badwordtwo 

OR

$sentence = "I dont give a badwordtwo"; 
$values = array("badwordone","badwordtwo","badwordthree","badwordfour"); 
$s = explode(" ",$sentence); 
var_dump(array_intersect($s, $values)); 

輸出

$ php test.php 
array(1) { 
    [4]=> 
    string(10) "badwordtwo" 
} 
+0

對不起,使輸出數字錯誤的幽靈。 「badwordtwo」當然不是4個字符長。 – 2010-01-06 02:26:14

0

難道你不喜歡php.net

實例#1基本str_replace()函數的例子

<?php 
// Provides: <body text='black'> 
$bodytag = str_replace("%body%", "black", "<body text='%body%'>"); 

// Provides: Hll Wrld f PHP 
$vowels = array("a", "e", "i", "o", "u", "A", "E", "I", "O", "U"); 
$onlyconsonants = str_replace($vowels, "", "Hello World of PHP"); 

// Provides: You should eat pizza, beer, and ice cream every day 
$phrase = "You should eat fruits, vegetables, and fiber every day."; 
$healthy = array("fruits", "vegetables", "fiber"); 
$yummy = array("pizza", "beer", "ice cream"); 

$newphrase = str_replace($healthy, $yummy, $phrase); 

// Provides: 2 
$str = str_replace("ll", "", "good golly miss molly!", $count); 
echo $count; 
?> 
1

我也有類似的問題,而回。這個答案完全適合你。

Is this efficient coding for anti-spam?

<?PHP 
$banned = array('bad','words','like','these'); 

$looksLikeSpam = false; 
foreach($banned as $naughty){ 
    if (strpos($string,$naugty) !== false){ 
     $looksLikeSpam=true; 
    } 
} 

if ($looksLikeSpam){ 
    echo "You're GROSS! Just... ew!"; 
    die(); 
} 
?> 
0

這是Kohana 3的一個片段。我總是發現它是一個有用的功能。它也允許你檢查部分單詞(或不單單)。

public static function censor($str, $badwords, $replacement = '#', $replace_partial_words = TRUE) 
{ 
    foreach ((array) $badwords as $key => $badword) 
    { 
    $badwords[$key] = str_replace('\*', '\S*?', preg_quote((string) $badword)); 
    } 

    $regex = '('.implode('|', $badwords).')'; 

    if ($replace_partial_words === FALSE) 
    { 
    $regex = '(?<=\b|\s|^)'.$regex.'(?=\b|\s|$)'; 
    } 

    $regex = '!'.$regex.'!ui'; 

    if (strlen($replacement) == 1) 
    { 
    $regex .= 'e'; 
    return preg_replace($regex, 'str_repeat($replacement, strlen(\'$1\'))', $str); 
    } 

    return preg_replace($regex, $replacement, $str); 
} 
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