如何使用模式是一个数组（PHP）的preg_match

Question

Php.net过这样的preg_replace片断

$string = 'The quick brown fox jumped over the lazy dog.'; 
$patterns = array(); 
$patterns[0] = '/quick/'; 
$patterns[1] = '/brown/'; 
$patterns[2] = '/fox/'; 
$replacements = array(); 
$replacements[2] = 'bear'; 
$replacements[1] = 'black'; 
$replacements[0] = 'slow'; 
echo preg_replace($patterns, $replacements, $string); 

有没有一种方法，以便做这样的事情

运行$图案的preg_match

如果preg_match在$ string中找到，那么preg_replace else回声没有匹配找到

谢谢。

Answer 1

这是你在哪里找？

$string = 'The quick brown fox jumped over the lazy dog.'; 
$patterns = array(); 
$patterns[0] = '/quick/'; 
$patterns[1] = '/brown/'; 
$patterns[2] = '/fox/'; 
$replacements = array(); 
$replacements[2] = 'bear'; 
$replacements[1] = 'black'; 
$replacements[0] = 'slow'; 

foreach ($patterns as $pattern) { 
    if (preg_match("/\b$pattern\b/", $string)) { 
    echo preg_replace($pattern, $replacements, $string); 
     } 
} 

Answer 2

似乎所有你想要做的是有一个preg_replace也提醒您的是没有发生的比赛？

下面会为你工作：

$string = 'The quick brown fox jumped over the lazy dog.'; 
$patterns = array(); 
$patterns[0] = '/quick/'; 
$patterns[1] = '/brown/'; 
$patterns[2] = '/pig/'; 
$replacements = array(); 
$replacements[2] = 'bear'; 
$replacements[1] = 'black'; 
$replacements[0] = 'slow'; 

for($i=0;$i<count($patterns);$i++){ 
    if(preg_match($patterns[$i], $string)) 
     $string = preg_replace($patterns[$i], $replacements[$i], $string); 
    else 
     echo "FALSE: ", $patterns[$i], "\n"; 
} 
echo "<br />", $string; 

/** 

Output: 

FALSE: /pig/ 
The slow black fox jumped over the lazy dog. 
*/ 

---

$string = preg_replace($patterns, $replacements, $string, -1, $count); 
if(empty($count)){ 
    echo "No matches found"; 
}