preg_grep不顯示用雙引號（「）

Question

我的代碼如下結果：

$escapedOperator = ":"; 
$operator['symbol'] = ":"; 
$string = 'title: "space before" and text breaks'; 
if(count(preg_grep('/\w["]*\s*'.$escapedOperator.'\s*["]*\w/',$string))){ 
     $search = "/\s*".$escapedOperator."\s*/"; 
     $string = preg_replace($search,$operator['symbol'],$string); 
}else{ 
     $string=str_replace($operator['symbol'],"",$string);     
} 

我得到的輸出：

title "space before" and text breaks 

但我需要：

title:"space before" and text breaks 

Answer 1

如上所述，preg_grep()需要將數組作爲第二個參數，而不是字符串。如果更改：

$string = 'title: "space before" and text breaks';

要：

$string = array('title: "space before" and text breaks');

你的代碼的工作，並echo $string[0];，將輸出title:"space before" and text breaks

如果目標是隻刪除空格周圍的冒號（:)，然後你能不能這樣做？

$string = 'title: "space before" and text breaks'; 
$string = preg_replace('/\s*:\s*/', ":", $string); 
echo $string[0]; 

這也將輸出title:"space before" and text breaks