preg_grep不显示双引号（“）的结果

Question

我的代码如下：

 $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);                 
 }

我正在输出：

title "space before" and text breaks 

但是我需要：

title:"space before" and text breaks 

Answer 1

如上所述， preg_grep()需要一个数组作为第二个参数，而不是字符串。 如果您更改：

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

至：

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

您的代码有效，并echo $string[0]; ，将输出title:"space before" and text breaks

如果目标是删除冒号（:)周围的空格，那么您是否可以这样做？

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

这还将输出title:"space before" and text breaks