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);                 
 }

我得到了输出：

title "space before" and text breaks

但我需要：

title:"space before" and text breaks

Answer 1

如上所述，preg_grep()需要一个数组作为第二个参数，而不是字符串。如果你改变了：

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

要：

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

您的代码有效，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，那么你能不能这样做吗？

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

这也会输出title:"space before" and text breaks

preg_grep不显示带双引号的结果（＆＃34;）

1 个答案: