用PHP中的preg_replace删除替换空间？

问题描述：

<a param1="false" class="param2" id="id#12" href="#1">toto</a>

I have made a preg_replace in PHP to remove certain params in a html text:

$re = "/(param1=\"false\"|class=\"param2\"|id=\"id#([^\"]+)\")/";

$text = preg_replace($re, " ", $data->text->value);

I obtain this html link:

<a href="#1">toto</a>

I would like to leave only one space beetween the and the href, is it possible to do that directly in the regex ?

 ＆lt; a param1 =“false”class =“param2”id =“id＃12”href  =“＃1”＆gt; toto＆lt; / a＆gt; 
  code>  pre> 
 
 我在 PHP  strong>中创建了 preg_replace  strong> 删除html文本中的某些参数： p> 
 
 
  $ re =“/（param1 = \”false \“| class = \”param2 \“| id = \”id  ＃（[^ \“] +）\”）/“; 
 
 $ text = preg_replace（$ re，”“，$ data-＆gt; text-＆gt; value）; 
  code>   pre> 
 
 我获得了 html  strong>链接： p> 
 
 
 ＆lt; a href =“＃1”＆gt; toto＆lt; /  a＆gt;  code>  p> 
 
 
我想在和href  strong>之间只留下一个空格，是否可以直接在正则表达式中执行此操作？ 
  div>

答

Change your regular regular in the following way,

$re = '/(param1="false"\s+|class="param2"\s+|id="id#([^\"]+)"\s+)/';

You need to pass \s+ in each of your component.

And change your preg_replace() function in the following way,

$text = preg_replace($re, "", $data->text->value);

Don't replace them with space, that's it.

答

Assuming you have to use all the backrefrences in your "preg_match". This is the only way you can accomplish

$re = "/<a.*(param1=\"false\"|class=\"param2\"|id=\"id#([^\"]+)\")/";
$text = preg_replace($re, "<a", $data->text->value);

答

You could try this:

$re = "/(param1=\"false\"|class=\"param2\"|id=\"id\#([^\"]+)\")([\ ]{1,}?)/";
$text = preg_replace($re, "", $data->text->value);

Regex101

用PHP中的preg_replace删除替换空间？

相关推荐