preg_replace留下两个空格

Question

我正在使用preg_replace从字符串中删除不需要的内容，它正在工作，但它留下2个空格，所以...

 $content = '[content I want deleted including square brackets] this would be content that I want left!';

 $result = preg_replace('`\[[^\]]*\]\`','',$content);

  echo $result;

以上产生以下内容（“__”代表两个空格）

__这将是我想要的内容！

结果应该是......

这将是我想要的内容！

如果没有前两个空格，有人可以告诉我哪里出错了。感谢。

Answer 1

不确定您的代码，但我测试了这个并且运行良好

$content = '[content __I want deleted including square brackets] this would be content that I want left!';

 $result = preg_replace('#\[([^\]]*)\]#is','',$content);

  echo $result;

（我使用＆＃34;是＆＃34;因为我总是忘记模式修饰符的含义，如果你不知道检查http://php-regex.blogspot.it/）它们总是运作良好

Answer 2

只需在正则表达式的末尾添加可选空格：

$content = '[content I want deleted including square brackets] this would be content that I want left!';
$result = preg_replace('`\[[^\]]*\]\s*`', '', $content);
echo $result;

这给出了：

this would be content that I want left!

在开头没有空格。