PHP-正则表达式如何按条件替换字符串

我有一个字符串：

{include "abc"}

{literal} 

function xyz() {

       "ok";

   }

{/literal}

{abc}

{123 }

我只想将所有

{

替换为

{{

和

}

替换为

}

而不是

{literal}

标记。结果将是：

{{include "abc"}}

{{literal}}

   function xyz() {

       "ok";

   }

   //... something contain { and }

{{/literal}}

{{abc}}

{123 }}

有人可以帮我，谢谢

(?s)(?<=\{literal\}).*?(?=\{\/literal\})(*SKIP)(*F)|([{}])

---

您可以使用以下模式：

$pattern = '~(?:(?<={literal})[^{]*(?:{(?!/literal})[^{]*)*+|[^{}]*)([{}])\K~'

$text = preg_replace($pattern, '$1', $text);

---

$pattern='~（？：（？你试过什么吗？是我一个人，还是这是另一个X-Y问题？你为什么要求助于正则表达式来解决这个问题？我觉得你好像在试图解析一些东西哦，我试着解析smarty:）非常感谢你，这对我来说很有用。我想你是正则表达式的大师：D
$pattern = '~(?:(?<={literal})[^{]*(?:{(?!/literal})[^{]*)*+|[^{}]*)([{}])\K~'

$text = preg_replace($pattern, '$1', $text);

~                       # pattern delimiter
(?:                     # non-capturing group
    (?<={literal})      # lookbehind: preceded by "{literal}"
                        # a lookbehind doesn't capture any thing, it is only a test
    [^{]*               # all that is not a {
    (?:
        {(?!/literal})  #/# a { not followed by "/literal}"
        [^{]*
    )*+                 # repeat as needed
  |                     # OR
    [^{}]*              # all that is not a curly bracket,
                        # (to quickly reach the next curly bracket)
)
([{}])                  # capture a { or a } in group 1
\K                      # discards all on the left from match result
                        # (so the whole match is empty and nothing is replaced,
                        # the content of the capture group is only added 
                        # with the replacement string '$1')
~