PHP字符串替换括号内的排除字符串

Question

$str = "PHP is a server-side {{scripting language}} designed for {{web development}} but also used as a general-purpose programming language";
$search = 'language';
$replace = 'english';

$new_str = str_replace($search, $replace, $str);

在上面的代码中，我们如何排除{{...}}中的字符串。这样输出应该是

"PHP is a server-side {{scripting language}} designed for {{web development}} but also used as a general-purpose programming english"

Answer 1

使用preg_replace()代替str_replace()：

$pattern = sprintf('/{{[^}]+}}(*SKIP)(*F)|%s/', preg_quote($search, '/'));
$new_str = preg_replace($pattern, $replace, $str);

{{[^}]+}}匹配{{...}}块内的内容，(*SKIP)(*F)跳过它，%s（用转义的搜索字符串替换）与所需的字匹配。

Demo