Php preg_replace - 正则表达式（删除标记）

Question

我对令牌很满意，想去除所有不以 test 开头的令牌。例如我有内容：

$content = 'Hello [test:username]. What are you [token:name]';
$string = preg_replace('/\[test(.*)\]/', '', $content);

这行得通，但用空字符串替换所有以 test 开头的标记。我想要相反 不启动 与 test 并替换所有其他。我应该如何更改正则表达式。我想在 preg_replace:

之后得到这个结果

$content = 'Hello [test:username]. What are you';

Answer 1

您可以使用以下正则表达式

(?:\[test:[^]]+\])(*SKIP)(*F)|(?:\[\w[^]]+\])

所以你的代码看起来像

preg_replace('/(?:\[test:[^]]+\])(*SKIP)(*F)|(?:\[\w[^]]+\])/', '', $content);

解释：

(?:\[test:[^]]+\]) // Will capture a group of tokens that have same 
                      pattern like as [test:...]

(*SKIP)(*F)        // This is supported by PCRE to skip the above captured group 

|(?:\[\w[^]]+\])   // This'll capture rest of the group that doesn't 
                     contains pattern like as [test:...]

Php preg_replace - 正则表达式（删除标记）

Php preg_replace - regular expression (remove tokens)

php

expression

preg-replace