使用preg_split作为模板引擎

正则表达式:

@(?> if | else (?>if)? ) \s*  # Match a control structure
(?> \( [^()]*+ \) \s*)?  # Match statements inside parentheses (if any)
\[  # Match start of block
(?:
    [^][@]*+  # Any character except `[`, `]` and `@`
    (?> \\.[^][@]* )* (@+ (?! (?> if | else (?>if)? ) ) )? # Match escaped or literal `@`s
    |  # Or
    (?R)  # Recurs whole pattern
)*  # As much as possible
\]\K\s*  # Match end of container block

Live demo

PHP:

print_r(preg_split("~@(?>if|else(?>if)?)\s*(?>\([^()]*+\)\s*)?\[(?:[^][@]*+(?>\\.[^][@]*)*(@+(?!(?>if|else(?>if)?)))?|(?R))*\]\K\s*~", $str, -1, PREG_SPLIT_NO_EMPTY));

Array
(
    [0] => @if(param1>=7) [ something here @if(param1>9)[ nested statement ] ]
    [1] => @elseif(param2==true) [ 2st condition true ]
    [2] => @else [ default condition ]
)

PHP live demo

要匹配嵌套括号,需要使用递归模式。

(?:(\[(?:[^[\]]|(?1))*])|[^@[\]])+

它将匹配每个段,不包括前导段 @ 它还将捕获组1中的最后一个括号,您可以忽略它。

使用模式 preg_match 或 preg_match_all .

感谢您的回复!雷沃的回答奏效了!

我自己无法提出正则表达式,而是构建了一个同样有效的解析器函数。也许它对某些人有用:

function parse_context($subject) { $arr = array(); // array to return

    $n = 0;  // number of nested sets of brackets
    $j = 0;  // current index in the array to return, for convenience

    $n_subj = strlen($subject);
    for($i=0; $i<$n_subj; $i++){
        switch($subject[$i]) {
            case '[':
                $n++;
                $arr[$j] .= '[';
                break;
            case ']':
                $n--;
                $arr[$j] .= ']';
                break;
            case '@':
                if ($n == 0) {
                    $j++;
                    $arr[$j] = '';
                    break;
                }
            default: $arr[$j].=$subject[$i];
        }
    }
    return $arr;

} // parse_context()