duanmie9682 2017-03-16 13:10 采纳率: 100%
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如何在正则表达式中匹配字符串后跟重复模式?

I'm looking for a regex to match the following pattern:

--header-- any text -pre- any text -/pre- -pre- any text -/pre-

For example: the following text

--comment--
list of comments
-pre-
comment 1
-/pre-

-pre-
comment 2
-/pre-

--Answers--
list of Answers
-pre-
Answer 1
-/pre-

-pre-
Answer 2
-/pre-

should be returned as following when apply the regex:

Array
(
    [comments] => Array
        (
            [0] => comment 1
            [1] => comment 2
        )

    [answers] => Array
        (
            [0] => answer 1
            [1] => answer 2
        )

)

I tried the following regex --(.*?)--.*?(-pre-(.*?)-\/pre-)+ but it only match comment 1 and answer 1

Example Code: https://regex101.com/r/59OKzs/1

  • 写回答

2条回答 默认 最新

  • dongli564510 2017-03-16 14:41
    关注

    Regex:

    (?:--([^-]+)--(?:(?!-pre-).)+|(?!\A)\G)\s*-pre-\s*((?:(?!-/pre-).)*)-/pre-\K
    

    Live demo

    PHP:

    preg_match_all('~(?:--([^-]+)--(?:(?!-pre-).)+|(?!\A)\G)\s*-pre-\s*((?:(?!-/pre-).)*)-/pre-\K~s', $str, $matches, PREG_SET_ORDER);
    

    To integrate appropriate keys and values:

    // Filter empty values
    $matches = array_map(function($v) {
        return array_values(array_filter($v));
    }, $matches);
    
    // Initialize two variables which we use them soon
    $index = null; $finalArray = [];
    
    // Iterate over recent matches in order to apply keys
    array_map(function($array) use (&$index, &$finalArray) {
        count($array) == 2 ? ($index = $array[0]) && ($finalArray[$index][] = $array[1]) : $finalArray[$index][] = $array[0];
    }, $matches);
    
    // Print out
    print_r($finalArray);
    

    PHP live demo

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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