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我正在尝试将一段 Javascript 转换为 .NET，但我似乎无法完全正确。

这是一个在 Express for NodeJs 中调用的方法。它将路径转换/test/:value1/:value2为可以在 URL 的一部分上使用的正则表达式。

/**
* Normalize the given path string,
* returning a regular expression.
*
* An empty array should be passed,
* which will contain the placeholder
* key names. For example "/user/:id" will
* then contain ["id"].
*
* @param {String|RegExp|Array} path
* @param {Array} keys
* @param {Boolean} sensitive
* @param {Boolean} strict
* @return {RegExp}
* @api private
*/

exports.pathRegexp = function(path, keys, sensitive, strict) {
  if (path instanceof RegExp) return path;
  if (Array.isArray(path)) path = '(' + path.join('|') + ')';
  path = path
    .concat(strict ? '' : '/?')
    .replace(/\/\(/g, '(?:/')
    .replace(/(\/)?(\.)?:(\w+)(?:(\(.*?\)))?(\?)?(\*)?/g, function(_, slash, format, key, capture, optional, star){
      keys.push({ name: key, optional: !! optional });
      slash = slash || '';
      return ''
        + (optional ? '' : slash)
        + '(?:'
        + (optional ? slash : '')
        + (format || '') + (capture || (format && '([^/.]+?)' || '([^/]+?)')) + ')'
        + (optional || '')
        + (star ? '(/*)?' : '');
    })
    .replace(/([\/.])/g, '\\$1')
    .replace(/\*/g, '(.*)');
  return new RegExp('^' + path + '$', sensitive ? '' : 'i');
}

我尝试将其转换为：

private Regex ConvertPathToRegex(string path, bool strict, out List<string> keys) {

    keys = new List<string>();

    List<string> tempKeys = new List<string>();

    string tempPath = path;

    if (strict)
        tempPath += "/?";

    tempPath = Regex.Replace(tempPath, @"/\/\(", delegate(Match m) {
        return "(?:/";
    });

    tempPath = Regex.Replace(tempPath, @"/(\/)?(\.)?:(\w+)(?:(\(.*?\)))?(\?)?(\*)?", delegate(Match m) {

        string slash = (!m.Groups[1].Success) ? "" : m.Groups[1].Value;
        bool formatSuccess = m.Groups[2].Success;
        string format = (!m.Groups[2].Success) ? "" : m.Groups[2].Value;
        string key = m.Groups[3].Value;
        bool captureSuccess = m.Groups[4].Success;
        string capture = m.Groups[4].Value;
        bool optional = m.Groups[5].Success;
        bool star = m.Groups[6].Success;

        tempKeys.Add(key);

        string expression = "/";
        expression += (optional ? "" : slash);
        expression += "(?:";
        expression += (optional ? slash : "");
        expression += (formatSuccess ? format : "");
        expression += (captureSuccess ? capture : (formatSuccess ? format + "([^/.]+?" : "([^/]+?)")) + ")";
        expression += (star ? "(/*)" : "");

        return expression;
    });

    tempPath = Regex.Replace(tempPath, @"/([\/.])", @"\$1");
    tempPath = Regex.Replace(tempPath, @"/\*", "(.*)");
    tempPath = "^" + tempPath + "$";

    keys.AddRange(tempKeys);

    return new Regex(tempPath, RegexOptions.Singleline | RegexOptions.IgnoreCase);
}

但问题是，这种方法不能正常工作。由于我不是正则表达式的超级明星，我想知道是否可以得到一些帮助。

当您还添加?param=1.

编辑：当我第一次从路径中删除查询参数时，它实际上工作得很好。

抱歉，问题的答案是从 URL 中删除查询参数。

4

1 回答 1

0

我已经在 PHP 中创建了这个端口，也许它可以以某种方式帮助您更好地理解 express 的作用，它返回一个正则表达式来匹配参数。

<?php
public static function pathRegexp($path, array &$keys, $sensitive = false, $strict = false)
{
    if (is_array($path)) {
        $path = '(' . implode('|', $path) . ')';
    }

    $path = (($strict === true) ? $path : (rtrim($path, '/ ') . '/?'));
    $path = preg_replace('/\/\(/', '(?:/', $path);
    $path = preg_replace_callback('/(\/)?(\.)?:(\w+)(?:(\(.*?\)))?(\?)?(\*)?/', function($matches) use(&$keys) {
                $slash = (isset($matches[1]) ? $matches[1] : "");
                $format = (isset($matches[2]) ? $matches[2] : false);
                $key = (isset($matches[3]) ? $matches[3] : null);
                $capture = (isset($matches[4]) ? $matches[4] : false);
                $optional = (isset($matches[5]) ? $matches[5] : false);
                $star = (isset($matches[6]) ? $matches[6] : false);

                array_push($keys, array("name" => $key, "optional" => (!!$optional)));

                return ''
                        . ($optional ? '' : $slash)
                        . '(?:'
                        . ($optional ? $slash : '')
                        . ($format ? $format : '')
                        . ($capture ? $capture : ($format ? '([^/.]+?)' : '([^/]+?)')) . ')'
                        . ($optional ? $optional : '')
                        . ($star ? '(/*)?' : '');
            }, $path);
    $path = preg_replace('/([\/.])/', '\\/', $path);
    $path = preg_replace('/\*/', '(.*)', $path);
    $path = '/^' . $path . '$/' . ($sensitive ? '' : 'i');
    return $path;
}
于 2013-03-13T12:35:27.600 回答