Question

我正在创建一个博客网站，我将允许用户在[代码]代码内容[/代码]

中输入代码

在一篇博文中会有多个[Code]块。

我想使用Regex找到每个[Code]块，然后用

替换它

<pre>command

另外，我想将pre tag中的<和>替换为＆lt; ＆GT;

现在我发现有用的代码可以帮助我解决这个问题，但我对正则表达式感到困惑，有人可以帮我解决这个问题。

    static string ProcessCodeBlocks(string value)
{
    StringBuilder result = new StringBuilder();

    Match m = Regex.Match(value, @"\[pre=(?<lang>[a-z]+)\](?<code>.*?)\[/pre\]");
    int index = 0;
    while( m.Success )
    {
        if( m.Index > index )
            result.Append(value, index, m.Index - index);

        result.AppendFormat("<pre class=\"{0}\">", m.Groups["lang"].Value);
        result.Append(ReplaceBreaks(m.Groups["code"].Value));
        result.Append("</pre>");

        index = m.Index + m.Length;
        m = m.NextMatch();
    }

    if( index < value.Length )
        result.Append(value, index, value.Length - index);

    return result.ToString();
}

Answer 1

..来自RegexBuddy的解释：

\[pre=(?<lang>[a-z]+)\](?<code>.*?)\[/pre\]

Match the character “[” literally «\[»
Match the characters “pre=” literally «pre=»
Match the regular expression below and capture its match into backreference with name     “lang” «(?<lang>[a-z]+)»
   Match a single character in the range between “a” and “z” «[a-z]+»
      Between one and unlimited times, as many times as possible, giving back as needed     (greedy) «+»
Match the character “]” literally «\]»
Match the regular expression below and capture its match into backreference with name     “code” «(?<code>.*?)»
   Match any single character that is not a line break character «.*?»
      Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the character “[” literally «\[»
Match the characters “/pre” literally «/pre»
Match the character “]” literally «\]»

要使其适用于[Code][/Code]，您可以将其更改为：

\[code\](?<code>.*?)\[/code\]

..请记住，这只适用于单行块。此外，只有code组...不再有lang组..所以从C＃中删除它..

替换多个[Code]块

1 个答案: