替换多个[Code]块

本文关键字：Code 替换 | 更新日期: 2023-09-27 17:50:26

我正在创建一个博客网站，在那里我将允许用户在[code]内输入代码代码内容[/code]

在一篇博文中会有多个这样的[Code]块。

我想用Regex找到每个[Code]块，然后用

替换它
<pre>command

我还想将前标签中的&lt;和&gt;替换为<>

现在我找到了有用的代码，可以帮助我通过，但我对Regex感到困惑，有人可以帮助我这个。

    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();
}

替换多个[Code]块

…RegexBuddy:

'[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，只有一个code组。没有lang组了…所以把它从c#中删除

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