php 下面的函数怎么改?
大家讲道理
大家讲道理 2017-04-10 15:17:16
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protected function doCodeBlocks($text) { # # Process Markdown `` blocks. # preg_match('/\{\{\{/i', $text, $_m_c_open); preg_match('/\}\}\}/i', $text, $_m_c_close); if (count($_m_c_open) == count($_m_c_close)) { $text = preg_replace_callback('/\{\{\{[ \n]*(.*?)\}\}\}/is', array( &$this, '_doCodeBlocks_callback' ), $text); } return $text; } protected function _doCodeBlocks_callback($matches) { $codeblock = $matches[1]; $codeblock = $this->outdent($codeblock); $codeblock = htmlspecialchars($codeblock, ENT_NOQUOTES); # trim leading newlines and trailing newlines $codeblock = preg_replace('/\A\n+|\n+\z/', '', $codeblock); $codeblock = str_replace(array( '&', "\t", ' ' ), array( '&', '    ', ' ' ), $codeblock); $codeblock = "
$codeblock\n
"; return $this->hashBlock($codeblock); }

这第一个个函数中的参数$text是markdown格式文章中的代码内容,markdown 格式的内容是这种形式的

{{{ 代码 }}}

如果markdown格式的代码块是这种形式的{{{javascript: 代码}}}
我想把javascript:用变量$lang_code的形式提交给$codeblock里面,之后返回下面的html内容:

$codeblock = "
$codeblock\n
";
大家讲道理
大家讲道理

光阴似箭催人老,日月如移越少年。

reply all (1)
Ty80

在这个后面

# trim leading newlines and trailing newlines $codeblock = preg_replace('/\A\n+|\n+\z/', '', $codeblock);

加上下面这两行试试:

$lang_pos = strpos($codeblock,":"); $lang_code_pre = substr($codeblock, 3,$lang_pos - 3); switch ($lang_code_pre) { case 'php': $lang_code = "php"; break; case 'javascript': $lang_code = "javascript"; break; default: $lang_code = "default value"; break; }
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