php preg_replace评论块

Question

patern就是这样

/* comment [comment goes here] */
/* comment please do not delete the lines below */
[I am a special line so I should not be changed ]
/* comment please do not delete the line above */

我想删除评论区块，但我不想查找/**/，而是希望评论为/* comment [] */，其中[]是实际评论。

这是为了避免任何应包含评论的文字。

以下是评论的条件

1. 以=＆gt;＆gt;开头/* comment
2. 后跟=＆gt;＆gt;任何
3. 后跟=＆gt;＆gt; */

Answer 1

这将删除注释块：

preg_replace('%/\*\s+comment\s+.*?\*/%s', '', $string)

这也可以摆脱过时的空白：

preg_replace('%/\s*\*\s+comment\s+.*?\*/\s*%s', '', $string)

这是一个测试脚本：

#!/usr/bin/php
<?php

$string = <<<EOS
/* comment [comment goes here] */
/* comment please do not delete the lines below */
[I am a special line so I should not be changed ]
/* comment please do not delete the line above */
EOS;

print $string;
print "\n---\n";
print preg_replace('%/\*\s+comment\s+.*?\*/%s', '', $string);
print "\n---\n";
print preg_replace('%/\s*\*\s+comment\s+.*?\*/\s*%s', '', $string);

?>

使用PHP 5.3.4输出：

/* comment [comment goes here] */
/* comment please do not delete the lines below */
[I am a special line so I should not be changed ]
/* comment please do not delete the line above */
---


[I am a special line so I should not be changed ]

---
[I am a special line so I should not be changed ]

Answer 2

似乎要做的工作：）

preg_replace("(\\/\\*[\s]*?comment[\\d\\D]*?[\s]*?\\*\\/)",'',$str)

我是怎么发现的？

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