我有 HTML 注释被包裹在 Li 和 P 标签中:(

发布于 2024-07-30 02:19:58 字数 490 浏览 6 评论 0原文

我的内容首先是 htmlentities，然后是 stripslashes，最后是 nl2br。

这意味着最后的水印最终会变成：

<li><p><!-- watermark --></p></li>

不是很有用。我有下面的代码来尝试删除 html 注释并停止显示，但它不太擅长！

$methodfinal = str_replace('<li><p><!--', '<!--', $method);
$methodfinal2 = str_replace('--></p></li>', '-->', $methodfinal);
echo $methodfinal2;

有人有什么想法吗？

原文

I have content that is first htmlentities and then stripslashes followed by nl2br.

This means a watermark at the end ends up as:

<li><p><!-- watermark --></p></li>

Not very useful. I have the code below to try and strip the html comments and stop it displaying but its not very good at it!

$methodfinal = str_replace('<li><p><!--', '<!--', $method);
$methodfinal2 = str_replace('--></p></li>', '-->', $methodfinal);
echo $methodfinal2;

anyone got any ideas?

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乱了心跳 2024-08-06 02:19:58

编辑：
根据 Zed 和您的评论，我已经做了一些测试，这就是您应该使用的：

$final = preg_replace('/<li><p>[\s]*?<\;!--(.*?)-->\;<\/p><\/li>/m', "<!--$1-->", $z);

这是 RE 的细分：

<li><p>

这很明显，

[\s]*?

因为您在

和注释，但我们想要最少数量的换行符，因此我们使用非贪婪的 *? （它也可以与 * 一起使用）

<\;

需要转义 ;

!--(.*?)--

我们再次使用 *? 所以我们只会匹配这一行（否则，如果你再次有相同的行，它会从第一行到最后一行匹配，

>\;<\/p><\/li>

与上面相同，

/m'

因此 php 会将换行符视为空格（我对此不确定，但似乎是在职的）

EDIT:
following Zed's and your comments I've done some testing and this is what you should use:

$final = preg_replace('/<li><p>[\s]*?<\;!--(.*?)-->\;<\/p><\/li>/m', "<!--$1-->", $z);

Here is a breakdown of the RE:

<li><p>

this is obvious

[\s]*?

because you have a few spaces and a newline between the <li> and the comment, but we want the least number of newlines so we use the non greedy *? (it sould work with * as well)

<\;

need to escape the ;

!--(.*?)--

again we use *? so we would match only this line (other wise if you had the same line again it wold match from the first one to the last one

>\;<\/p><\/li>

same as above

/m'

so php would treat newlines as whitespace (i am not sure about this but it seems to be working)

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唐婉 2024-08-06 02:19:58

像这样的东西吗？

$final = preg_replace("/<li><p>(<!--.*-->)<\/p><\/li>/", "$1", $original);

Something like this?

$final = preg_replace("/<li><p>(<!--.*-->)<\/p><\/li>/", "$1", $original);

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素染倾城色 2024-08-06 02:19:58

@Zed：

让我们更加关心：

$final = preg_replace("/<li><p>(<!--.*?-->)<\/p><\/li>/", "$1", $original);
# use .*? every time over .* unless you specificly want what it does
# .*? matches as less as it can
# .* matches as much as it can

甚至更好：

$final = preg_replace("/<li><p>(<!--[^\-\>]+-->)<\/p><\/li>/", "$1", $original);
# [^\-\>]+ will look for any character that is not - or > 
# so will perform faster

只是试图提倡更好的正则表达式实践。希望这可以帮助。

@Zed:

Lets be more caring:

$final = preg_replace("/<li><p>(<!--.*?-->)<\/p><\/li>/", "$1", $original);
# use .*? every time over .* unless you specificly want what it does
# .*? matches as less as it can
# .* matches as much as it can

even better:

$final = preg_replace("/<li><p>(<!--[^\-\>]+-->)<\/p><\/li>/", "$1", $original);
# [^\-\>]+ will look for any character that is not - or > 
# so will perform faster

Just trying to advocate better regex practice. Hope this helps.

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