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当我尝试通过PHP imap functions. 我想提取正文的文本,但没有 HTML 链接(如 mailto:xxxx)或编码问题。我想我已经尝试了(几乎,我猜)一切。但是,我接近所需最终结果的唯一代码是下一个:

$bodyText = imap_fetchbody($inbox,$email_number,1.2);

if(!strlen($bodyText)>0)
{
   $bodyText = imap_fetchbody($inbox,$email_number,1);
}

var_dump($bodyText);

关键是我遇到了最终编码的问题(我认为这是问题所在)。所以 var_dump 输出类似于:

> From: SomeOne <someone=40domain.com>
> To: Someone Else <someoneelse=40domain.com>
> =Date: lunes 23 de julio de 2012 13:04:43
> Subject: =46wd: =46W: URGE=21=21=21=21 Ley de Reforma del Congreso de 20=11
> =20
> Some text here, blah, blah, blah
> =20
> ---------- =46orwarded message ----------
> From: Whatever <whatever=40domain.com (mailto:whatever=40domain.com)>
> Date: 23 de julio de 2012 12:53
> Subject: =46wd: =46W: URGE=21=21=21=21 Ley de Reforma del Congreso de 20=11
> To: Someone <someone=40domain.com (mailto:someone=40domain=.com)>
> =20
> =20
> Some stuff=21=21
> =20
> ---------- =46orwarded message ----------
> =46rom: samuel l jackson <sanvuco=40domain.com (mailto:sanvuco=40domain.com)>
> Date: 2012/7/23
> Subject: =46W: URGE=21=21=21=21 Ley de Reforma del Congreso de 2011
> To: =22...Scary Monster=C2=B7=C2=B7=C2=B7 =C3=B2=5F=5F=C3=B3=22 <eowyn2=
6=40domain.com (mailto:eowyn26=40domain.com

我的意思是,出现了坏而=40不是块@mailto:xxxx@domain.com

谢谢

4

1 回答 1

2

运行身体字符串

quoted_printable_decode($bodyText);

你会得到你正在寻找的可打印的、未编码的字符串。然后,您可以继续使用一些正则表达式来摆脱 mailto 并进行您喜欢的任何其他解析。这些任务的适当功能是:

preg_replace();

http://www.php.net/manual/en/function.preg-replace.php

于 2012-07-24T14:11:31.203 回答