Java Regular Expression用于捕获电子邮件编码的附件字符串

Question

我想编写一个正则表达式，用于捕获电子邮件MIME消息字符串（eml）中的编码字部分。例如，这是电子邮件的一部分：

<div dir=3D"ltr"><br clear=3D"all"><div><div dir=3D"ltr"><div style=3D"dire=
ction:rtl">-------------</div><div style=3D"direction:rtl">=D7=91=D7=91=D7=
=A8=D7=9B=D7=94,</div><div style=3D"direction:rtl">=D7=90=D7=91=D7=99=D7=A2=
=D7=93 =D7=9B=D7=94=D7=9F</div></div></div>
</div>

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--20cf3003bc2e044e9d0500f755de
Content-Type: text/plain; charset=US-ASCII; name="EhudBanay.txt"
Content-Disposition: attachment; filename="EhudBanay.txt"
Content-Transfer-Encoding: base64
X-Attachment-Id: f_hz0z4us30

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--20cf3003bc2e044e9d0500f755de--

我只想抓住这一部分：

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我写了下面的正则表达式： （？＆LT = \ r \ n \ r \ n）的（。+ \ r \ n）的+（？= - ） 这提供了我想要的结果。我认为： *正则表达式应该返回所有这些部分的出现。多 *在想要的部分之前总是有一条空行。 *所需部分后面跟着一行＃34; - ＆＃34; （它们之间可能存在分隔线。）

这个解决方案的问题是我在线程＆＃34; main＆＃34;中得到了Exception。 java.lang.StackOverflowError当输入字符串很长时，是什么让我想到了正则表达式的效率。 我想使正则表达式更精确。我们可以利用这个事实，即在想要的部分之前的空行之前的某个地方，始终有一行以＆＃34; Content-Transfer-Encoding开头：＆＃34;。 有人可以帮忙吗？

Answer 1

您可以尝试增加堆栈大小：-Xss2048k