Regex Pattern to Extract Email Data

Question

I'm retrieving raw text (includes header, and message) from a POP server. I need to capture everything after the header which is terminated by a blank line between it and the user message.

At the same time I'm wanting to ignore anything from original messages if it's a reply. The start of a reply for the emails I'm parsing start with

------Original Message------

An example email might look like this

Return-Path: ...
...
More Email Metadata: ...

Hello from regex land, I'm glad to hear from you.
------Original Message------
Metadata: ...
...

Hey regex dude, can you help me? Thanks!

Sincerely, Me.

I need to extract "Hello from regex land, I'm glad to hear from you." and any other text/lines prior to the original message.

I'm using this regex right now (C# in multiline mode)and it seems to work except it's capturing ------Original Message------ if the body is blank. I'd rather just have a blank string instead.

^\s*$\n(.*)(\n------Original Message------)?

Edit
I haven't down voted anyone and if you happen to downvote, it's usually helpful to include comments.

Answer 1

Why not replace "\n------Original Message------" after the capture?

Answer 2

Why don't you not use DotnetOpenMail? Using a regex to do this is a wrong approach, you'd be better off using a dedicated email handler instead....

Answer 3

The reason for this is that you have an extra \n inside the parenthesis. If the body is blank, there is no extra newline there. Therefore, try this:

^\s*$\r\n(.*)(^------Original Message------$)?

If you don’t want the newline at the end of the body, you can still use string.Trim() on the matched part.

Note: This assumes that the input uses \r\n line terminators (which is required in e-mail headers according to the MIME standard).

Answer 4

You need to replace (\n------Original Message------) with (?=(\n------Original Message------)) lookahead to not return that part, just to ensure it's there