Using StringRegExpReplace To Replace Section of a File

[TripleDES 0]

I'm trying to use StringRegExpReplace to replace/remove a section of a file that includes carriage returns.

File contents:

interface Vlan2
ip address 192.168.21.1 255.255.255.0
##DBB##
ip nat inside
no shut
exit
##DBB##

Expression:

StringRegExpReplace($chars, "##DBB##(.*?)##DBB##", "")

In this case, I'd like to replace anything in between the two delimiters "##DBB##" with an empty "". However, (.*?) doesn't play nicely with carriage returns. I've tried variations of \v and \s to no avail. Can someone let me know what else I need to try?

[PsaltyDS 27]

You want: "(?s)##DBB##.*##DBB##"

Modifier "(?s)" specifies that newlines also match ".".

P.S. If you want to keep the delimiters:

$sChars = StringRegExpReplace($sChars, "(?s)(##DBB##)(.*)(##DBB##)", "\1" & @CRLF & "\3")

Edited June 16, 2010 by PsaltyDS

[TripleDES 0]

You want: "(?s)##DBB##.*##DBB##"

Modifier "(?s)" specifies that newlines also match ".".

Thanks! Please help me understand why (?s) is added to the beginning of the delimiter. I was experimenting with ?s and \v but I was placing them everywhere except the beginning...

[PsaltyDS 27]

That's a control switch, changing how the rest of the pattern is interpreted. It doesn't match or return anything on its own. In this case it changes how "." is interpreted from that point forward.

Another example is "(?i)" for case insensitive.

[jchd 990]

Beside the brief introduction found in AutoIt helpfile, you can refer to here and here as reference documentations.