.NET Regex issue: end-of-string char matching the dollar of the string

Question

I'm having an issue with a Regex. I basically need to split a string with Chr(1) as a delimiter and each group is formatted as key=value (where key is numeric). The beginning of the string is a 3-char code and is not gathered with the Regex. Example:

031|0=2013/12/04 00:03:35|400=lr5ulz1jxg8ss|3=SFE|4=2$1NNR|6=1

So I used the following Regexp:

Private Shared _RegexDeserialize As New Regex(String.Format("{0}(\d+)=([^${0}]*)", Chr(1)))

But when there is a dollar ($) in the value part, the rest is not matched.

Then I used the following:

Private Shared _RegexDeserialize As New Regex(String.Format("{0}(\d+)=([^{0}]*)", Chr(1)))

I don't understand why my previous Regex doesn't work properly and why VB.Net is matching the end-of-string delimiter ($) to the dollar ($) of the string. If I wanted to match to a dollar, I should have escaped it: \$

Any help will be appreciated. Thanks very much.

If I wanted to match to a dollar, I should have escaped it: \$. Yes you answered yourself. — Siva Charan
– Siva Charan, Commented Feb 4, 2014 at 16:21
All characters inside a character class are treated as literal characters and not as regex meta characters, except the characters ^ and - which take different meanings depending on their position in the character class. — Jerry
– Jerry, Commented Feb 4, 2014 at 16:46
i think the correct pattern should be: "(\d+)=([^|]*)" in your example takes 5 matches and for each two groups before and after "=" — Francesco Milani
– Francesco Milani, Commented Feb 4, 2014 at 16:50

The Guy with The Hat · Accepted Answer · 2014-02-04 16:59:23Z

3

[...] is a character class. Normally, $ (along with many other characters such as *+?)[.(|) are metacharacters. However, when they are inside a character class, they become literal characters. This means that they lose their special meaning, so a $ inside a character class matches $, not and end-of-string. Instead of $, you can use \Z which is the equivalent, but still works inside character classes.

I'm not sure exactly what you are trying to accomplish, but I think the regex you want is

\d+=[^|]+

Demonstration here: http://regex101.com/r/rI4tM3

edited Feb 4, 2014 at 16:59

answered Feb 4, 2014 at 16:51

The Guy with The Hat

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2 Comments

user3271580 Over a year ago

Thanks a lot for your help. I didn't know this specificity.

Alan Moore Over a year ago

Are you saying [\Z] matches the end of a string, just like \Z? That's not true. It's just a syntax error.

Luaan · Accepted Answer · 2014-02-04 16:51:37Z

2

Well, your $ is inside of [^...]. It can't mean end-of-string in that location, so it thinks you mean a literal $.

The reason this "usually" works for you is because you don't actually need the end-of-string at all :)

"{0}(\d+)=([^{0}]*)"

The * is greedy, it will eat up as many characters as it can, so everything works just fine :)

answered Feb 4, 2014 at 16:51

Luaan

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1 Comment

user3271580 Over a year ago

Thanks a lot for your help and to confirm me that the fix-version will work properly. Much appreciated.

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