Time for another game of "Fix My Rube"

[GregFreeman]

When I went to impliment this, I realized because the delimiters are actually telling me something about the following data, I need to know which delimiter was at the beginning of each match. Is there any easy way to get this? There doesn't seem to be (i.e. it's not an output of the function).

[altenbach]

Use my VI, but wire the array of delimiters to the "operators" input instead (leave delimiters unwired), and also autoindex the "token Index" output.

You'll get two arrays, one with small strings and one with indices. Whenever the index is -1, you get the elements of the original solution. If the index is 0 or larger, it points to the element in the delimiter array. 

(but wait a few minutes, maybe somebody will post a regex solution :D)

[GregFreeman]

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@altenbach wrote:

(but wait a few minutes, maybe somebody will post a regex solution :D)

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I guess we'll see! Thanks though, works great.

[Darin.K]

Simply change the replace string to:

\n$1\t

Change the output to a 2D array and the first column is your token and the second is the string.

[crossrulz]

Darin, you're slacking.  That was more than just a few minutes...

[GregFreeman]

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@Darin.K wrote:

Simply change the replace string to:

\n$1\t

Change the output to a 2D array and the first column is your token and the second is the string.

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Oh regex's...

[GregFreeman]

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@crossrulz wrote:

Darin, you're slacking.  That was more than just a few minutes...

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He must actually be working today 

[Charles_CLA]

Darin, I'm sure you've seen this already and probably already have the shirt but I couldn't resist posting it as you appear to be the foremost expert on regex in LabVIEW.

[Mark_Yedinak]

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@Charles_CLA wrote:

Darin, I'm sure you've seen this already and probably already have the shirt but I couldn't resist posting it as you appear to be the foremost expert on regex in LabVIEW.

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I like the shirt. My kids would give me lots of grief if I got it though.