topic Re: How do I append a literal digit (e.g., "0") after a regex capture group in JMP? in Discussions

How do I append a literal digit (e.g., "0") after a regex capture group in JMP?

BHarris — Thu, 14 Aug 2025 13:20:19 GMT

I'm using regular expressions in JMP Recode's "Replace String" function to transform strings like "en127" into "1270". My approach is to capture the two digits using:

Find:    en(\d+)

But when I try to use:

Replace: \10

…it is interpreted as a reference to group 10, not group 1 followed by a literal "0".

Is there a way in JMP to append a literal "0" (or any literal digit) after a capture group without it being misinterpreted as a higher-numbered group?

Other engines (like Perl or JavaScript) let you disambiguate with braces or dollar signs (e.g., ${1}0), but JMP doesn't seem to support those.

I’d prefer a one-step solution, not one that requires a second replace operation.

Any insight on how to do this cleanly in JMP?

Thanks!

Re: How do I append a literal digit (e.g., "0") after a regex capture group in JMP?

Craige_Hales — Thu, 14 Aug 2025 14:48:45 GMT

Away from computer so I can't look it up right now. I think you can use another escape after the \1 that won't do anything special. I think there's some escapes for upper casing and lower casing and things of that nature. Edit: if there is a code to upcase the next char that would be perfect since it would not change the next digit.

Re: How do I append a literal digit (e.g., "0") after a regex capture group in JMP?

jthi — Thu, 14 Aug 2025 16:04:14 GMT

So for example, using \L (single character that is not lowercase [^a-z])

Scripting Guide > Types of Data > Regular Expressions > Escaped Characters in Regular Expressions

Re: How do I append a literal digit (e.g., "0") after a regex capture group in JMP?

Craige_Hales — Fri, 15 Aug 2025 00:11:24 GMT

\e (lower case e) was designed for this in a replacement. (not in a pattern match where it matches the escape character.)

https://community.jmp.com/t5/Discussions/Regex-backreference-followed-by-a-number-unambiguous/m-p/714744/highlight/true#M89807

and

@shannon_conners The doc @jthi points to has unfortunately mixed the parameter 2 matching grammar with the parameter 3 replacement grammar; \L etc has different usage in the two places. \L matches a not-lower-case character in parameter 2 and begins lower casing in a replacement string. Same for \u \U and \l. Also \E \Q only belong in parameter 3 replacements.

Re: How do I append a literal digit (e.g., "0") after a regex capture group in JMP?

BHarris — Fri, 15 Aug 2025 15:32:27 GMT

Yes, excellent, this appears to be the most canonical response, e.g. replace "en(\d+)" with "\1\e0". That works, and follows the same pattern as "\U" and other special characters in replace-strings.

(For the probably-less-than-3 people who read this and are wondering what "\U" does, it allows you to do case changes on the replace string, e.g. if you had a list of names in a column, and wanted to change "Harris, Bryan" to "HARRIS, Bryan", you could do by with finding "(.+)(,.+)" and replacing it with "\U\1\E\2". "\U" and "\L" are uppercase and lowercase respectively, and they operate until the next "\E". You can also do a case-change on just the first letter with "\l" and "\u", terminating with "\e". I cannot find any mention of these in the documentation, but it may be that I just don't know where to look.)

Thanks @Craige_Hales and @jthi !