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Question Jonathan Harris · Mar 7, 2023

Is it possible to use a regular expression in a ..ReplaceStr function

Messages will contain fields with expressions like "REASON->Blood(1.23)" or "REASON->Bone(4.56)" or "REASON->Muscle Mass(7.89)". The word after the "->" can vary. I would like the outputs to be; "REASON(1.23)" or "REASON(4.56)" or "REASON(7.89)", basically removing "->" and the word or words that follow but leaving the parens and the numeric characters within.

Thanks,

Jonathan Harris

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Jeffrey Drumm · Mar 7, 2023

Something like this, perhaps?

Class User.Util.StringFunctions Extends Ens.Util.FunctionSet
{

ClassMethod ReReplace(pStr As %String, pPat As %String, pRepl As %String = "") As %String
{
    Set tStrt = $LOCATE(pStr,pPat,,tEnd) - 1
    // in case the pattern isn't found, return source string
    Return:(tStrt < 0) pStr
    Set tPrefix = $EXTRACT(pStr,1,tStrt)
    Set tSuffix = $EXTRACT(pStr,tEnd,*)
    Return tPrefix_pRepl_tSuffix
}

}

USER> set mystr = "REASON->Blood(1.23)"
USER> set newstr = ##class(User.Util.StringFunctions).ReReplace(mystr,"->\w+")
USER> write newstr
REASON(1.23)
USER> set altstr =  ##class(User.Util.StringFunctions).ReReplace(mystr,"->\w+","-CODE")
USER> write altstr
REASON-CODE(1.23)

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Timo Lindenschmid · Mar 8, 2023

Hi,

if you make some assumptions

1. numbers are always enclosed in curved brackets

2. you want to always return string starting with REASON

this could be as easy as just:

set mystring="REASON->Blood(1.23)"

w "REASON"_$extract(mystring,$find(mystring,"(")-1,*)

or if you really want to use regex:

IRISHEALTH:USER>set mystring="REASON->Blood(1.23)"
IRISHEALTH:USER>set regex=##class(%Regex.Matcher).%New("^([A-Z]*)->.*(\([0-9]*.[0-9]*\))",mystring)
IRISHEALTH:USER>zw regex.Locate()                                               1
IRISHEALTH:USER>zw regex.Group(1)
"REASON"
 IRISHEALTH:USER>zw regex.Group(2)
"(1.23)"

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