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Jonathan Harris · Mar 7

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

#InterSystems IRIS for Health

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

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

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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