Trying different fields when Maplex labelling?

Question

Let's say, a road has the name "William-Bill-Gates-Road".

If the road is too short, the label can only be abbreviated by using a dictionary in Maplex. Is it possible to make Maplex try several label fields, if one is not fitting?

Structure:

name | name_1 | name_2 | ... William-Bill-Gates-Road | William-Bill-Gates-Rd. | W.-Bill-Gates-Rd. | W.-B.-Gates-Rd.

Answer 1

The short answer is yes.
here is some code that should work in 10.0 and previous.
Grant it doesn't exactly use an alternate field but shortens the word.
I figure you might be able to recreate it with the loop you need.

Function FindLabel ( [Name] )

Dim strArray
Dim outStr
Dim x
Dim maxWordLength
Dim minLengthLength
Dim everyother
Dim isFirst
Dim lastWordLong
lastWordLong = false

maxWordLength = 10
minWordLength = 3
everyother = False
isFirst = True
strArray = Split( [Name], " ")

For x = 0 To UBound(strArray)

    If ((Not isFirst) And (Len(strArray(x)) <= minWordLength)) Then
        'keep short word on existing line
        outStr = outStr & " " & strArray(x)
        everyother = False
    ElseIf (everyother) Then
        'word x is slated to be second word on line
        If (Len(strArray(x)) > maxWordLength) Then
            outStr = outStr & vbNewLine & strArray(x)
            everyother = False
        Else
            outStr = outStr & " " & strArray(x)
            everyother = False
        End If
    Else
        'word x is slated to be first word on line
        If (isFirst) Then
            outStr = strArray(x)
            everyother = True
            isFirst = False
        Else
            If (Len(strArray(x)) > maxWordLength) Then
                lastWordLong = True
                everyother = False
            Else
                lastWordLong = False
                everyother = True
            End If

            outStr = outStr & vbNewLine & strArray(x)

        End If
    End If
Next

FindLabel = outStr
End Function

I didn't write it but inherited a map with that vbscript in it.