labeling a layer with the values of stand-alone table (that is associated to the layer with 1-Many relationship),
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Dear Richard Fairhurst,
Thank you very much.
I want to clarify about my second question, actually I create the relationship based on a field “Ref_B”, and in my case I need my label to be (Ref_B from the feature class, and the “Name” from related table) as shown below,
so I tired the following code and I got the following error.
# Initialize a global dictionary for a related feature class/table
relateDict = {}
def FindLabel ( [Ref_B], [Ref_B]):
# declare the dictionary global so it can be built once and used for all labels
global relateDict
# only populate the dictionary if it has no keys
if len(relateDict) == 0:
# Provide the path to the relate feature class/table
relateFC = r"D:\RelateLabeling\s.gdb\Buildings_T"
# create a field list with the relate field first (Ref_B),
# followed by sort field(s) (Ref_B), then label field(s) (NAME)
relateFieldsList = ["Ref_B", "NAME"]
# process a da search cursor to transfer the data to the dictionary
with arcpy.da.SearchCursor(relateFC, relateFieldsList) as relateRows:
for relateRow in relateRows:
# store the key value in a variable so the relate value
# is only read from the row once, improving speed
relateKey = relateRow[0]
# if the relate key of the current row isn't found
# create the key and make it's value a list of a list of field values
if not relateKey in relateDict:
# [searchRow[1:]] is a list containing
# a list of the field values after the key.
relateDict[relateKey] = [relateRow[1:]]
else:
# if the relate key is already in the dictionary
# append the next list of field values to the
# existing list associated with the key
relateDict[relateKey].append(relateRow[1:])
# delete the cursor, and row to make sure all locks release
del relateRows, relateRow
# store the current label feature's relate key field value
# so that it is only read once, improving speed
labelKey = [Ref_B]
# start building a label expression.
# My label has a bold key value header in a larger font
expression = u'<FNT name="Times New Roman" size="14"><BOL>' + Ref_B +'</BOL></FNT>'
# determine if the label key is in the dictionary
if labelKey in relateDict:
# sort the list of the list of fields
sortedList = sorted(relateDict[labelKey])
# add a record count to the label header in bold regular font
# expression += u'\n<FNT name="Times New Roman" size="12"><BOL>NAME Count = {}</BOL></FNT>'.format(len(sortedList))
# process the sorted list
for fieldValues in sortedList:
# append related data to the label expression
# my label shows a list of related
# cross streets and measures sorted in driving order
expression += u'\n' + fieldValues[0]
# clean up the list variables after completing the for loop
del sortedList, fieldValues
else:
expression += '\n<FNT name="Arial" size="12"><BOL>NAME Count = 0</BOL></FNT>'
# return the label expression to display
return expression
But I found workaround to solve this, I add a new field in the feature class with different name but have the same value as Ref_B and used this field in the code and that working fine.
And for the third question I found that if we remove all null values the code works fine.
Best,
Rawan
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