Can Representations be used to Spatially Offset (Disperse) XY points for Web Mapping Application?

Question

So I have a bit of an issue of points with the same x and y that I need to display separately from themselves. Here is some history of this issue:

I have a database of incidents that happen over a county. These incidents join to a parcel fabric with a one-to-many relationship. So several incidents occur on one parcel. I need the incidents to be shown as a point feature class for a web mapping application. So I added the x and y centroid of the associated parcel and created the point feature class. This resulted in several points being drawn on top of each other as shown in this picture:

I converted the symbology to a representation with a 10 meter minimum gap between points which resulted in exactly what I was looking for as show in this picture:

Now this works great in ArcMap, but when uploaded to my web mapping app the points can only be identified at there original x and y. So when you identify points that have been spaced you get no result, but when you identify the center of the parcel you get several results.

What I would love to happen is to somehow return the xy of these representations and change xy of each point to that. I've tried a few programming solutions, but I haven't been able to solve it yet.

Answer 1

I ended up finding a pretty good programming way to work with this. I used an array to alter the x and y in the table before converting the table to point. This array simple placed the points around the centroid of the parcel. Heres the code:

for x in output:                  # output is a list of unique ID's of the table
    expr =  "UID like '" + str(x) + "'")
    Z = 0
    B = 1
    X = [0, 0,10,10, 10,  0,-10,-10,-10]
    Y = [0,10,10,00,-10,-10,-10,  0, 10]
    rows = arcpy.UpdateCursor("Table", expr)
    for row in rows:
        row.Northing = row.Northing + (Y[Z]*B)
        row.Easting = row.Easting + (X[Z]*B)
        Z = Z + 1
        if Z == 9:
            Z = 0
            B = B + 1
        rows.updateRow(row)

Here is what it ended up looking like:

I hope this helps anyone else. If you have some questions let me know.