reduce left and right side of poly separately

Please read the Forum Guidelines. You have on two occasions posted to existing threads, for things that are not directly related. Also, this question is very similar to one you have asked earlier.

I will split this into a separate thread (since the subject is different from the thread you appended to, to make matters worse)

The simple answer is no, but you could prevent the top and bottom being stretched by doing a dbLayerAnd with a rectangle which is a slightly stretched version of the bounding box. I had thought of putting together an example, but since I'm still having to guess what precisely you're doing (as each post seems to have contradicting requirements), I'm going to wait until you've clarified what you're trying to do (as I asked for in your other thread).

Please bear in mind that everyone on these forums is a volunteer, and are doing so in their spare time, and so you need to ensure that you minimize any calls on their time by maximizing the amount of information you provide.

Kind Regards,

Andrew.

One quick thought as to how you could do this assymmetric sizing:

1. Foreach polygon, find the bBox (it's an attribute on the shape itself)
2. Looking along the point list, you can identify points which are on the top or bottom edge (because the y-coordinates will match the yCoord(upperRight(bBox)) or yCoord(lowerLeft(bBox))).
3. From this, you should be able to determine which points form the left or right edges, because they will start from the smaller xCoord on the top to the smaller xCoord on the bottom (and similar for the right side).
4. Any points on the left side you could update the xCoord to subtract the shift you want, and similarly for the right.
5. The resulting point list could then be updated on the polygon.

Needs a bit of manipulation of the point list, but it shouldn't be too difficult if the polygons are reasonably consistent.

So you don't even need to use the dbLayer.* functions in this case.

Andrew.