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Find if objects are on top of each other (text, attributes)

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T.Willey:
I was trying to find a way to find what objects are on top of each other, so I went with the method
'BoundingBoxIntersectWith' to test two entities.  It looked like it worked, but upon more testing it doesn't do what I want.  Then I went with just the 'IntersectWith' method, but that doesn't work either.  I'm at a lose, as I can't use points because they don't always share a point.

Any help is appreciated.  Attached is a pic showing my problem; as it returns they do not intersect.

LE:
Tim;

Do not know if we can use from C# the ARX class AcDbMPolygon - in there you will find methods that work for what you are trying to implement.

T.Willey:

--- Quote from: LE on October 29, 2007, 06:55:44 PM ---Tim;

Do not know if we can use from C# the ARX class AcDbMPolygon - in there you will find methods that work for what you are trying to implement.

--- End quote ---
Not sure Luis.  Let me do some research, and get back to you.  Thanks.

Edit:  Looks like you can, now to just find out how to use it.  Thanks.

LE:
The other, there is one function I wrote in C# it is available in show your stuff to verify if a point is inside of a closed area, if you are looking to check for four points, then, it can be easy to adapt...

Let me check where is that - I'll be back

Here:

http://www.theswamp.org/index.php?topic=17222.msg209140#msg209140

T.Willey:

--- Quote from: LE on October 29, 2007, 06:58:43 PM ---The other, there is one function I wrote in C# it is available in show your stuff to verify if a point is inside of a closed area, if you are looking to check for four points, then, it can be easy to adapt...

Let me check where is that - I'll be back

Here:

http://www.theswamp.org/index.php?topic=17222.msg209140#msg209140

--- End quote ---
I just downloaded it, and was looking at the stuff you and Alexander Rivilis were doing here.  I think I will need more time, so I will put this off till tomorrow when I might... might be able to dig into it.  Thanks again Luis.

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