Re: Label postions on item.
drawing process relies on a relative object as start point. in normal example, first and last symbols, and lines are drawn first. now you have three options to select as a relative point. in second example of start/end time being the same, you have only one relative point to select.
Kuphryn
Re: Label postions on item.
Quote:
Originally Posted by kuphryn
drawing process relies on a relative object as start point. in normal example, first and last symbols, and lines are drawn first. now you have three options to select as a relative point. in second example of start/end time being the same, you have only one relative point to select.
Kuphryn
Thanks for your reply Kuphryn, my problem is not necessarily the point of reference, but how do I go about determining the label position based on the point of reference, the index of the label, and the text alignment specified.
In the example bitmap, label 2, 3, and 4 have the connector as referece, but, what if 4 had a vertical alignment of "TOP" specified, that would mean 4 must be drawn above the connector, and labels 2 and 3 have to be moved up to accomadate 4.
This gets even tougher since each label, must be able to accomadate a different font and font size.
One solution I thought of, but not sure if it is overkill, is to create 3 - 3X3X5 matrix's of "Label positions", where the first 3 is Left symbol, connector, and right symbol labels, then the 3X3X5 matrix represents all the different alignment values and the 5 deep is because there is 5 lables.
This would probably not be very efficient for drawing however?
Mike B
Re: Label postions on item.
This depends on design specifically drawing. one case you draw all objects. now select a different object based on its position attribute. another case each object draws itself. key is no collision via keeping track of the set of position attributes.
Kuphryn
Re: Label postions on item.
Sorry Kuphrin, see my previous post as it was edited while you were replying.
Mike B
Re: Label postions on item.
array is one solution. inefficiency would be something like bunch of if/else
Kuphryn