Hi Ryan, You can get the coordinates of the point by using coordinate geometry and trigonometry. The key would be to determine the angle (or slope) of the line segment and the distance to the event point. Or, if you wanted to make the segments square, use form="step":
<mx:LineSeries form="step"/> -TH --- In [email protected], "Ryan" <rmarples+flexcod...@...> wrote: > > I've got a line chart showing a dollar value over time. Think of it as the > balance of a savings account. It goes up and down every month or so. What I'm > trying to do is overlay little widgets on top of that line on the chart to > represent events that may or may not have affected the line's value. This is > actually similar to the Yahoo Finance stock chart with the event overlays. > > Anyway, it's easy enough to add a CartesianDataCanvas as an annotation > element to my line chart and then add children to it. It has a nice little > addDataChild method that allows you to specify an x axis value and a y axis > value and it figures out what the corresponding x and y coordinates are and > places the child there. Great so far. > > Say my line series data provider looks like this: > {date:new Date(2009, 1, 1), balance:100} > {date:new Date(2009, 2, 8), balance:140} > {date:new Date(2009, 3, 5), balance:95} > > If I have an event for 1/1/2009, I can look up that the balance was 100 on > that day and place the event widget at [1/1/2009, 100]. However, if I have an > event on 1/15/2009, from looking at the data provider above, I know that the > balance was still 100, but because the line is diagonally vectoring towards > the next value of 140, my data point is misplaced if I put it at [1/15/2009, > 100]. > > Is there a way to get the line's y value given an x value? > or > Can I change my line chart to not draw diagonal lines and instead draw square > lines so I won't have this problem? > > Ryan >
