Hi Regina, Regina Henschel <rb.hensc...@t-online.de> 於 2019年4月10日 週三 上午2:38寫道：
> Hi Mark, > > Mark Hung schrieb am 09-Apr-19 um 11:27: > > Hi Regina, > > > > Regina Henschel <rb.hensc...@t-online.de > > <mailto:rb.hensc...@t-online.de>> 於 2019年4月9日 週二 上午7:06寫道： > > > > Hi all, > > > > I'm looking at > > https://bugs.documentfoundation.org/show_bug.cgi?id=115813 > > "Position shifted while moving the handle of OOXML Callout shape". > > > > Unfortunately not only the Callout shape is affected. > > There are 106 shapes in MS Office, which have handles. 38 are OK with > > the current implementation and 68 have at least one handle, which > > reacts > > wrongly. These are only those from ooxml, not from binary format. > > > > The calculation is in method > > EnhancedCustomShape2d::SetHandleControllerPosition in > > svx\source\customshapes\EnhancedCustomShape2d.cxx. The relevant part > > starts about #1230 > > > > > > It's not clear to me what makes control point of Impress callout shape > > attach to the pointer, > > The definition of the handle position has a direct reference to an > adjustment value or an indirect reference via a formula and is > calculated from that. A user handle-drag requires a new adjust value, > which will then produce a new handle position. If the calculation is > wrong, then the new position of the handle is not there, where the user > has released the mouse. Calculating an adjustment value from the > drag-end position is done in method SetHandleControllerPosition. > > and > > what are the differences between the Impress one and the PowerPoint one. > > The PowerPoint shape uses other formulas, even if you convert from pptx > to odp. The used formulas are direct translations of the formulas in the > OOXML spec. > > For example (only for the x-coordinate): > OOXML has > gdRefX="adj1" and x="xPos" in the handle description > and the formulas > xPos: fmla="+- hc dxPos 0" means: horizontal center + dxPos - 0 > dxPos: fmla="*/ w adj1 100000" means: width * adj1 / 100000 > > That is translated by the part Miklos has mentioned and results in these > formulas, when written as odp: > handle-position="?f3 ?f5" > f3: formula="?f2 +?f0 -0" > f2: formula="logwidth/2" > f0: formula="logwidth*$0 /100000" > > A ? means a reference to a formula. > A $ means a reference to an adjust value. > > Our own callout has > handle-position ="$0 $1" > The handle position is directly bind to the adjustment values. > In that case only scaling is needed. > > Thanks for explaining in detail. > > > > Have you checked how custom shapes are imported already? > Relevant > codes are in > > /oox/source/drawingml/customshapes/customshapeproperties.cxx > > Yes, the generated formulas look correct for me. > > The problem in the example above is this: > If you put the formulas together you get > handlePositionX = logwidth/2 + logwidth * $0 /100000 > Solve the equation for $0. You get > $0 = (handlePositionX - logwidth/2) * 100000 / logwidth > > But current calculation has only * 100000 / logwidth without the > subtraction. > > I'm thinking if we can use delta(handlePosition) as an estimation of delta(adjustment), in case the position is a formula instead of an adjustment. Might be a naive idea. > I see no general rule. I think, I need to look at each individual handle > to get its correct formula. > > Kind regards > Regina > -- Mark Hung
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