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How to check if ComboBox dropdown list is shown or not shown?
thanks.
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GetDroppedState()
You may be right I may be crazy -- Billy Joel --
Within you lies the power for good, use it!!!
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For a limited user on Windows XP, will regsvr32.exe be able to register a namespace extension? I tried and got error 0x80040201, but I'm not sure if this is an issue to do with the account being a limited user or not.
Any help would be greatly appreciated.
Thanks in advance!
Dave Kerr
codechamber@hotmail.com
http://www.codechamber.com
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Dave Kerr wrote: I tried and got error 0x80040201, but I'm not sure if this is an issue to do with the account being a limited user or not.
It resolves to, "An event was unable to invoke any of the subscribers."
"The largest fire starts but with the smallest spark." - David Crow
"Judge not by the eye but by the heart." - Native American Proverb
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Hello,
I have some problems with the PlotXY functions in the NTGraph activeX, Here is the article about this ActiveX:
http://www.codeproject.com/miscctrl/ntgraph_activex.asp[^]
I'll upload for you the picture of the problem here:
http://img433.imageshack.us/img433/4972/untitled2ee.jpg[^]
the function I wrote there the code is: BOOL CGraphsXYAxisDlg::OnInitDialog()(This is the function in the general Dialog)
I sure the problem is there because when I earase the line it works fine.
I tried a few combinations of the point(x,y) but it doesn't work and the errer occurs.
Do you know how can I fix it?
Thank you all for your great help!
SnaidiS(Semion)
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Incidentally the correct place to post these questions would actually be at the relevant article[^] itself.
The error message says Element not found . Do you actually have 3 elements for your graph? (ie. did you add 2 extra elements as the 1st was created automatically)
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OK thank you! I think I found the problem..
SnaidiS(Semion)
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How can I draw a line with this ActiveX.
I tried to use the functions:
OLE_COLOR ElementLineColor
OLE_COLOR ElementPointColor
LineType ElementLinetype
but it doesn't draw..
Do you have any ideas how can I draw a line between all the points?
SnaidiS(Semion)
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Provide some source code of what you are doing. I don't see what is so difficult.
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I know what you want to do. I want to know what steps you have taken to attempt to achieve this and your existing code because it looks to me like the SetElementXXX functions have everything you need.
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I haven't tried to add the line because I totaly dont know how. I just know how to add points to the board, if you want the code of adding the points: take:
<br />
void CGraphsXYAxisDlg::OnBnClickedButton2()<br />
{<br />
m_Graph.AddElement();<br />
m_Graph.SetElementWidth(10);<br />
m_Graph.PlotXY(X,m_Y,X);<br />
X++;
UpdateData(FALSE);<br />
<br />
}<br />
Thank you!
SnaidiS(Semion)
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Semion_N wrote: I just know how to add points to the board
You didn't even get that right...
An element is not a data point, it's a data set. If you only have one data set, which you do, you are supposed to just m_Graph.PlotXY(x,y,0);
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You mean that when I'm adding element so its not a point it can be a line too?
but how can I use the line and not the point-you wrote this m_Graph.PlotXY(x,y,0);
I used it but as a point draw.
How can I make lines?
SnaidiS(Semion)
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From the article...
m_Graph.SetElementLineColor(RGB(255,0,0));
m_Graph.SetElementLinetype(0);
m_Graph.SetElementWidth(1);
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Yes-I tri9ed this option but when I'm typing it here for instance:
<br />
void CGraphsXYAxisDlg::OnBnClickedButton2()<br />
{<br />
UpdateData(TRUE);<br />
m_Graph.AddElement();<br />
m_Graph.SetElementWidth(10);<br />
m_Graph.PlotXY(X,m_Y,X);<br />
X++;<br />
m_Graph.SetElementLineColor(RGB(0,0,0));<br />
m_Graph.SetElementLinetype(0);<br />
m_Graph.SetElementWidth(1);<br />
}<br />
The program do nothing it dont draw even only the points...
TY!
SnaidiS(Semion)
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You don't seem to understand my previous message. An element is NOT a data point. An element contains a set of data points along with the parameters of how to draw the data set. Lets say I would like to plot the stock market performance of a few companies, the performance data of Yahoo would be one element and the performance data of Google would be another element. So I have 2 elements. I can set the color for Yahoo to be yellow and the color for Google to be blue.
<br />
void CGraphsXYAxisDlg::OnBnClickedButton2()<br />
{<br />
UpdateData(TRUE);<br />
m_Graph.AddElement();<br />
m_Graph.SetElementWidth(10);<br />
m_Graph.PlotXY(X,m_Y,X);<br />
m_Graph.PlotXY(X,m_Y,0);<br />
X++;<br />
}<br />
If it still doesn't work, set a breakpoint at the PlotElement() function in NTGraphCtl.cpp and see why it's not drawing the line.
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Yes it works!!
First of all-thank you!
but I don't understand- an element contains number of points,so why don't you add elements here?
and then draw them?
I don't really understand why PlotXY draws a set of point if you give to this function only one point?
SnaidiS(Semion)
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By default the graph starts with one element. It is reasonable to assume that you must at least have one data set if you want to plot a graph.
PlotXY doesn't draw anything. The element contains an array of points, PlotXY simply adds a point to the array. When there's a WM_PAINT message, the control iterates thru each element and draws all the points based on the per element settings.
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OK I understoon TY but PlotXY adds into the array only one point on each button press in my case so why does it draws a line and not point?
SnaidiS(Semion)
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I don't get the question. With 2 points you can draw a line.
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yeah... you are right!
Thank you man! you helped me alot!
SnaidiS(Semion)
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class RTI_EXPORT RTIambassador {<br />
public:<br />
#include "RTIambServices.hh"<br />
RTIambPrivateData* privateData;<br />
private:<br />
RTIambPrivateRefs* privateRefs;<br />
};
I didn't know the meaning of having two names beside the name Class like this:
class RTI_EXPORT RTIambassador
Someone gave me an answering saying the class is intended to be exported from a DLL.
What is meant by a class being exported from a DLL. Does this mean that this specific code here is a a little chunk of a code from a DLL file and is being used by the current file Im working with? Or does it mean, tell the DLL linker to add this specific class onto the linker as an addition function.
I'm very confused here. Someone help me out. Thanks in advance
PS: I've already posted this before. Sorry for posting it again but I haven't got much reply. I beleive its because of the time I posted. Hope someone can help me out
Cheers,
Jay -
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Jay03 wrote: class RTI_EXPORT RTIambassador
RTI_EXPORT is a preprocessor macro. When dealing with DLLs, there are some compiler directives you can use (instread of having *.def files) to say you are exporting (or importing) classes, variables, functions, etc. If you do a search for RTI_EXPORT, you should find it defined something like the following:
#ifdef _WINDLL
#define RTI_EXPORT __declspec(dllexport)
#else
#define RTI_EXPORT __declspec(dllimport)
#endif
What that does is allow the you to compile the DLL stating that this class is being exported while you can use the same header file in the application and state that the class is being imported.
Now, when writing DLLs for Windows applications (as opposed to so's for *Nix systems), you have to tell the compiler which symbols (variables, functions, classes, etc) are being exposed to be used by anyone wishing to use the library. At the same time, when you write an application that needs to use a DLL, you need to know which symbols are being imported so the linker knows what to do with them.
If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week
Zac
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