I tried with a component TTable and TQuery BDE) with the employee table of DBDEMOS and
actually it works!

Do I have to conclude that the CHART does not work with all QUERY?

For my part, I use TIBOQuery.

I am the same problem with the version 7.12...

I am very sad..

I download the 7.12..

hou hou..

I bought this version in June 2007.. and I read into the mail :

Thanks for Ordering TeeChart Pro 7.0 VCL/CLX version !

At that time the PIE not working?

there is a problem..

Can I get the updated 7.12?

Thank you

Didier

here is my version :

Teechart pro v7
@1995-2004 by David Berneda

is it ok ?

When I choose BAR, that's good but
if I choose PIE, nothing
only the legend
I do not see the chart!
???
Thank you

Didier

It's a very simple exercise:

I put a TDBChart, I choose "PIE" and I connects
the series to a dataset is working well.

thank you for your help

Didier

Hi narcis,

The Chart type "BAR" work BUT not those type "PIE".
It does not just see the legend..
Why?

thank you for your help

Didier

Hi Narcis,

First of all, thank you for your help !

I have taken careful note of subtlety:
AddField ( 'X', 1);
AddField ( 'KN', 2);

Whether we can write:
AddField ( 'X', 1);
AddField ( 'Y', 2);

Everything works well now!

Didier

hé hé, too easy..
actually, I put X, Y for understanding
but I have in my file:

mm; KN
0; 0
0; -0.02
0; -0.02
0; -0.02
....
and so on...

and I always have the same problem.

Thank you for your help

Didier

Narcis, here is the code : procedure TForm1.BitBtn1Click(Sender: TObject); begin with SeriesTextSource1 do begin Fields.Clear; AddField('mm', 1); AddField('KN', 2); HeaderLines := 1; FieldSeparator := ';'; DecimalSeparator := ','; FileName := 'perrier.csv'; Series := Series1; Chart1.Legend.Visible :...

Hi Narcis,

Thank you for your help
I had already noted that example.
the graph displayed
but I can not keep the real values on the axes.


Didier

Here is a list of point X, Y in a textfile. I would like to draw the graph with a TEECHART after reading the file and while maintaining the real values in the two axes. thank you for me to proceed Didier x;y 0;0 0;-0,02 0;-0,02 0;-0,02 0;0 0;-0,02 0;-0,02 0;-0,02 0;-0,02 0;-0,02 0;0 0;-0,02 0;-0,02 ...