Email config setting. - C / C++ / MFC Discussion Boards

ShilpiP18-Aug-10 1:17

18-Aug-10 1:17

If your setting is "Not using ATL" than dont change the setting if your setting is "Dynamic Link to ATL" than change it to "Static link to ATL" and what about MFC setting.

Yes U Can ...If U Can ,Dream it , U can do it ...ICAN

Re: Code-Generation Setting

gothic_coder18-Aug-10 1:27

gothic_coder

18-Aug-10 1:27

MFC Setting is "Use Standard Windows Libraries"

Re: Code-Generation Setting

ShilpiP18-Aug-10 2:21

ShilpiP

18-Aug-10 2:21

I am not sure please change the settings of "Use Standard Windows Library" to "Use MFC in a Static Library" and compile.
if u receive any error of start up than update the setting of Linker Entry point and than check if it is working fine.

Yes U Can ...If U Can ,Dream it , U can do it ...ICAN

Re: Code-Generation Setting

gothic_coder18-Aug-10 2:42

gothic_coder

18-Aug-10 2:42

I added msvcrt.lib and it's working Smile | :)

Re: Code-Generation Setting

ShilpiP18-Aug-10 3:10

ShilpiP

18-Aug-10 3:10

Thats Gr8

Yes U Can ...If U Can ,Dream it , U can do it ...ICAN

Re: Code-Generation Setting

Rick York19-Aug-10 20:19

Rick York

19-Aug-10 20:19

You, sir, are tied for having the world's lamest signature.

Re: Code-Generation Setting

gothic_coder19-Aug-10 22:37

gothic_coder

19-Aug-10 22:37

Who me??

Email config setting.

Samarjeet Singh@india17-Aug-10 21:40

Samarjeet Singh@india

17-Aug-10 21:40

Hi All,
In my website there is facility that user can send mail using it own email SMTP setting. some user is sending mail success fully but some facing problem for this there is option in config page in which user have to enter his SMTP setting. And also there is test button when users press a test button and if any wrong entry (username, password, smtpServer) then I want to show pop message accordingly.
Is any way in which we can identified that what is wrong entry (username, password or mail server) because I want to give message like "wrong username or wrong SMTP server.

Thanks

Re: Email config setting.

Richard MacCutchan17-Aug-10 22:27

Richard MacCutchan

17-Aug-10 22:27

Samarjeet Singh@india wrote:
Is any way in which we can identified that what is wrong

You could start by posting your question in the correct forum.

It's time for a new signature.

Re: Email config setting.

Samarjeet Singh@india17-Aug-10 22:31

Samarjeet Singh@india

17-Aug-10 22:31

Realy I am so sorry.
is there any way by which we can change this post in correct(asp.net)forum?
thanks

Re: Email config setting.

Richard MacCutchan18-Aug-10 1:15

Richard MacCutchan

18-Aug-10 1:15

Don't worry about this as it has now been marked. Just repost in the ASP.NET forum and hope you get a good answer.

It's time for a new signature.

Rotate rerctangle in Bitmap

raju_shiva17-Aug-10 21:32

raju_shiva

17-Aug-10 21:32

Hi sir,
I have origin axis of bitmap and x1,x2,y1,y2.
I want to draw a rectangle and rotate it with the given degree.I am trying with this code,but i am not getting.
Can some one help me

double x1,x2,y1,y2;
sf=(1024/31);  //Its a bitmap image
A1 = 10.00; //these are values i am reading from hardware
A2 = 10.00;
B1 = 10.00;
B2 = 10.00;

x1 = (A1 * sf);
x2 = (A2 * sf);
y1 = (B1 * sf);
y2 = (B2 * sf);
//To find the center origin of Bitmap
center.x=(((pCellInfo->rcBitmapRect.right-pCellInfo->rcBitmapRect.left)/2)+pCellInfo->rcBitmapRect.left);
center.y=(((pCellInfo->rcBitmapRect.bottom-pCellInfo->rcBitmapRect.top)/2)+pCellInfo->rcBitmapRect.top);

int a = 45; 
float Angle = ( 3.142 * a ) / 180; 

x1 =  x1 * cos(Angle) + y1 * sin(Angle); 
y1 = -x1 * sin(Angle) + y1 * cos(Angle);
x2 =  x2 * cos(Angle) + y2 * sin(Angle); 
y2 = -x2 * sin(Angle) + y2 * cos(Angle);

//I am using the right formula to claculate the axis for rotation
Origin(525,454);  // center origin(center.x,center.y);
MoveToEx(pCellInfo->hDC,center.x-x1,center.y-y2,NULL);

LineTo(pCellInfo->hDC,center.x-x1,center.y+y1);
LineTo(pCellInfo->hDC,center.x+x2,center.y+y1);
LineTo(pCellInfo->hDC,center.x+x2,center.y-y2);
LineTo(pCellInfo->hDC,center.x-x1,center.y-y2);


If i test with this hard code values,i am getting the rectangle rotated
MoveToEx(pCellInfo->hDC,525,289,NULL);// Angle 45 for reference
	LineTo(pCellInfo->hDC,360,454);
	LineTo(pCellInfo->hDC,525,619);
	LineTo(pCellInfo->hDC,690,454);
	LineTo(pCellInfo->hDC,525,289);

how can i get the same values

Thanks
Raj

Re: Rotate rerctangle in Bitmap

CPallini17-Aug-10 21:58

CPallini

17-Aug-10 21:58

raju_shiva wrote:
i am not getting

What does it mean, exactly (i.e. please elaborate)?

raju_shiva wrote:
A1 = 10.00; //these are values i am reading from hardware
A2 = 10.00;
B1 = 10.00;
B2 = 10.00;

How do you hope to get a rectangle from these values?
Smile | :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke

[My articles]

Re: Rotate rerctangle in Bitmap

raju_shiva17-Aug-10 22:38

raju_shiva

17-Aug-10 22:38

CPallini wrote:
. please elaborate)?

These are left,top,right,bottam values for the rectangle

i.e Rectangle(__in HDC hdc, __in int left, __in int top, __in int right, __in int bottom);

Thanks
Raj

Re: Rotate rerctangle in Bitmap

CPallini17-Aug-10 22:49

CPallini

17-Aug-10 22:49

{10,10,10,10} as {left,top,right,bottom} values give an empty rectangle.
Smile | :)

Re: Rotate rerctangle in Bitmap

raju_shiva17-Aug-10 23:13

raju_shiva

17-Aug-10 23:13

CPallini wrote:
{10,10,10,10} as {left,top,right,bottom} values give an empty rectangle

I am not passing the values directly.

As it is a bitmap image,on that i am drawing the rectangle
I am multiplying it :

x1 = (A1 * sf); //i.e (10 * 33.03)
x2 = (A2 * sf); //i.e (10 * 33.03)
y1 = (B1 * sf); //i.e (10 * 33.03)
y2 = (B2 * sf); //i.e (10 * 33.03)

where sf is

sf=(1024/31); //Its a bitmap image

Then while drawing the rectangle,i start as (center.x-left)

MoveToEx(pCellInfo->hDC,center.x-x1,center.y-y2,NULL); //(center.x-left)
                                                       //(center.y-bottom)

I hope you got it.

Thanks
Raj

Re: Rotate rerctangle in Bitmap

CPallini17-Aug-10 23:19

CPallini

17-Aug-10 23:19

raju_shiva wrote:
x1 = (A1 * sf); //i.e (10 * 33.03)
x2 = (A2 * sf); //i.e (10 * 33.03)
y1 = (B1 * sf); //i.e (10 * 33.03)
y2 = (B2 * sf); //i.e (10 * 33.03)

I see no rectangle here.
Smile | :)

Re: Rotate rerctangle in Bitmap

raju_shiva17-Aug-10 23:45

raju_shiva

17-Aug-10 23:45

CPallini wrote:
I see no rectangle here

After getting all the values
Here i am drawing the rectangle

MoveToEx(pCellInfo->hDC,center.x-x1,center.y-y2,NULL);
LineTo(pCellInfo->hDC,center.x-x1,center.y+y1);
LineTo(pCellInfo->hDC,center.x+x2,center.y+y1);
LineTo(pCellInfo->hDC,center.x+x2,center.y-y2);
LineTo(pCellInfo->hDC,center.x-x1,center.y-y2);

Thanks
Raj

Re: Rotate rerctangle in Bitmap

CPallini17-Aug-10 23:54

CPallini

17-Aug-10 23:54

OK (sorry if I didn't get you).
Now, what is the problem with your code (expected behaviour vs observed one)?.
Smile | :)

Re: Rotate rerctangle in Bitmap

raju_shiva18-Aug-10 0:16

raju_shiva

18-Aug-10 0:16

Now i want to rotate it for the given degree.How can i do it.

I am bit confused

please help me

Thanks
Raj

Re: Rotate rerctangle in Bitmap

CPallini18-Aug-10 0:36

CPallini

18-Aug-10 0:36

You are defining the rectangle via offsets from center, that is

P0={-x1,-y2}, P1{-x1, y1}, P2={x2,y1}, P3={x2,-y2}

hence, if you wan't rotate with angle phi around the center, than you should compute:

Pr = { x * cos(phi) + y * sin(phi), -x * sin(phi) + y * cos(phi)}

i.e.:

P0R= { -x1 * cos(phi) -y2*sin(phi), x1 * sin(phi) - y2 * cos(phi)}
P1R =...
P2R =...
P3R =...

and then connect the center+PiR points the way you did before.
Smile | :)

Re: Rotate rerctangle in Bitmap

raju_shiva18-Aug-10 1:18

raju_shiva

18-Aug-10 1:18

CPallini wrote:
P0R= { -x1 * cos(phi) -y2*sin(phi), x1 * sin(phi) - y2 * cos(phi)}

I am confused with it???

Now suppose my values are:

x1 = 330,x2 = 330,y1=330,y2=330
Origin(center.x,cebter.y) = Origin(525,454);

I have to do the get the new x1,x2,y1,y2 before calling MoveToEx and LineTo

Am i right??

i.e

Pr = { x * cos(phi) + y * sin(phi), -x * sin(phi) + y * cos(phi)}

for each x1,x2,y1,y2.

Then call the function

MoveToEx(pCellInfo->hDC,center.x-x1,center.y-y2,NULL); // where x1 = Pr1,x2=Pr2....y2 = Pr4
LineTo(pCellInfo->hDC,center.x-x1,center.y+y1);
LineTo(pCellInfo->hDC,center.x+x2,center.y+y1);
LineTo(pCellInfo->hDC,center.x+x2,center.y-y2);
LineTo(pCellInfo->hDC,center.x-x1,center.y-y2);

Thanks
Raj

Re: Rotate rerctangle in Bitmap

CPallini18-Aug-10 1:47

CPallini

18-Aug-10 1:47

Something like this

C++

// rotate around center
double phi = atan(1.0) * 2/3; // 30 degrees
int x[4];
int y[4];
x[0] = -x1 * cos(phi) - y2 * sin(phi);
y[0] =  x1 * sin(phi) - y2 * cos(phi);

x[1] = -x1 * cos(phi) + y1 * sin(phi);
y[1] =  x1 * sin(phi) + y1 * cos(phi);

x[2] =  x2 * cos(phi)  + y1 * sin(phi);
y[2] = -x2 * sin(phi) + y1 * cos(phi);

x[3] =  x2 * cos(phi) - y2 * sin(phi);
y[3] = -x2 * sin(phi) - y2 * cos(phi);

for (int i=0; i<4; i++)
{
  x[i] += center.x;
  y[i] += center.y;
}

MoveToEx(pCellInfo->hDC,x[3],y[3],NULL); 
for (int i=0; i<4; i++)
{
  LineTo(pCellInfo->hDC, x[i],y[i]);
}

I suppose.
Smile | :)

Re: Rotate rerctangle in Bitmap

raju_shiva18-Aug-10 20:54

raju_shiva

18-Aug-10 20:54

Thanks a lot ,its working fine.

Thanks
Raj

Re: Rotate rerctangle in Bitmap

CPallini18-Aug-10 20:59

CPallini

18-Aug-10 20:59

You are welcome.
Smile | :)

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