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Here it is. I use it for making DTMs of items in my inventory that there could be alot of (Ex. There are so many types of fish and food, i don't want to make dtms for all of them, so i just made this. The script will make the DTM for me :D).

function CreateDTMFromClient(x1, y1, x2, y2, maxsubpoints: integer): integer;
var
  DDTMDef: TDTMPointDef;
  DDTMSub: array of TDTMPointDef;
  DDTMMain: TDTM;
  a, b, c, d, e, bitmap: integer;
  first: boolean;
begin
  if x2 <= x1 then Swap(x2, x1);
  if y2 <= y1 then Swap(y2, y1);
  CopyClientToBitmap(bitmap, x1, y1, x2, y2);
  c:= x2 - x1;
  d:= y2 - y1;
  first:= true;
  for a:= 0 to c do
    begin
    for b:= 0 to d do
      begin
        if FastGetPixel(bitmap, a, b) = 65536 then
          begin
            if First then
              begin
                DDTMDef.x:= x1 + a;
                DDTMDef.y:= y1 + b;
                DDTMDef.AreaSize:= 1;
                DDTMDef.AreaShape:= 0;
                DDTMDef.Color:= 65536;
                DDTMDef.Tolerance:= 0;
                First:= false;
                wait(500);
              end else
              begin
                SetArrayLength(DDTMSub, Length(DDTMSub) + 1);
                e:= Length(DDTMSub) - 1;
                DDTMSub[e].x:= x1 + a;
                DDTMSub[e].y:= y1 + b;
                DDTMSub[e].AreaSize:= 1;
                DDTMSub[e].AreaShape:= 0;
                DDTMSub[e].Color:= 65536;
                DDTMSub[e].Tolerance:= 0;
                wait(500);
                if e >= maxsubpoints then
                begin
                   DDTMMain.MainPoint:= DDTMDef;
                   DDTMMain.SubPoints:= DDTMSub;
                   result:= AddDTM(DDTMMain);
                   exit;
                end;
              end;
          end;
      end;
    end;
  DDTMMain.MainPoint:= DDTMDef;
  DDTMMain.SubPoints:= DDTMSub;
  result:= AddDTM(DDTMMain);
end;

1). It probably took more time to make that than the DTM's,

2). large bitmaps are waste full, you can cut it down somehow, and

3). setting the main point as the edge isn't condoned by me. It would be better to use the center.

From: r0b0t1
1). It probably took more time to make that than the DTM's,

2). large bitmaps are waste full, you can cut it down somehow, and

3). setting the main point as the edge isn't condoned by me. It would be better to use the center.

1. Probably was, that isnt the point..

2. the amount is trivial. It doesnt save it and it releases it at the end.

3. considering 9/10 the static color is on the outside, there is nothing i can do. I made it the first so users can say

If FindDTM(dtm,x,y,bla)then
  Mouse(x + 15, y + 15, 0, 0, false);

Thats not what I mean! You can just set the color as the middle of the object, and make its tolerance 255. Another problem is the the main point of the DTM is the place where the X and Y coordinates are returned from. Odd to see someone clicking on the edge every time, eh?

its not…

If FindDTM(dtm,x,y,bla)then
  Mouse(x + 15, y + 15, 4, 4, false);

What Pwnaz0r is saying it finds the edge then it clicks 15 extra pixels to make it in the middle.

Thank you buckley lol

Cant taht go wrong so if say the first point is on the bottom edge? surely there must be a way to locate a center point from teh surrounding ones?


EDIT: I have an idea… make all the sub points first, then with a bit of maths get the smallest X and Y and the biggest X and Y and then get the mid point, which then becomes the Mpoint, i may be able to write it uot for you if you want

EDIT:

procedure MPoint;
var Mx1,My1,Mx2,My2:integer;
begin
  Mx1 := x2;
  Mx2 := x1;
  My1 := y2;
  My2 := y1;
    for e:= 0 to MaxSubPoints do
    begin
        if(DDTMSub[e].x < Mx1) then
          Mx1 := DDTMSub[e].x;

        if(DDTMSub[e].x > Mx2) then
          Mx2 := DDTMSub[e].x;

        if(DDTMSub[e].y < My1) then
          My1 := DDTMSub[e].y;

        if(DDTMSub[e].y > My2) then
          My2 := DDTMSub[e].y;
    end;
  DDTMDef.x:= Mx1 + (Mx2-Mx1);
  DDTMDef.y:= My1 + (My2-My1);
  DDTMDef.AreaSize:= 1;
  DDTMDef.AreaShape:= 0;
  DDTMDef.Color:= 65536;
  DDTMDef.Tolerance:= 255;
end;

Hows That? Placed in the script that is

function CreateDTMFromClient(x1, y1, x2, y2, maxsubpoints: integer): integer;
var
  DDTMDef: TDTMPointDef;
  DDTMSub: array of TDTMPointDef;
  DDTMMain: TDTM;
  Mx1,My1,Mx2,My2,a, b, c, d, e, bitmap: integer;
begin
  if x2 <= x1 then Swap(x2, x1);
  if y2 <= y1 then Swap(y2, y1);
  CopyClientToBitmap(bitmap, x1, y1, x2, y2);
  c:= x2 - x1;
  d:= y2 - y1;
  for a:= 0 to c do
    begin
    for b:= 0 to d do
      begin
        if FastGetPixel(bitmap, a, b) = 65536 then
          begin
            SetArrayLength(DDTMSub, Length(DDTMSub) + 1);
            e:= Length(DDTMSub) - 1;
            DDTMSub[e].x:= x1 + a;
            DDTMSub[e].y:= y1 + b;
            DDTMSub[e].AreaSize:= 1;
            DDTMSub[e].AreaShape:= 0;
            DDTMSub[e].Color:= 65536;
            DDTMSub[e].Tolerance:= 0;
              if e >= maxsubpoints then
              begin
                Mx1 := x2;
                Mx2 := x1;
                My1 := y2;
                My2 := y1;
                  for e:= 0 to MaxSubPoints do
                  begin
                      if(DDTMSub[e].x < Mx1) then
                        Mx1 := DDTMSub[e].x;

                      if(DDTMSub[e].x > Mx2) then
                        Mx2 := DDTMSub[e].x;

                      if(DDTMSub[e].y < My1) then
                        My1 := DDTMSub[e].y;

                      if(DDTMSub[e].y > My2) then
                        My2 := DDTMSub[e].y;
                  end;
                DDTMDef.x:= Mx1 + (Mx2-Mx1);
                DDTMDef.y:= My1 + (My2-My1);
                DDTMDef.AreaSize:= 1;
                DDTMDef.AreaShape:= 0;
                DDTMDef.Color:= 65536;
                DDTMDef.Tolerance:= 255;
                exit;
              end;
          end;
      end;
  DDTMMain.MainPoint:= DDTMDef;
  DDTMMain.SubPoints:= DDTMSub;
  result:= AddDTM(DDTMMain);
end;

I would rather not use the midpoint because

1. I don't like using non static colors in dtm's. don't work well (if in center if the box).

2. Just b/c it is in the middle of the array, doesn't mean it is in the middle of the object. Its going to be close to the same spot(look at the way it looks for colors. Left to right and there will prolly be more than 1 color in each line. So it would be the same anyways except you will just have to make different adjustments.