cond:=c<=ref(CLOSE,10);
ma10:=ma(c,10);
if cond then aa:ma10;
else bb:ma10;
partline(c<=ref(CLOSE,10),ma10, COLORGREEN),linethick2;
partline(c>ref(CLOSE,10),ma10,COLORred),linethick2;
cond1:=c<=ref(CLOSE,22);
ma22:=ma(c,22);
if cond1 then aa1:ma22;
else bb1:ma22;
partline(c<=ref(CLOSE,22),ma22, COLORGREEN),linethick3;
partline(c>ref(CLOSE,22),ma22,COLORred),linethick3;
cond2:=c<=ref(CLOSE,51);
ma51:=(ma(c,30)+ma(c,72))/2;
if cond2 then aa2:ma51;
else bb2:ma51;
partline(c<=ref(CLOSE,51),ma51, COLORGREEN),linethick4;
partline(c>ref(CLOSE,51),ma51,COLORred),linethick4;
drawtext(c<=ref(CLOSE,10) and c<=ref(CLOSE,22) and c<=ref(CLOSE,51),c*1.01,'空');
drawtext(c>ref(CLOSE,10) and c>ref(CLOSE,22) and c>ref(CLOSE,51),c*0.99,'多');
帮忙让‘多’和‘空’只出现一次,第一次全部满足的时候出现 后面的过滤掉,谢谢
满足只要第一次时候你做个 i:=i+1 计数器就好了
开仓时候只要 I>0 的必定就不开了。
然而问题是你要什么时候把I归零?
cond:=c<=ref(CLOSE,10);
ma10:=ma(c,10);
if cond then aa:ma10;
else bb:ma10;
partline(c<=ref(CLOSE,10),ma10, COLORGREEN),linethick2;
partline(c>ref(CLOSE,10),ma10,COLORred),linethick2;
cond1:=c<=ref(CLOSE,22);
ma22:=ma(c,22);
if cond1 then aa1:ma22;
else bb1:ma22;
partline(c<=ref(CLOSE,22),ma22, COLORGREEN),linethick3;
partline(c>ref(CLOSE,22),ma22,COLORred),linethick3;
cond2:=c<=ref(CLOSE,51);
ma51:=(ma(c,30)+ma(c,72))/2;
if cond2 then aa2:ma51;
else bb2:ma51;
partline(c<=ref(CLOSE,51),ma51, COLORGREEN),linethick4;
partline(c>ref(CLOSE,51),ma51,COLORred),linethick4;
variable:duo=0,kong=0;
if c<=ref(CLOSE,10) and c<=ref(CLOSE,22) and c<=ref(CLOSE,51) then kong:=kong+1;
if c>ref(CLOSE,10) and c>ref(CLOSE,22) and c>ref(CLOSE,51) then duo:=duo+1;
drawtext(cross(kong=1,0.5),c*1.01,'空');
drawtext(cross(duo=1,0.5),c*0.99,'多');
cond:=c<=ref(CLOSE,10);
ma10:=ma(c,10);
if cond then aa:ma10;
else bb:ma10;
partline(c<=ref(CLOSE,10),ma10, COLORGREEN),linethick2;
partline(c>ref(CLOSE,10),ma10,COLORred),linethick2;
cond1:=c<=ref(CLOSE,22);
ma22:=ma(c,22);
if cond1 then aa1:ma22;
else bb1:ma22;
partline(c<=ref(CLOSE,22),ma22, COLORGREEN),linethick3;
partline(c>ref(CLOSE,22),ma22,COLORred),linethick3;
cond2:=c<=ref(CLOSE,51);
ma51:=(ma(c,30)+ma(c,72))/2;
if cond2 then aa2:ma51;
else bb2:ma51;
partline(c<=ref(CLOSE,51),ma51, COLORGREEN),linethick4;
partline(c>ref(CLOSE,51),ma51,COLORred),linethick4;
variable:i=0,w=0;
kong:=c<ref(CLOSE,10)and c<ref(CLOSE,22)and c<ref(CLOSE,51);
duo:=c>ref(CLOSE,10)and c>ref(CLOSE,22)and c>ref(CLOSE,51);
if kong then begin
i:=i+1;
w:=0;
end
if duo then begin
w:=w+1;
i:=0;
end
drawtext(cross(i,0.5),c*1.001,'空');
drawtext(cross(w,0.5),c*0.999,'多');
多可以实现,只出现一个而且是三个条件同时满足的 但空实现不了 会因为其中一个条件发生变化就出现一次。