aobv:="obv.obv;amacd:="macd.macd;adif:="macd.dif;adea:="macd.dea;aa1:=h=hhv(aobv,120);aa2:=amacd>0 and adif>adea;aa3:=ma(c,5)>ma(c,10)>ma(c,20) 条件成立:aa1 and aa2 and aa3;
x1:=finance(7)<=300000000 and dyna**(7)>5;x2:=macd.macd>ref(macd.macd,1) and macd.dif<macd.dea ;x1 and count(x2,3)=3 and macd.dif>=0;快要金叉是个模糊的概念,公式无法准确表达出来,所以只能用白线连续3天向上来表达。
t1:=c(c,20)*1.1 and c>o;a1:=(h-max(c,o));a2:=h-l;t2:=a1/ref(c,1)*100>=2 and a1*2>=a2;t3:=v>ref(v,1);xg:t1 and t2 and t3;参考资料:http://tieba.baidu.com/f?kw=%b9%c9%c6%b1%b9%ab%ca%bd%d7%a8%bc%d...
上影:=if(close>open,high-close,high-open);下影:=if(close>open,open-low,close-low);长上影:=abs(close-open)*1.7<上影 and abs(close-open)>下影*2;长下影:=abs(close-open)*1.7<下影 and abs(close-open)>上影*...
试这个: a1:=ma(c,5);a2:=ma(c,10);ab:=c>a1 and c>a2;长上影:=(h-max(c,o))>=(max(c,o)-l)and c>o;实体1:=(max(c,o)-min(c,o))/min(c,o)*100;长上影选:ab and 长上影 and 实体1>1.5;