問題一:一根K棒可能帶出3,5,7或4,6,8三個數字若只要出現MAX的數字,要如何寫?問題二:新低黑K3給-3(紅K給+3),新低黑K4給-4(紅K給+4),依此類推,則將每根K棒的分數加總要如何寫呢?有勞總版主了,謝謝!
源碼如下:
新低黑K1:=C<O AND REF(C,1)>REF(O,1);
新低黑K1低:=REF(L,BARSLAST(新低黑K1)),COLORBLUE;
新低黑K2:=CROSS(新低黑K1低-0.1,L) AND C<O;
新低黑K2低:=REF(L,BARSLAST(新低黑K2)),COLORRED;
新低黑K3:=CROSS(新低黑K2低-0.1,L) AND C<O;
新低黑K3低:=REF(L,BARSLAST(新低黑K3)),COLORWHITE;
新低黑K4:=CROSS(新低黑K3低-0.1,L) AND C<O;
新低黑K4低:=REF(L,BARSLAST(新低黑K4)),COLORRED;
新低黑K5:=CROSS(新低黑K4低-0.1,L) AND C<O;
新低黑K5低:=REF(L,BARSLAST(新低黑K5)),COLORRED;
新低黑K6:=CROSS(新低黑K5低-0.1,L) AND C<O;
新低黑K6低:=REF(L,BARSLAST(新低黑K6)),COLORWHITE;
新低黑K7:=CROSS(新低黑K6低-0.1,L) AND C<O;
新低黑K7低:=REF(L,BARSLAST(新低黑K7)),COLORRED;
新低黑K8:=CROSS(新低黑K7低-0.1,L) AND C<O;

新高紅K1:=C>O AND REF(C,1)<REF(O,1);
新高紅K1高:=REF(H,BARSLAST(新高紅K1)),COLORWHITE;
新高紅K2:=CROSS(H,新高紅K1高+0.1) AND C>O;
新高紅K2高:=REF(H,BARSLAST(新高紅K2)),COLORRED;
新高紅K3:=CROSS(H,新高紅K2高+0.1) AND C>O;
新高紅K3高:=REF(H,BARSLAST(新高紅K3)),COLORWHITE;
新高紅K4:=CROSS(H,新高紅K3高+0.1) AND C>O;
新高紅K4高:=REF(H,BARSLAST(新高紅K4)),COLORRED;
新高紅K5:=CROSS(H,新高紅K4高+0.1) AND C>O;
新高紅K5高:=REF(H,BARSLAST(新高紅K5)),COLORRED;
新高紅K6:=CROSS(H,新高紅K5高+0.1) AND C>O;
新高紅K6高:=REF(H,BARSLAST(新高紅K6)),COLORWHITE;
新高紅K7:=CROSS(H,新高紅K6高+0.1) AND C>O;
新高紅K7高:=REF(H,BARSLAST(新高紅K7)),COLORRED;
新高紅K8:=CROSS(H,新高紅K7高+0.1) AND C>O;

DRAWTEXT(新低黑K3,H,'3'),COLORBLUE,ALIGN0,PXUP25;
DRAWTEXT(新低黑K4,H,'4'),COLORBLUE,ALIGN0,PXUP50;
DRAWTEXT(新低黑K5,H,'5'),COLORBLUE,ALIGN0,PXUP75;
DRAWTEXT(新低黑K6,H,'6'),COLORBLUE,ALIGN0,PXUP100;
DRAWTEXT(新低黑K7,H,'7'),COLORBLUE,ALIGN0,PXUP125;
DRAWTEXT(新低黑K8,H,'8'),COLORBLUE,ALIGN0,PXUP125;

DRAWTEXT(新高紅K3,L,'3'),COLORRED,ALIGN0,PXDN5;
DRAWTEXT(新高紅K4,L,'4'),COLORRED,ALIGN0,PXDN30;
DRAWTEXT(新高紅K5,L,'5'),COLORRED,ALIGN0,PXDN55;
DRAWTEXT(新高紅K6,L,'6'),COLORRED,ALIGN0,PXDN80;
DRAWTEXT(新高紅K7,L,'7'),COLORRED,ALIGN0,PXDN105;
DRAWTEXT(新高紅K8,L,'8'),COLORRED,ALIGN0,PXDN105;

問題一:使用MAX函數就可以了..
問題二:將
DRAWTEXT(新低黑K3,H,'3'),COLORBLUE,ALIGN0,PXUP25;
DRAWTEXT(新低黑K4,H,'4'),COLORBLUE,ALIGN0,PXUP50;

DRAWTEXT(新高紅K3,L,'3'),COLORRED,ALIGN0,PXDN5;
DRAWTEXT(新高紅K4,L,'4'),COLORRED,ALIGN0,PXDN30;

改成...

DRAWTEXT(新低黑K3,H,'-3'),COLORBLUE,ALIGN0,PXUP25;
DRAWTEXT(新低黑K4,H,'-4'),COLORBLUE,ALIGN0,PXUP50;

DRAWTEXT(新高紅K3,L,'+3'),COLORRED,ALIGN0,PXDN5;
DRAWTEXT(新高紅K4,L,'+4'),COLORRED,ALIGN0,PXDN30;

使用MAX函數?可否進一步指導,SORRY,我完全沒有寫程式的底子與觀念,奇狐都是熱心的PETER大哥與總版主與九兩兄一點一點教的,感恩!

先確定一下.....一根K棒中會出現的數字只有3,5,7或4,6,8.....可否貼張圖出來看看..

大大:請容我補充說明
一,一根K棒可能會出現357其中任一或任二或三個數字全部出現但出現單數357就不會出現雙數468.
二.原諒我不會貼圖,您只要將源碼COPY建立一個新公式.就可以看到圖了,謝謝您的協助!

說錯了,單雙數會混合出現.
因為有可能出現很多數字,所以我想做到只出現最大值的那個數字
另外這些出現的數字我想把他們加總!

程式改為以下.....直接複製貼上就好囉.....



新低黑K1:=C<O AND REF(C,1)>REF(O,1);
新低黑K1低:=REF(L,BARSLAST(新低黑K1)),COLORBLUE;
新低黑K2:=CROSS(新低黑K1低-0.1,L) AND C<O;
新低黑K2低:=REF(L,BARSLAST(新低黑K2)),COLORRED;
新低黑K3:=CROSS(新低黑K2低-0.1,L) AND C<O;
新低黑K3低:=REF(L,BARSLAST(新低黑K3)),COLORWHITE;
新低黑K4:=CROSS(新低黑K3低-0.1,L) AND C<O;
新低黑K4低:=REF(L,BARSLAST(新低黑K4)),COLORRED;
新低黑K5:=CROSS(新低黑K4低-0.1,L) AND C<O;
新低黑K5低:=REF(L,BARSLAST(新低黑K5)),COLORRED;
新低黑K6:=CROSS(新低黑K5低-0.1,L) AND C<O;
新低黑K6低:=REF(L,BARSLAST(新低黑K6)),COLORWHITE;
新低黑K7:=CROSS(新低黑K6低-0.1,L) AND C<O;
新低黑K7低:=REF(L,BARSLAST(新低黑K7)),COLORRED;
新低黑K8:=CROSS(新低黑K7低-0.1,L) AND C<O;

新高紅K1:=C>O AND REF(C,1)<REF(O,1);
新高紅K1高:=REF(H,BARSLAST(新高紅K1)),COLORWHITE;
新高紅K2:=CROSS(H,新高紅K1高+0.1) AND C>O;
新高紅K2高:=REF(H,BARSLAST(新高紅K2)),COLORRED;
新高紅K3:=CROSS(H,新高紅K2高+0.1) AND C>O;
新高紅K3高:=REF(H,BARSLAST(新高紅K3)),COLORWHITE;
新高紅K4:=CROSS(H,新高紅K3高+0.1) AND C>O;
新高紅K4高:=REF(H,BARSLAST(新高紅K4)),COLORRED;
新高紅K5:=CROSS(H,新高紅K4高+0.1) AND C>O;
新高紅K5高:=REF(H,BARSLAST(新高紅K5)),COLORRED;
新高紅K6:=CROSS(H,新高紅K5高+0.1) AND C>O;
新高紅K6高:=REF(H,BARSLAST(新高紅K6)),COLORWHITE;
新高紅K7:=CROSS(H,新高紅K6高+0.1) AND C>O;
新高紅K7高:=REF(H,BARSLAST(新高紅K7)),COLORRED;
新高紅K8:=CROSS(H,新高紅K7高+0.1) AND C>O;

黑:=IF(新低黑K8,8,IF(新低黑K7,7,IF(新低黑K6,6,IF(新低黑K5,5,IF(新低黑K4,4,IF(新低黑K3,3,0))))));
紅:=IF(新高紅K8,8,IF(新高紅K7,7,IF(新高紅K6,6,IF(新高紅K5,5,IF(新高紅K4,4,IF(新高紅K3,3,0))))));


DRAWTEXT(黑=3,H,'-3'),COLORBLUE,ALIGN0,PXUP25;
DRAWTEXT(黑=4,H,'-4'),COLORBLUE,ALIGN0,PXUP50;
DRAWTEXT(黑=5,H,'-5'),COLORBLUE,ALIGN0,PXUP75;
DRAWTEXT(黑=6,H,'-6'),COLORBLUE,ALIGN0,PXUP100;
DRAWTEXT(黑=7,H,'-7'),COLORBLUE,ALIGN0,PXUP125;
DRAWTEXT(黑=8,H,'-8'),COLORBLUE,ALIGN0,PXUP125;

DRAWTEXT(紅=3,L,'+3'),COLORRED,ALIGN0,PXDN5;
DRAWTEXT(紅=4,L,'+4'),COLORRED,ALIGN0,PXDN30;
DRAWTEXT(紅=5,L,'+5'),COLORRED,ALIGN0,PXDN55;
DRAWTEXT(紅=6,L,'+6'),COLORRED,ALIGN0,PXDN80;
DRAWTEXT(紅=7,L,'+7'),COLORRED,ALIGN0,PXDN105;
DRAWTEXT(紅=8,L,'+8'),COLORRED,ALIGN0,PXDN105;


還有....我就是九兩...哈哈哈

加總的處理方式又不一樣囉

感謝九兩兄鼎力相助,加總的東西我再努力一下,尋找解答.感恩!

加總的寫法...貼上去就好囉

新低黑K1:=C<O AND REF(C,1)>REF(O,1);
新低黑K1低:=REF(L,BARSLAST(新低黑K1)),COLORBLUE;
新低黑K2:=CROSS(新低黑K1低-0.1,L) AND C<O;
新低黑K2低:=REF(L,BARSLAST(新低黑K2)),COLORRED;
新低黑K3:=(CROSS(新低黑K2低-0.1,L) AND C<O)*-3;
新低黑K3低:=REF(L,BARSLAST(新低黑K3)),COLORWHITE;
新低黑K4:=(CROSS(新低黑K3低-0.1,L) AND C<O)*-4;
新低黑K4低:=REF(L,BARSLAST(新低黑K4)),COLORRED;
新低黑K5:=(CROSS(新低黑K4低-0.1,L) AND C<O)*-5;
新低黑K5低:=REF(L,BARSLAST(新低黑K5)),COLORRED;
新低黑K6:=(CROSS(新低黑K5低-0.1,L) AND C<O)*-6;
新低黑K6低:=REF(L,BARSLAST(新低黑K6)),COLORWHITE;
新低黑K7:=(CROSS(新低黑K6低-0.1,L) AND C<O)*-7;
新低黑K7低:=REF(L,BARSLAST(新低黑K7)),COLORRED;
新低黑K8:=(CROSS(新低黑K7低-0.1,L) AND C<O)*-8;

新高紅K1:=C>O AND REF(C,1)<REF(O,1);
新高紅K1高:=REF(H,BARSLAST(新高紅K1)),COLORWHITE;
新高紅K2:=CROSS(H,新高紅K1高+0.1) AND C>O;
新高紅K2高:=REF(H,BARSLAST(新高紅K2)),COLORRED;
新高紅K3:=(CROSS(H,新高紅K2高+0.1) AND C>O)*3;
新高紅K3高:=REF(H,BARSLAST(新高紅K3)),COLORWHITE;
新高紅K4:=(CROSS(H,新高紅K3高+0.1) AND C>O)*4;
新高紅K4高:=REF(H,BARSLAST(新高紅K4)),COLORRED;
新高紅K5:=(CROSS(H,新高紅K4高+0.1) AND C>O)*5;
新高紅K5高:=REF(H,BARSLAST(新高紅K5)),COLORRED;
新高紅K6:=(CROSS(H,新高紅K5高+0.1) AND C>O)*6;
新高紅K6高:=REF(H,BARSLAST(新高紅K6)),COLORWHITE;
新高紅K7:=(CROSS(H,新高紅K6高+0.1) AND C>O)*7;
新高紅K7高:=REF(H,BARSLAST(新高紅K7)),COLORRED;
新高紅K8:=(CROSS(H,新高紅K7高+0.1) AND C>O)*8;

黑:=新低黑K8+新低黑K7+新低黑K6+新低黑K5+新低黑K4+新低黑K3;
紅:=新高紅K8+新高紅K7+新高紅K6+新高紅K5+新高紅K4+新高紅K3;


DRAWNUMBER(黑<>0,H,黑,0),COLORBLUE,ALIGN0,PXUP50;
DRAWNUMBER(紅<>0,L,紅,0),COLORRED,ALIGN0,PXDN25;

就原創意加上些條件
應該股票+期貨都可使用

peter109 附帶上了此圖片:

__________________
至誠能勝天下至偽
至拙能勝天下至巧

PETER大哥您貼的圖,有用到這個部分嗎?
很久沒見到大哥了,您也知道我是很遲鈍的!讓我在想想大哥的話中含意,感謝指導!

因為好奇
所以 看了下內容
然後加上些條件(可比對您源碼)
就有了貼圖的內容

不過 坦白說
這只是初稿
應該還有改善的空間(股票部分)
但就出稿而言
期貨的結果比股票好

思考邏輯和我幾天前的東西 類似度挺高
差別在使用的條件

__________________
至誠能勝天下至偽
至拙能勝天下至巧

感謝!我再想想!