跨周期引用數(shù)據(jù)導(dǎo)致的信號(hào)閃爍問題

小周期引用大周期,當(dāng)時(shí)15MIN的K線還沒有最終形成,導(dǎo)致有了未來

引用15MIN周期的上個(gè)周期的數(shù)據(jù),就不會(huì)有未來了

如果采用把每一分鐘的數(shù)值先賦值，再引用，這樣在小周期引用時(shí)是否能夠逐一應(yīng)用到對(duì)應(yīng)的分鐘數(shù)數(shù)據(jù)？

例如

//IF MINUTE()=1 THEN BEGIN
//kaiduo:=開多18 or (開多15 and (開多12 and 開多13 and 開多14 and 開多17 and 開多19));
//END
//ELSE if minute()=2 then BEGIN
//kaiduo:=開多18 or (開多15 and (開多12 and 開多13 and 開多14 and 開多17 and 開多19));
//END
//ELSE if minute()=3 then BEGIN
//kaiduo:=開多18 or (開多15 and (開多12 and 開多13 and 開多14 and 開多17 and 開多19));
//END
//ELSE if minute()=4 then BEGIN
//kaiduo:=開多18 or (開多15 and (開多12 and 開多13 and 開多14 and 開多17 and 開多19));
//END
//ELSE if minute()=5 then BEGIN
//kaiduo:=開多18 or (開多15 and (開多12 and 開多13 and 開多14 and 開多17 and 開多19));
//END
//ELSE if minute()=6 then BEGIN
//kaiduo:=開多18 or (開多15 and (開多12 and 開多13 and 開多14 and 開多17 and 開多19));
//END
//ELSE if minute()=7 then BEGIN
//kaiduo:=開多18 or (開多15 and (開多12 and 開多13 and 開多14 and 開多17 and 開多19));
//END
//ELSE if minute()=8 then BEGIN
//kaiduo:=開多18 or (開多15 and (開多12 and 開多13 and 開多14 and 開多17 and 開多19));
//END

不能

http://www.weistock.com/bbs/dispbbs.asp?boardid=10&Id=15388

看看這個(gè)帖子,里面的方法是否有幫助