有人能帮我写点蜡烛剧本吗？
有人能帮我写点蜡烛剧本吗？
提问于 2022-10-13 18:08:51
我想用蜡烛计数器绘制ema8、ema20和sma20 (即1，2，3.)当交叉发生和计数结束时，根据给定的条件，以另一个交叉或交叉结束。
//@version=5
indicator(title='candle count', overlay=true,max_labels_count=10)
ema1 = ema(close,8)
ema2 = ema(close,20)
sma1 = sma(close,20)
// plot //
plot(ema1, color=red, linewidth=1, title="EMA 1")
plot(ema2, color=green, linewidth=1, title="EMA 2")
plot(sma1, color=black, linewidth=1, title="SMA 1")
count = 1
crossover = crossover(ema1,ema2)
crossunder = crossunder(ema2,sma1)
if crossover == crossover(ema1,ema2)
for i = 1 to 10
if crossunder == crossunder(ema2,sma1)
break
count := count + 1
label.new(bar_index,high, str.tostring(count), textcolor=color.blue, style=label.style_none)回答 1
Stack Overflow用户
发布于 2022-10-14 06:15:10
//@version=5
indicator(title='candle count', overlay=true,max_labels_count=10)
ema1 = ema(close,8)
ema2 = ema(close,20)
sma1 = sma(close,20)
// plot //
plot(ema1, color=red, linewidth=1, title="EMA 1")
plot(ema2, color=green, linewidth=1, title="EMA 2")
plot(sma1, color=black, linewidth=1, title="SMA 1")
var int count_over = 0
var int count_under = 0
crossover = ta.crossover(ema1,ema2)
crossunder = ta.crossunder(ema2,sma1)
if crossover
count_over += 1
count_under := 0
else if crossunder
count_under += 1
count_over := 0
plot(count_over)
plot(count_under)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/74060052
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