如何进行超过 40 次 request.security() 调用？

Question

Request.Security 达到 40，我不知道如何解决。

我只是想给我的桌子上色，虽然我尝试了不同的函数和元组，但最终我只是解决了 4 行。

// data(sym,timeframee,variablee) =>
//     ind=0.0
//     emaData=0.0
//     wmaData=0.0
//     hmaData=0.0
//     rmaData=0.0
//     sma_short_Data=0.0
//     sma_long_Data=0.0
//     if variablee==emaData
//         emaData  := request.security(sym, input_timeframe,ema_for_table)
//     if variablee==wmaData
//         wmaData := request.security(sym, input_timeframe,wma_for_table)
//     if variablee==hmaData
//         hmaData:=request.security(sym,input_timeframe,hma_for_table)
//     if variablee==rmaData
//         rmaData:=request.security(sym,input_timeframe,rma_for_table)
//     if variablee==sma_short_Data
//         sma_short_Data:=request.security(sym,input_timeframe,sma_short_term)
//     if variablee==sma_long_Data
//         sma_long_Data:=request.security(sym,input_timeframe,sma_long_term)


[sma_s_if,sma_l_if,e_if,h_if,r_if,w_if] = request.security(symbol1,input_timeframe,[sma_short_if,sma_long_if,ema_if,hma_if,rma_if,wma_if])
[sma_s2_if,sma_l2_if,e2_if,h2_if,r2_if,w2_if] = request.security(symbol2,input_timeframe,[sma_short_if,sma_long_if,ema_if,hma_if,rma_if,wma_if])
[sma_s3_if,sma_l3_if,e3_if,h3_if,r3_if,w3_if] = request.security(symbol3,input_timeframe,[sma_short_if,sma_long_if,ema_if,hma_if,rma_if,wma_if])
[sma_s4_if,sma_l4_if,e4_if,h4_if,r4_if,w4_if] = request.security(symbol4,input_timeframe,[sma_short_if,sma_long_if,ema_if,hma_if,rma_if,wma_if])
[sma_s5_if,sma_l5_if,e5_if,h5_if,r5_if,w5_if] = request.security(symbol5,input_timeframe,[sma_short_if,sma_long_if,ema_if,hma_if,rma_if,wma_if])
[sma_s6_if,sma_l6_if,e6_if,h6_if,r6_if,w6_if] = request.security(symbol6,input_timeframe,[sma_short_if,sma_long_if,ema_if,hma_if,rma_if,wma_if])

[sma_short_term1,sma_long_term1,ema_for_table_1,hma_for_table_1,rma_for_table_1,wma_for_table_1] = request.security(symbol1,input_timeframe,[sma_short_term,sma_long_term,ema_for_table,hma_for_table,rma_for_table,wma_for_table])
[sma_short_term2,sma_long_term2,ema_for_table_2,hma_for_table_2,rma_for_table_2,wma_for_table_2] = request.security(symbol2,input_timeframe,[sma_short_term,sma_long_term,ema_for_table,hma_for_table,rma_for_table,wma_for_table])
[sma_short_term3,sma_long_term3,ema_for_table_3,hma_for_table_3,rma_for_table_3,wma_for_table_3] = request.security(symbol3,input_timeframe,[sma_short_term,sma_long_term,ema_for_table,hma_for_table,rma_for_table,wma_for_table])
[sma_short_term4,sma_long_term4,ema_for_table_4,hma_for_table_4,rma_for_table_4,wma_for_table_4] = request.security(symbol4,input_timeframe,[sma_short_term,sma_long_term,ema_for_table,hma_for_table,rma_for_table,wma_for_table])
[sma_short_term5,sma_long_term5,ema_for_table_5,hma_for_table_5,rma_for_table_5,wma_for_table_5] = request.security(symbol5,input_timeframe,[sma_short_term,sma_long_term,ema_for_table,hma_for_table,rma_for_table,wma_for_table])
[sma_short_term6,sma_long_term6,ema_for_table_6,hma_for_table_6,rma_for_table_6,wma_for_table_6] = request.security(symbol6,input_timeframe,[sma_short_term,sma_long_term,ema_for_table,hma_for_table,rma_for_table,wma_for_table])

Answer 1

你不能。 40 是极限。没有解决办法。

您可以多次添加相同的脚本（或类似的脚本）以获得类似的结果，具体取决于您的用例。