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Friday/Sunday Problem

  • #1
  • First Post: Oct 15, 2007 7:30pm Oct 15, 2007 7:30pm
I am writing an EA in which I am wanting the Friday bar and the two hour Sunday bar, to be treated as one single bar.

In other words, I have a formula that operates only once every bar. If a trade opens on Friday, I don't want another trade to open on the 2 hour Sunday bar.

How can I treat the entire Friday bar, and the two hour Sunday bar as a single bar? Thank you in advance.
  • #2
  • Last Post: Oct 16, 2007 2:44pm Oct 16, 2007 2:44pm
There are a variety of ways to get around the Sunday bars... some people look for brokers who don't have Sunday bars, others deal with it programmatically. Check out the code below for basic Fibonacci levels, and how it 'skips' Sunday. On Monday's it runs a bit of extra logic to get past the Sunday bar and find the Friday bar. A little bit of extra work, but it does the trick.

PHP Code
   double rates[1][6],yesterday_close,yesterday_high,yesterday_low;
   RefreshRates();
   ArrayCopyRates(rates, strSymbol, PERIOD_D1);

   if(DayOfWeek() == 1)
   {
      if(TimeDayOfWeek(iTime(strSymbol,PERIOD_D1,1)) == 5)
      {
          yesterday_close = rates[1][4];
          yesterday_high = rates[1][3];
          yesterday_low = rates[1][2];
      }
      else
      {
         for(int d = 5;d>=0;d--)
         {
            if(TimeDayOfWeek(iTime(strSymbol,PERIOD_D1,d)) == 5)
            {
                yesterday_close = rates[d][4];
                yesterday_high = rates[d][3];
                yesterday_low = rates[d][2];
            }
         
         }  
      
      }
   }
   else
   {
       yesterday_close = rates[1][4];
       yesterday_high = rates[1][3];
       yesterday_low = rates[1][2];
   }


   //---- Calculate Pivots

   //Comment("\nYesterday quotations:\nH ",yesterday_high,"\nL ",yesterday_low, "\nC ",yesterday_close);
   double R = yesterday_high - yesterday_low;//range
   double p = (yesterday_high + yesterday_low + yesterday_close)/3;// Standard Pivot
   double r3 = p + (R * 1.000);
   double r2 = p + (R * 0.618);
   double r1 = p + (R * 0.382);
   double s1 = p - (R * 0.382);
   double s2 = p - (R * 0.618);
   double s3 = p - (R * 1.000);
Pippin' ain't easy.
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