ToTime

[Edgar V.]

When using the following ToTime condition:

if (ToTime(Time[0]) >= ToTime(13,00,00)&& ToTime(Time[0]) <= ToTime(14,30,00 ));

For some reason is not working within the time requested. Order either gets placed before time or not placed at all. If I take out the ToTime condition then strategy works fine, but I'll like it to work in a specific Time frame.
Am I missing something?

Thanks,

[whitmark]

Try this:

if (ToTime(Time[0]) >= 130000 && ToTime(Time[0]) <= 143000 );
{
// do something
}

Remember, the ToTime() merely converts a time in standard format to an integer. HTH.

Whitmark

[Edgar V.]

Still behaves the same. Orders are getting entered before time. My Time Zone is EST.

[NinjaTrader_Ray]

You will have to add some Print() statements to debug your code. In case you have not seen it before, here is a tip that provide some guidance.

http://www.ninjatrader-support.com/vb/showthread.php?t=3418

[whitmark]

Hard to determine what is going on without seeing the code as the logic/syntax is good and is the same as what is posted on the help as follows:

// Only trade between 7:45 AM and 1:45 PM
if (ToTime(Time[0]) >= 74500 && ToTime(Time[0]) <= 134500)
{
// Strategy logic goes here
}

There must be some issue in the logic so you may want to do a Print("Before Time Check") and Print("After Time Check") above and below and monitor in the output window to ensure this if statement is getting executed. HTH

Regards,

Whitmark

[Edgar V.]

Where is it that ToTime needs to be declared? within the OnBarUpdate()?
This is a sample of how I have it:

protectedoverridevoid OnBarUpdate()

{ //Only run on real-time data
if (Historical)
return;
if (ToTime(Time[0]) >= 142000 && ToTime(Time[0]) <= 173000 );

[NinjaTrader_Ray]

Yes however, your code sample for sure would cause problems since you have a semi colon after your if statement which is incorrect.

Should be

if (ToTime(Time[0]) >= 142000 && ToTime(Time[0]) <= 173000 )
{
// Do something
}

[whitmark]

Good catch Ray. My apologies, Edgar, for offering an initial solution while hastily cutting and pasting and not noticing the semicolon. No good deed goes unpunished

I will say that this type of coding error is one hardest to detect in that it 1) syntax checks okay, 2) does not throw any runtime error and 3) executes the "then logic" in {} regardless of the conditions in the if() statement.

Regards,

Whitmark