RE: [amibroker] How to detect if 'foreign' is successful or not ?

[Salil V Gangal <salil_gangal@xxxxxxxxx>]

Herman,
 

>>Cum(Foreign(...))==0<<
 
Thanks for the suggestion.
 
Regards,
- Salil V GangalHerman vandenBergen <psytek@xxxxxxxx> wrote:


Salil, I'd love an answer to this as i am struggling to detect the existence of Composites. 
 
I thought of using IsEmpty(Foreign(...)) but I couldn't get it to work well, another way may be to use Cum(Foreign(...))==0 to indicate there is no data... but there must be a better way.
 
herman

-----Original Message-----From: Salil V Gangal [mailto:salil_gangal@xxxxxxxxx]Sent: April 30, 2003 12:00 PMTo: amibroker@xxxxxxxxxxxxxxxSubject: [amibroker] How to detect if 'foreign' is successful or not ?

Friends,
 
If a 'foreign' function is executed, then how to detect programatically if it is successful or not ?
 
I thought IsNull will be useful, but it was not.  I executed the following code (and there is MSFT in the database I'm using) and it shows 'IsNull is true' for both.  I tried the code without 'Ref' as well ... Same result.
 


security=foreign("MSFT","C");
if ( IsNull(Ref(security,0)) ) {
    message = "IsNull is true";
}
else {
    message = "IsNull is false";
}
message;
 

security=foreign("DoesNotExist","C");
if ( IsNull(Ref(security,0)) ) {
    message = "IsNull is true";
}
else {
    message = "IsNull is false";
}
message;
 
Regards,
- Salil V Gangal


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