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RPdAnjou
Posts: 29
Registered: ‎06-06-2011
Message 1 of 5 (256 Views)

VB.Net - How do I determine whether a Positional Representation is failing?

256 Views, 4 Replies
08-09-2012 01:39 AM

Hi,

 

I'm looping trough positional representations in a assembly and I would like to determine if one of them is failing.

Can some one give me a nice hint on a solution?

 

Thanks.

Valued Mentor
jcneal
Posts: 322
Registered: ‎04-10-2010
Message 2 of 5 (240 Views)

Re: VB.Net - How do I determine whether a Positional Representation is failing?

08-13-2012 10:47 AM in reply to: RPdAnjou

loop through all positional representations and activate each. fail = error

see if it helps.

 

on error resume next

if err then

err clear

counter = counter +1

end if

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Contributor
RPdAnjou
Posts: 29
Registered: ‎06-06-2011
Message 3 of 5 (237 Views)

Re: VB.Net - How do I determine whether a Positional Representation is failing?

08-13-2012 01:19 PM in reply to: RPdAnjou

Thanks for the obvious solution!

Will try that.

 

Some times you just don't see the simple solutions :smileyindifferent:

Contributor
RPdAnjou
Posts: 29
Registered: ‎06-06-2011
Message 4 of 5 (219 Views)

Re: VB.Net - How do I determine whether a Positional Representation is failing?

08-16-2012 06:34 AM in reply to: jcneal

Well, I tried the solution but this will not give me the right result.

This is because the positional representation does activate correctly but has errors.

 

So now the question is more like: How do I determine whether a positional representation has errors?

 

Well, I did find a half solution and don't know whether there is a better one.

For now, I activate each positional representation and then search for failing constraints.

When one fails, I know the positional representation also fails.

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Vladimir.Ananyev
Posts: 458
Registered: ‎08-14-2012
Message 5 of 5 (194 Views)

Re: VB.Net - How do I determine whether a Positional Representation is failing?

08-22-2012 04:45 AM in reply to: RPdAnjou

There is no API for this task. Your workaround is a good idea.

Cheers,


Vladimir Ananyev
Developer Technical Services
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