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Solaras
Wanderer
Joined: 11 Mar 2002
Posts: 93
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 Posted: Thu Sep 29, 2005 11:21 pm
Evaluate Function
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I can't seem to get this evaluate to work right. If anyone can help I would appreciate it.
| Code: |
| %eval( (!%ismember( "broken left arm", @Affliction) and !%ismember( "broken right arm", @Affliction)) and !%ismember( paralysed, @Affliction) and !%ismember( "pinned left leg", @Affliction) and !%ismember( "pinned right leg", @Affliction) and !%ismember( entangled, @Affliction) and !%ismember( webbed, @Affliction) and !%ismember( impaled, @Affliction)) |
What I want is if I have a broken left arm AND a broken right arm, to evaluate as 0. However it seems to evaluate as 0 if I have either or. |
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Vijilante
SubAdmin
Joined: 18 Nov 2001
Posts: 5182
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Posted: Fri Sep 30, 2005 8:05 am
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Perhaps you should describe the full logic you want in words. As you have it now if any 1 of those conditions is true then the %eval will result in false. If this is what you want with the exception of the arm things then this would work.
| Code: |
| %eval( !((%ismember( "broken left arm", @Affliction) and %ismember( "broken right arm", @Affliction)) or %ismember( paralysed, @Affliction) or %ismember( "pinned left leg", @Affliction) or %ismember( "pinned right leg", @Affliction) or %ismember( entangled, @Affliction) or %ismember( webbed, @Affliction) or %ismember( impaled, @Affliction))) |
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Solaras
Wanderer
Joined: 11 Mar 2002
Posts: 93
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Posted: Fri Sep 30, 2005 8:57 pm
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| Vijilante wrote: |
Perhaps you should describe the full logic you want in words. As you have it now if any 1 of those conditions is true then the %eval will result in false. If this is what you want with the exception of the arm things then this would work.
| Code: |
| %eval( !((%ismember( "broken left arm", @Affliction) and %ismember( "broken right arm", @Affliction)) or %ismember( paralysed, @Affliction) or %ismember( "pinned left leg", @Affliction) or %ismember( "pinned right leg", @Affliction) or %ismember( entangled, @Affliction) or %ismember( webbed, @Affliction) or %ismember( impaled, @Affliction))) |
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That's exactly what I want. If any one of those is true then evaluate as false, but both arms must be broken, not just a single arm.
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Vijilante
SubAdmin
Joined: 18 Nov 2001
Posts: 5182
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Posted: Fri Sep 30, 2005 9:32 pm
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The requirement that both arms be broken, but only when other things, actually make a very big difference on the logic. In my previous post there actually was no requirement that either arm be broken to make the statment evaluate to the value you want. All that is required is one of the other conditions, arms are only a factor if both are broken. Trying to be a mind reader here, but I am guessing you are looking at conditions for using a tree tatoo. I am also guessing that my previous post does not really provide the logic you want. Try this on for size:
| Code: |
| %eval( !(%ismember( "broken left arm", @Affliction) and %ismember( "broken right arm", @Affliction) and (%ismember( paralysed, @Affliction) or %ismember( "pinned left leg", @Affliction) or %ismember( "pinned right leg", @Affliction) or %ismember( entangled, @Affliction) or %ismember( webbed, @Affliction) or %ismember( impaled, @Affliction)))) |
I would suggest playing with it offline and manually setting different afflictions to make sure it evaluates how you want under ever condition. If you find any problem spots please post back the specific set of afflictions and how you want it to come out. |
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