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Check horse by python


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  • Former Staff

You are need to create new cpp files for this if you are planning to create new function for pet. I think you can make something via character manager.

PythonCharacterModule.cpp

Kind Regards

Zerelth ~ Ellie

mm ok i'll see ...

thanks

 

client-game/game-client communication with packets will help me ??

 

Best Regards

   FlyGun

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  • Bronze

 

You are need to create new cpp files for this if you are planning to create new function for pet. I think you can make something via character manager.

PythonCharacterModule.cpp

Kind Regards

Zerelth ~ Ellie

mm ok i'll see ...

thanks

 

client-game/game-client communication with packets will help me ??

 

Best Regards

   FlyGun

 

 

You can use Server - Client communication with python files. it's first way. second way is via packets.

 

Kind Regards

Zerelth ~ Ellie

Do not be sorry, be better.

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  • Former Staff

 

 

You are need to create new cpp files for this if you are planning to create new function for pet. I think you can make something via character manager.

PythonCharacterModule.cpp

Kind Regards

Zerelth ~ Ellie

mm ok i'll see ...

thanks

 

client-game/game-client communication with packets will help me ??

 

Best Regards

   FlyGun

 

 

You can use Server - Client communication with python files. it's first way. second way is via packets.

 

Kind Regards

Zerelth ~ Ellie

 

okay i will see now ... thanks

, wrong subject

no problem !!

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  • Former Staff

you don't check with direct python(without adding new module) but easy check with lua you can use the communication as "if horse.is_summon == true then" 

look the point is i'm making a GUI and there is button that calls the horse i want to change the summon to unsummon  if the horse is already summoned

 

8voV1iQ.jpg?1

Edited by Metin2 Dev
Core X - External 2 Internal
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you don't check with direct python(without adding new module) but easy check with lua you can use the communication as "if horse.is_summon == true then" 

look the point is i'm making a GUI and there is button that calls the horse i want to change the summon to unsummon  if the horse is already summoned

 

8voV1iQ.jpg?1

 

Edited by Metin2 Dev
Core X - External 2 Internal
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  • Former Staff

 

game.py:

self.is_summoned_horse = self.affectShower.horseImage <> None

Or try with this:

import uiAffectShower
summoned = uiAffectShower.AffectShower().horseImage <> None

 

 

 

 

game.py:

self.is_summoned_horse = self.affectShower.horseImage <> None

Or try with this:

import uiAffectShower
summoned = uiAffectShower.AffectShower().horseImage <> None

 

That's a good idea, but sometimes the icons are bugged and still shown without a summoned horse.

 

where i must put this ...

 

the funny part is i removed the image from up there O.O

 

i will put it back and try :) thx

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