Hi everybody,
i was wondering how to collect a list of the parameters from a modifier by scripting?
for example, if you have a “bend” modifier on your object, you have parameters like (“center”, “gizmo”, “angle”…), and if you do $.modifers[1].center, you can access the value. But if you dunnot know especially that the modifier is bend, and you don’t know what could be the parameters, how to have a list of them for a random modifier?
thx for helping,
remi
try:
show $.modifiers['yourmodifier']
-Johan
eh eh thanks, it works but except for attribute holders, or custom attributes. it returns a blank list. maybe there is no way for this kind of modifiers to have them?
getPropNames $.bend.custom_Attributes
getSubAnimNames $.bend.custom_Attributes
First gets all the names of the properties on the object. Second gets all the animation tracks. You need to know when to use one or the other.
getPropNames you can use with get/setProperty.
getSubAnimNames you can use to get just the animation tracks as well use it when you need to know the name of tracks for wiring.
thanks for your answer paul, it will help
but for the “bend” modifier, it was just an example that i have taken
maybe you will be well-informed about that, my real problem is :
i have some personal controllers for a rig like for example “R_foot_control”, with a custom attribute holder on it named “footctrl”. With this CA, i can control some sliders like “footroll”, “bank”…
i’m building a script about this, and as for each controllers, it wouldn’t be the same names for these sliders, how can i obtain the list of these sliders for each of my nodes?
i don’t know if you seee what i mean, but to illustrate, if i write
$.modifiers[1].footroll, it will recognize the existing property and it will return to me the current value. But it works only coz i know the name of this property, that’s why i wanna obtain a list of these properties. To use them after in a loop.
Thks,
remi
I should also mention, to find out if there is a custom attribute def on something you can do…
numAttribs=custAttributes.count $.bend
and then then loop through all the defs on that object with
for i = 1 to numAttribs do print (custAttributes.get $.bend i)
i have tried something like that :
bb=custattributes.getdefs $.modifiers[1]
for k = 1 to bb.count do (
print bb[k].source
)
and it results :
#(<AttributeDef:footCtrl1>, <AttributeDef:footCtrl2>)
"attributes footCtrl1
(
parameters footP rollout:footR
(
footroll type:#float ui:(footrollSp, footrollSl)
bank type:#float ui:(bankSp, bankSl)
)
rollout footR "Foot Controls"
(
spinner footrollSp "" range:[-1,1,0]
slider footrollSl "Foot Roll" range:[-1,1,0] offset:[0,-20]
spinner bankSp "" range:[-1,1,0]
slider bankSl "Bank" range:[-1,1,0] offset:[0,-20]
)
)
"
"attributes footCtrl2
(
parameters footP rollout:footR
(
Footext type:#float ui:(footextSp, footextSl)
)
rollout footR "Foot Controls +"
(
spinner footextSp "" range:[-1,1,0]
slider footextSl "Foot Toes" range:[-1,1,0] offset:[0,-20]
)
)
"
OK
i have my 3parameters “footroll”, “bank” and ‘footext” that appears and the definition of all of my CA, but i don’t know what function to use to catch their names?
no other idea?
Sorry I will try and find time to respond later to this.
Remi,
I’m working on this now as well. I think Paul answered the question earlier.
The trick here is knowing what the name of the Custom Attributes definition is. I’m rewriting my definitions to always have the same name. So footRcntrl and footLcntrl will always use the same definition name like “animation_props” or something to that effect. This way I can always get my cntrl object properties (getPropNames $Node_Name.animation_props).
EDIT:
Actually after looking at it a little longer, here’s what you can do so that you can get any custom attribute (Paul also mentioned this):
for i = 1 to (custAttributes.count $.modifiers[1]) do
(
defName = custAttributes.get $.modifiers[1] i
propNames = getPropNames defName
for p in propNames do
(
print p
)
)
propNames will return all the propertied in the custom attributes definition on your object.
Seems like that makes the most sense.
dan
eeeh thanks to paul and dmak
Indeed, the functions was good. I just had not used the right syntax, that’s why it wasn’t work!!
thx for helping me
THanks for jumping in Dan.