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use lod and reset lod made simpler

This commit is contained in:
Asad M. Zaman 2006-08-18 02:14:37 +00:00
parent 758947d603
commit e4dd8745bf
1 changed files with 18 additions and 14 deletions
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32
direct/src/actor/Actor.py
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@ -471,17 +471,20 @@ class Actor(DirectObject, NodePath):
"""
# make sure we don't call this twice in a row
# and pollute the the switches dictionary
self.resetLOD()
# store the data in the switches for later use
sortedKeys = self.switches.keys()
sortedKeys.sort()
for eachLod in sortedKeys:
index = sortedKeys.index(eachLod)
# set the switches to not display ever
self.__LODNode.node().setSwitch(index, 0, 10000)
# turn the given LOD on 'always'
index = sortedKeys.index(lodName)
self.__LODNode.node().setSwitch(index, 10000, 0)
self.__LODNode.node().forceSwitch(index)
## self.resetLOD()
## # store the data in the switches for later use
## sortedKeys = self.switches.keys()
## for eachLod in sortedKeys:
## index = sortedKeys.index(eachLod)
## # set the switches to not display ever
## self.__LODNode.node().setSwitch(index, 0, 10000)
## # turn the given LOD on 'always'
## index = sortedKeys.index(lodName)
## self.__LODNode.node().setSwitch(index, 10000, 0)
def printLOD(self):
sortedKeys = self.switches.keys()
@ -501,12 +504,13 @@ class Actor(DirectObject, NodePath):
def resetLOD(self):
"""
Restore all switch distance info (usually after a useLOD call)"""
sortedKeys = self.switches.keys()
sortedKeys.sort()
for eachLod in sortedKeys:
index = sortedKeys.index(eachLod)
self.__LODNode.node().setSwitch(index, self.switches[eachLod][0],
self.switches[eachLod][1])
self.__LODNode.node().clearForceSwitch()
## sortedKeys = self.switches.keys()
## sortedKeys.sort()
## for eachLod in sortedKeys:
## index = sortedKeys.index(eachLod)
## self.__LODNode.node().setSwitch(index, self.switches[eachLod][0],
## self.switches[eachLod][1])
def addLOD(self, lodName, inDist=0, outDist=0):
"""addLOD(self, string)