11

我面临一个相当容易解决的问题,但我只是不知道该怎么做。我希望搅拌机列出所有选择为字符串的对象。例如。如果我运行:

selection_names = bpy.context.selected_objects
print (selection_names)

它给了我这一行:

[bpy.data.objects['Cube.003'], bpy.data.objects['Cube.002'], bpy.data.objects['Cube.001'], bpy.data.objects['Cube']]

但我想要的selection_names是打印为:

['Cube.001','Cube.002','Cube.003','Cube']
4

2 回答 2

10

解决这个问题的最快方法是通过列表理解:

selection_names = [obj.name for obj in bpy.context.selected_objects]

这完全等同于:

selection_names = []
for obj in bpy.context.selected_objects:
    selection_names.append(obj.name)
于 2017-03-09T11:46:16.427 回答
9
>> selection_names = bpy.context.selected_objects
>>> print (selection_names)
[bpy.data.objects['Armature 05.04 p'], bpy.data.objects['Armature 04.08 l'], bpy.data.objects['Armature 04.07 p'], bpy.data.objects['Armature 04.07 l'], bpy.data.objects['Armature 04.04 p'], bpy.data.objects['Armature 04.05 p'], bpy.data.objects['Armature 04.05 l']]

>>> for i in selection_names:
...     print(i.name)
...     
Armature 05.04 p
Armature 04.08 l
Armature 04.07 p
Armature 04.07 l
Armature 04.04 p
Armature 04.05 p
Armature 04.05 l

如果您希望它们成为数组中的对象,您可以这样做:

>>> SelNameArr=[]
>>> for i in selection_names:
...     SelNameArr.append(i.name)
...     
>>> SelNameArr
['Armature 05.04 p', 'Armature 04.08 l', 'Armature 04.07 p', 'Armature 04.07 l', 'Armature 04.04 p', 'Armature 04.05 p', 'Armature 04.05 l']
于 2015-02-22T15:11:32.880 回答