Question

我有两套骷髅（A和B），并试图将一组关节设为父约束。

我所做的是将A选择放入A数组，B选择放入B数组。但是如何按距离进行父约束A到B，距离基本为0，所以脚趾关节不会用手指连接。比如波纹管。

如果有人能告诉我，我会很感激。

firstSel = []
secondSel = []

def firstSelection(*args):

    sel = mc.select(hi = True)
    joints = mc.ls(sl=True, type = "joint")
    firstSel.append(joints)
    print "this is first selection"


def secondSelection(*args):

    tsmgRoot = mc.select(hi = True)
    tsmgJoints = mc.ls(sl=True, type = "joint")
    secondSel.append(tsmgJoints)

Answer 1

以下是一个示例，说明如何通过世界位置将对象从一个列表约束到另一个列表。可能有更聪明的方法，比如将值添加到Sub Choose_Country() If (c2 = "Germany") Then Sheet8.Visible = True Sheet9.Visible = False Sheet10.Visible = False ElseIf (C2 = Australia) Then Sheet8.Visible = False Sheet9.Visible = True Sheet10.Visible = False ElseIf (C2 = Austria) Then Sheet8.Visible = False Sheet9.Visible = False Sheet10.Visible = True End if End sub然后对其进行排序，但这样会感觉更具可读性。

dict

另外作为附注，当您获得选择时，不需要import math # Add code here to get your two selection lists # Loop through skeleton A for obj in first_sel: closest_jnt = None # Name of the closest joint closest_dist = 0 # Distance of the closest joint obj_pos = cmds.xform(obj, q=True, ws=True, t=True) # Loop through skeleton B to find the nearest joint for target in second_sel: target_pos = cmds.xform(target, q=True, ws=True, t=True) # Calculate the distance between the 2 objects dist = math.sqrt((target_pos[0]-obj_pos[0])**2 + (target_pos[1]-obj_pos[1])**2 + (target_pos[2]-obj_pos[2])**2) # If this is the first object we loop through, automatically say it's the closest # Otherwise, check to see if this distance beats the closest distance if closest_jnt is None or dist < closest_dist: closest_jnt = target closest_dist = dist cmds.parentConstraint(closest_jnt, obj, mo=True)（您实际上是在列表中添加列表！）。你可以像firstSel.append(joints)一样直接分配它，或者如果列表中已存在某些项目并且你想添加它，你可以使用firstSel = mc.ls(sl=True, type = "joint")

玛雅蟒蛇。 parentConstraint按距离

1 个答案: