Question

我正在尝试从列表的另一部分中调用对象，这显然不起作用。有人可以解释我如何正确地做到这一点？先谢谢。

def doSLM(btn, *args):
    if btn == 'btnBuild':
        allObj = buildSnowLayers()
        # Let's say that:
        # allObj = [ground1, snow1, ground2, snow2, ground3, snow3]
        allGrounds = allObj[::2]
        # So we have:
        # allGrounds = [ground1, ground2, ground3]
        allSnowLayers = allObj[1::2]
        # And:
        # allSnowLayers = [snow1, snow2, snow3]
    elif btn == 'btnClearScene':
        # I need to pass 'allSnowLayers' in this part of
        # the condition, but how?
        for obj in allSnowLayers:
            clearLayer(obj) # This function simply delete the object

修改：添加一些细节。对不起，如果不清楚，我还在学习......

Answer 1

究竟是什么阻止您在条件之外创建列表？如果您需要访问它，无论条件如何解决，您也可以在正确的范围内定义它。

def doSLM(btn, *args):
    allObj = buildSnowLayers()
    allGrounds = allObj[::2]
    allSnowLayers = allObj[1::2]
    if btn == 'btnClearScene':
        # I need to pass 'allSnowLayers' in this part of
        # the condition, but how?
        for obj in allSnowLayers:
            clearLayer(obj) # This function simply delete the object

如果需要，可以在完成后清空中间列表，为其分配[]或“无”。如果您确实只需要在某些情况下填充列表，则可以将其定义为条件之外的空列表，然后将其填充到内部。

def doSLM(btn, *args):
    allSnowLayers = []
    if btn == 'btnBuild':
        allObj = buildSnowLayers()
        # Let's say that:
        # allObj = [ground1, snow1, ground2, snow2, ground3, snow3]
        allGrounds = allObj[::2]
        # So we have:
        # allGrounds = [ground1, ground2, ground3]
        allSnowLayers = allObj[1::2]
        # And:
        # allSnowLayers = [snow1, snow2, snow3]
    elif btn == 'btnClearScene':
        # I need to pass 'allSnowLayers' in this part of
        # the condition, but how?
        for obj in allSnowLayers:
            clearLayer(obj) # This function simply delete the object

Answer 2

我猜测您第一次了解scope。一般来说，描述变量范围超出了简单的StackOverflow问题范围，但是您需要确定从此函数范围之外访问allSnowLayers对象的方法。例如，考虑使用类和类属性。然后，每个按钮的行为可以通过单独的方法处理，而不是大的if / else开关。

例如：

class SLM:
    def __init__(self):
        layers = buildSnowLayers()
        self.layers = layers
        self.grounds = layers[::2]
        self.snow_layers = layers[1::2]
    def clearScene(self):
        for obj in self.snow_layers:
            clearLayer(obj)

在上面的示例中，实例化SLM类会创建一个SLM对象，该对象始终了解其＆＃34;层＆＃34;，＆＃34;地面＆＃34;的状态。和＃34; snow_layers＆＃34;属性，直到它被摧毁。如果你像

那样调用clearScene

slm = SLM()
slm.clearScene()

然后slm会对clearLayer属性的每个成员应用snow_layers。它仍然知道该属性，因为它是对象的成员，而不仅仅是函数变量。

在另一个条件下调用参数

2 个答案: