实例没有属性 [英] instance has no attribute

问题描述

我是python的新手，正在努力理解为什么我在调用main()时不断收到"AttributeError:worker实例不具有'workerNo'属性."

im new to python and am struggling to understand why i keep getting "AttributeError: worker instance has no attribute 'workerNo'" when i call main().

beltLength = 5

class worker:
    def __init__(self, wId):
        workerNo = wId

def main():
    production_Line = productionLine()
    workers = []
    for index in xrange(0, beltLength*2):
        workers.append(worker(index))  
        print workers[index].workerNo 

我的想法是，它应该在10个新的worker实例后面附加一个workerNo属性，该属性等于列表中的索引. 谢谢

My thinking is that it should append 10 new worker instances with a workerNo attribute which is equal to index in a list. Thanks

推荐答案

这里的问题是，当您应该使用实例变量时，您正在使用局部变量.

The issue here is that you are using a local variable when you should be using an instance variable.

调用函数或方法将创建一个新的名称空间，该名称空间仅在调用期间存在.为了使workerNo(worker_no的值根据 PEP 8 ，这是Python代码的标准)，它必须保存在不会消失的命名空间中，以超出__init__()方法的调用范围.

Calling a function or method creates a new namespace, which exists only for the duration of the call. In order for the value of workerNo (worker_no would be better according to PEP 8, the standard for Python code) to persist beyond the __init__() method call it has to be stored in a namespace that doesn't evaporate.

每个实例都有一个这样的名称空间(在实例化其类时创建)，并且每个方法调用的self(第一个)参数都可以访问该名称空间.因此，在您的__init__()方法中，您应该编写

Each instance has such a namespace (which is created when its class is instantiated), and the self (first) argument to every method call gives access to that namespace. So in your __init__() method you should write

self.workerNo = wId

，然后您可以从其他方法访问它(因为它们也收到引用同一个名称空间的self参数.对实例的外部引用也可以访问实例属性.

and then you can access it from the other methods (since they also receive a self argument referring to the same namespace. External references to the instance can also access the instance attributes.

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